What is MySQL Error #1064; How to fix it

What is MySQL Error #1064 – How to fix it?

MySQL error messages may seem very cryptic at first, but upon closer analysis, you’ll notice that they are informative enough in that they provide sufficient detail in terms of what the problem is. Understanding how to interpret a MySQL error code will help in fixing any future problems of this nature. MySQL error codes have a 4-digit numeric value (from 1000 to 2051) that reflect a certain type of error that occurred.

What is MySQL Error 1064?

The MySQL error 1064 is a syntax error, meaning that the MySQL is unable to understand the command you are issuing, because your command is not a valid one within the Structured Query Language or SQL. Syntax errors are just like grammar errors in linguistics, but unlike in the English language, where violations of the rules of grammar may still be able to produce an understandable sentence even though it’s not a grammatically valid one, syntax errors in programming will result in the inability of the parser to understand the command and fail to do anything with it.

Possible causes and how to resolve MySQL 1064 Errors

When a 1064 error code appears, it will usually state the problem that caused it and the solution to fix it. Carefully reading the error message will reveal how your command may have violated the syntax rules in the MySQL. If you see words like “near” or “at line”, you need to check for problems in those lines of the code before the command ends.

MySQL errors will not only indicate where the problem might have occurred, but also what to do to fix it. For example, it may recommend that you compare the grammatical rule violation against the manual that corresponds to your MySQL version for the right syntax to use.

Here are some of the possible causes of a MySQL 1064 Error:

Mistyping of commands

A 1064 Error can occur when you misspell a command (e. g. you write UDPATE instead of UPDATE). This can occur quite frequently since typos are so easy to miss. To prevent errors because of mistyped commands, make sure that you review your command for any typos before running it. If you have trouble with using the right syntax, refer to the manual of the MySQL version you’re using to search for the specific command(s) you’re having troubles with. Clearing up the typos or correcting the syntax will make the error disappear. Nowadays, a lot of IDEs and SQL tools come with SQL syntax highlighter or even parsers, which will alert you of a syntax error even before you run your query. If your IDE does not have this feature you might look around to see if there are any plugins which would do the job. There are a lot of online syntax checkers which can help you debug your queries.

Error caused by obsolete commands

Another possible reason for the MySQL 1064 error is the use of outdated commands. A number of commands and keywords have been deprecated, meaning that they are due for removal, but they are still allowed for a period of time before they turn obsolete. One example of a deprecated command that went obsolete is the ‘TYPE’ command, which was deprecated with the debut of MySQL 4.1 and was completely removed in version 5.1. Therefore, if you use the TYPE command in version 5.1, then it will return a 1064 syntax error. The ‘TYPE’ keyword was replaced with the ‘ENGINE’ keyword as of version 5.1. If you are switching hosting companies or you just happen to have an older backup of a MySQL database that you might want to import, a quick solution is to just search and replace “TYPE=InnoDB” with “ENGINE=InnoDB”.

Missing data

When data is missing in the database and that particular data is required by a query, the 1064 error code will be returned by the MySQL engine. The dashboard interface of your application will allow you to enter the missing data, or you can go into the database usually via phpMyAdmin or MySQL Workbench and manually add the data in the right table row.

If your application runs a query at every page load, for example, and you are not following the recommendations and you’re still constructing queries basically by string concatenation, you might have the following query in PHP for example: “SELECT * FROM customers WHERE customer_id = “ . $_SESSION[‘user_id’]. This assumes that the user is logged in and the $_SESSION[‘user_id’] has a value, however, sessions can expire and in that case the query which will run would become “SELECT * FROM customers WHERE customer_id = ” and MySQL’s response to this query will be the 1064 error code. There can be quite a lot of places where people do not use default values, another example would be the sorting parameters, which are usually column names but sometimes the application might receive empty values, which in turn still receive a 1064 error code.

Reserved words

Reserved words are words that are different from MySQL version to MySQL version and serve specific purposes or are used to perform specific functions in the MySQL engine. You might receive a 1064 error when using a reserved word, either because you’re using a reserved word that is not specific for the MySQL version you have or either because you fail to meet the specific requirements to use it (e. g. using it with quotation marks or backticks). A full list of the reserved words specific for each MySQL version and their usage requirements are available at MySQL. com.

WordPress MySQL database transfer to another server

The 1064 error can also be caused by a WordPress database export to another server. Choosing the compatibility mode and changing the database version to the current version you’re using can help resolve the error. Remember to select the compatibility mode under the advanced tab when performing a backup and click the auto-detect file character set when restoring the MySQL database.

Conclusion

MySQL errors are pretty straightforward, signaling both exactly where the error was encountered by the parser and making suggestions for fixing them. Reading the error message and following the recommendation set forth in the message will resolve the error.

MySQL 1064 Error: You have an error in your SQL syntax

So, you’re creating a custom SQL query to perform a task in the database. After putting the code together and running it in PHPmyAdmin it responds with a 1064 error. It may look similar to this:

The 1064 error displays any time you have an issue with your SQL syntax, and is often due to using reserved words, missing data in the database, or mistyped/obsolete commands. So follow along and learn more about what the 1064 error is, some likely causes, and general troubleshooting steps.

Note: Since syntax errors can be hard to locate in long queries, the following online tools can often save time by checking your code and locating issues:

Causes for the 1064 error

This may seem cryptic since it is a general error pointing to a syntax issue in the SQL Query statement. Since the 1064 error can have multiple causes, we will go over the most common things that will result in this error and show you how to fix them. Follow along so you can get your SQL queries updated and running successfully.

Using Reserved Words

Every version of MySQL has its own list of reserved words. These are words that are used for specific purposes or to perform specific functions within the MySQL engine. If you attempt to use one of these reserved words, you will receive the 1064 error. For example, below is a short SQL query that uses a reserved word as a table name.

How to fix it:

Just because the word alter is reserved does not mean it cannot be used, it just has special requirements to use it as the MySQL engine is trying to call the functionality for the alter command. To fix the issue, you will want to surround the word with backticks, this is usually the button just to the left of the “1” button on the keyboard. The code block below shows how the code will need to look in order to run properly.

Missing Data

Sometimes data can be missing from the database. This causes issues when the data is required for a query to complete. For example, if a database is built requiring an ID number for every student, it is reasonable to assume a query will be built to pull a student record by that ID number. Such a query would look like this:

If the $id is never properly filled in the code, the query would look like this to the server:

Since there is nothing there, the MySQL engine gets confused and complains via a 1064 error.

How to fix it:

Hopefully, your application will have some sort of interface that will allow you to bring up the particular record and add the missing data. This is tricky because if the missing data is the unique identifier, it will likely need that information to bring it up, thus resulting in the same error. You can also go into the database (typically within phpMyAdmin) where you can select the particular row from the appropriate table and manually add the data.

Mistyping of Commands

One of the most common causes for the 1064 error is when a SQL statement uses a mistyped command. This is very easy to do and is easily missed when troubleshooting at first. Our example shows an UPDATE command that is accidentally misspelled.

How to fix it:

Be sure to check your commands prior to running them and ensure they are all spelled correctly.

Below is the syntax for the correct query statement.

Obsolete Commands

Some commands that were deprecated (slated for removal but still allowed for a period of time) eventually go obsolete. This means that the command is no longer valid in the SQL statement. One of the more common commands is the ‘TYPE‘ command. This has been deprecated since MySQL 4.1 but was finally removed as of version 5.1, where it now gives a syntax error. The ‘TYPE‘ command has been replaced with the ‘ENGINE‘ command. Below is an example of the old version:

This should be replaced with the new command as below:

Error 1064 Summary

As you can see there is more than one cause for the 1064 error within MySQL code. Now, you know how to correct the issues with your SQL Syntax, so your query can run successfully. This list will be updated as more specific instances are reported.

Thoughts on “ MySQL 1064 Error: You have an error in your SQL syntax ”

I created a drop down list of states in the HTML form using data set present in the table named cool in my database in phpmyadmin. Names of states are present in the table “cool”. Now I created another table called “data” to store the user input values of name, address, contact and the email ID. I created a third table named “tab” which consists of ID and the stat columns to store the name of the state when the user selects from the drop down list. Now i tried joining tables “data” and “tab” using inner join in SQL. When I fetched the data and tried printing it in another web page , the name, email, contact and address are getting printed but the ID of the state in the table “cool” is getting printed instead of the name of state. I tried a lot to get the name of the state but always in vain.
I request you to help me fix this problem and print the name of the state instead of it’s ID in the web page !!

Your issue is a coding one with your database. We do not provide support for coding, so I recommend that you speak with an experienced developer or programmer. You can also find many tutorials for MySQL online that may provide the information you seek. Apologies that we cannot provide a direct solution for your issue.

I get this error when trying to import a database to phpmyadmin

3 errors were found during analysis.

Variable name was expected. (near ” ” at position 5)
Variable name was expected. (near ” ” at position 26)
Unrecognized statement type. (near “Host” at position 0)

Host: localhost (Version: 5.6.16) # Date: 2015-06-03 23:46:52 # Generator: MySQL-Front 5.3 (Build 4.122) /*!40101 SET NAMES utf8 */

MySQL said: Documentation
#1064 – You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ‘Host: localhost (Version: 5.6.16)
# Date: 2015-06-03 23:46:52
# Generator: My’ at line 1

phpmy admin seemingly has a problem with this:

# Host: localhost (Version: 5.6.16)
# Date: 2015-06-03 23:46:52
# Generator: MySQL-Front 5.3 (Build 4.122)

/*!40101 SET NAMES utf8 */;

ERROR 1064 (42000) at line 6: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ‘

<link rel' at line 1

Operation failed with exitcode 1

It looks like there is an issue with how the query is coded. Unfortunately, we are unable to provide direct support for coding questions, so I recommend that you speak with an experienced developer or programmer. That said, I’d start by examining the early link syntax to see if it might be something simple like a missing comma or semi-colon. Apologies that we cannot provide a direct solution for your issue.

Hi I got this error when trying to import mysql database into phpmyadmin using a local XAMPP server. Please what probably go wrong?

Error:
Static analysis:

10 errors were found during analysis.

Unexpected character. (near “ <” at position 224)
Unexpected beginning of statement. (near “DOCTYPE” at position 2)
Unexpected beginning of statement. (near “HTML” at position 10)
Unexpected beginning of statement. (near “html” at position 16)
Unexpected beginning of statement. (near “lang” at position 21)
Unexpected beginning of statement. (near “‘en’” at position 26)
Unexpected beginning of statement. (near “dir” at position 31)
Unexpected beginning of statement. (near “‘ltr’” at position 35)
Unexpected beginning of statement. (near “meta” at position 42)
Unrecognized statement type. (near “charset” at position 47)

MySQL said: Documentation
#1064 – You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ‘<meta name="ref' at line 1

The error indicates there is an issue with your syntax. This depends on your version of MySQL. I recommend that you review the documentation for the version of MySQL you are using to ensure you have the proper syntax in your code.

I got this error and I tried so many ways to fix it but I could not so can any one help me to figure it out !

1 errors were found during analysis.

Unexpected beginning of statement. (near “id11897681_admin” at position 0)
SQL query:

id11897681_admin SET time_zone = “+00:00”

MySQL said: Documentation

#1064 – You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ‘id11897681_admin
SET time_zone = “+00:00″‘ at line 1

Sorry for the problem that you’re getting with the SQL Syntax. I’m not 100% sure of the issue but it appears to be possible syntax near the ID you’re using. The documentation shows three backticks before the timezone. Check it out in the MariaDB’s official documentation: https://mariadb. com/kb/en/library/time-zones/.

Hi Is there any possibilities to get a error code 00088x from SQL database, if there pls help me to fix it

Please provide more information and we can look into the issue further. Describe where/when the error occurs and provide the query.

Hello
mysql error but Apache is running xampp is a problem ( Error: MySQL shutdown unexpectedly.
This may be due to a blocked port, missing dependencies,
improper privileges, a crash, or a shutdown by another method.
Press the Logs button to view error logs and check
the Windows Event Viewer for more clues
If you need more help, copy and post this
entire log window on the forums)

Someone can help me.

Hello – A MySQL shutdown can happen for any number of reasons. We would advise following the recommendation of using the log files to determine when the shutdown occurred and if the event can be duplicated. You may need to submit a ticket with our live technical support team, or you may need to speak with an experienced MySQL developer for further assistance. If you are getting a specific error due to Syntax, we would need more information on the MySQL query where the error is happening. If you have any further questions or comments, please let us know.

Executing SQL script in server

ERROR: Error 1064: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ‘(16) NOT NULL,

`password` (32) NOT NULL,

`create_time`’ at line 5

CREATE TABLE IF NOT EXISTS `mydb`.`user_1` (

`username` (16) NOT NULL,

`password` (32) NOT NULL,

`create_time` NULL DEFAULT CURRENT_TIMESTAMP)

SQL script execution finished: statements: 9 succeeded, 1 failed

Fetching back view definitions in final form.

Executing SQL script in server

ERROR: Error 1064: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ‘(16) NOT NULL,

`password` (32) NOT NULL,

`create_time`’ at line 5

CREATE TABLE IF NOT EXISTS `mydb`.`user_1` (

`username` (16) NOT NULL,

`password` (32) NOT NULL,

`create_time` NULL DEFAULT CURRENT_TIMESTAMP)

SQL script execution finished: statements: 9 succeeded, 1 failed

Fetching back view definitions in final form.

Thanks for post on the MYSQL syntax error. The error indicates that the variable you’re building is not being properly defined. You need to indicated what each item is going to be. For example:

CREATE TABLE pet (name VARCHAR(20), owner VARCHAR(20),
species VARCHAR(20), sex CHAR(1), birth DATE, death DATE);

Specify what your variables are supposed to be in the table, then it should work.

I am getting this error when using SQl Developer and trying to “monitor sessions” in a mysql database. I can access the mysql database and look at tables, run queries, etc., and I have increased max_allowed_packets per a few other posts.

Sorry for the problem with the error when trying monitor sessions. We would need to see the MySQL query in order to troubleshoot the problem. Otherwise, we recommend speaking with an experienced database developer to resolve the issue.

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ‘default(Cus_name, Rs_status, Acc_type, Acc_class, Currency, Init_deposit, Dob, Gender, M’ at line 1.
this error how to solve plz help me

Hello Sandar – sorry for the syntax error. We would need to see more of the MySQL query in order to determine the problem. I would recommend that you speak with an experienced developer or review your code to find the syntax issue that is causing the error to appear.

hello I’am getting an error through my workbench while I’m trying to forward engineer.

CREATE TABLE IF NOT EXISTS `electronics`.`customer_order` (

`id` INT UNSIGNED NOT NULL AUTO_INCREMENT,

`amount` DECIMAL(6,2) NOT NULL,

`date_created` TIMESTAMP NOT NULL,

`confirmation_number` INT NOT NULL,

`customer_id` INT UNSIGNED NOT NULL,

INDEX `fk_customer_order_customer` (`customer_id` ASC) VISIBLE,

FOREIGN KEY (`customer_id`)

REFERENCES `electronics`.`customer` (`id`)

ON DELETE NO ACTION

ON UPDATE NO ACTION)

COMMENT = ‘Maintain the details of the customer order’

SQL script execution finished: statements: 9 succeeded, 1 failed

ERROR: Error 1064: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ‘

FOREIGN KEY (`customer_id`)

According to the MySQL checker provided above, the following line needs to be checked:
INDEX `fk_customer_order_customer` (`customer_id` ASC) VISIBLE,

when i import a database from wamp i get an error that ther is an error in line one

create databse student_mgt charset utf-8;

Depends on how you’re doing the import. If you’re logged into WAMP and you have already created that database, then that line would not be needed and you can just import the structure. If you have any further questions or comments, please let us know.

hii there ,,

for the following database im getting these errors, could you please help me encounter teh errors.

<?php
$c_album=”CREATE TABLE IF NOT EXISTS album (
aid int(11) NOT NULL auto_increment,
aperformer_id int(11) NOT NULL default ‘0’,
aname varchar(200) NOT NULL,
bio_short text,
bio_long text,
PRIMARY KEY (aid),
KEY aperformer_id (aperformer_id),
KEY aname (aname)
) ENGINE=MyISAM;”;

$c_fav=”CREATE TABLE IF NOT EXISTS fav (
id int(11) NOT NULL auto_increment,
track_id int(11) NOT NULL default ‘0’,
performer_id int(11) default NULL,
album_id int(11) default NULL,
name varchar(200) NOT NULL,
duration varchar(6) default NULL,
last_played varchar(20) default NULL,
times_played int(11) NOT NULL default ‘0’,
year varchar(4) default NULL,
user_id int(11) NOT NULL default ‘0’,
fav_name varchar(80) NOT NULL default ”,
PRIMARY KEY (id),
KEY fav_name (fav_name),
KEY user_id (user_id),
KEY track_id (track_id)
) ENGINE=MyISAM;”;

$c_fav_shares=”CREATE TABLE IF NOT EXISTS fav_shares (
id int(11) NOT NULL auto_increment,
owner_id int(11) NOT NULL default ‘0’,
fav_name varchar(80) NOT NULL,
share_id int(11) NOT NULL default ‘0’,
PRIMARY KEY (id)
) ENGINE=MyISAM”;

$c_performer=”CREATE TABLE IF NOT EXISTS performer (
pid int(11) NOT NULL auto_increment,
pname varchar(200) NOT NULL,
bio_short text,
bio_long text,
PRIMARY KEY (pid),
KEY pname (pname)
) ENGINE=MyISAM;”;

$c_queue=”CREATE TABLE IF NOT EXISTS queue (
qid bigint(20) NOT NULL auto_increment,
user_name varchar(80) NOT NULL,
track_id bigint(20) NOT NULL default ‘0’,
PRIMARY KEY (qid),
KEY user_name (user_name)
) ENGINE=MyISAM;”;

$c_track=”CREATE TABLE IF NOT EXISTS `track` (
`id` bigint(20) NOT NULL AUTO_INCREMENT,
`performer_id` int(11) DEFAULT NULL,
`album_id` int(11) DEFAULT NULL,
`track_no` smallint(6) DEFAULT NULL,
`name` varchar(200) NOT NULL,
`duration` varchar(6) DEFAULT NULL,
`last_played` varchar(20) DEFAULT NULL,
`times_played` int(11) NOT NULL DEFAULT ‘0’,
`year` varchar(4) DEFAULT NULL,
`path` text COLLATE latin1_danish_ci,
`echonest_id` varchar(20) NOT NULL DEFAULT ‘-1’,
`echonest_tempo` varchar(8) NOT NULL DEFAULT ‘-1’,
`echonest_loudness` varchar(8) NOT NULL DEFAULT ‘-1’,
`echonest_danceability` varchar(8) NOT NULL DEFAULT ‘-1’,
`echonest_energy` varchar(8) NOT NULL DEFAULT ‘-1’,
`echonest_mode` varchar(2) NOT NULL DEFAULT ‘-1’,
`echonest_key` varchar(2) NOT NULL DEFAULT ‘-1’,
`echonest_time_signature` varchar(2) NOT NULL DEFAULT ‘-1’,
`echonest_status` varchar(10) NOT NULL DEFAULT ‘-1’,
`echonest_liveness` varchar(8) NOT NULL DEFAULT ‘-1’,
`echonest_speechiness` varchar(8) NOT NULL DEFAULT ‘-1’,
`echonest_acousticness` varchar(8) NOT NULL DEFAULT ‘-1’,
`echonest_valence` varchar(8) NOT NULL DEFAULT ‘-1’,
PRIMARY KEY (`id`),
KEY `performer_id` (`performer_id`),
KEY `album_id` (`album_id`),
KEY `name` (`name`),
KEY `year` (`year`),
KEY `echonest_tempo` (`echonest_tempo`),
KEY `echonest_loudness` (`echonest_loudness`),
KEY `echonest_danceability` (`echonest_danceability`),
KEY `echonest_energy` (`echonest_energy`),
KEY `echonest_mode` (`echonest_mode`),
KEY `echonest_key` (`echonest_key`),
KEY `echonest_time_signature` (`echonest_time_signature`),
KEY `echonest_liveness` (`echonest_liveness`),
KEY `echonest_speechiness` (`echonest_speechiness`),
KEY `echonest_acousticness` (`echonest_acousticness`),
KEY `echonest_valence` (`echonest_valence`)
) ENGINE=MyISAM;”;

$c_user=”CREATE TABLE IF NOT EXISTS `user` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(80) NOT NULL,
`email` varchar(80) NOT NULL DEFAULT ‘0’,
`password` varchar(80) NOT NULL,
`password_salt` varchar(40) NOT NULL DEFAULT ‘0’,
`last_login` varchar(40) NOT NULL DEFAULT ”,
`last_ip` varchar(40) NOT NULL DEFAULT ”,
`admin` tinytext NOT NULL,
`lang` char(2) NOT NULL DEFAULT ”,
`count` mediumint(9) NOT NULL DEFAULT ‘0’,
`enqueue` char(1) DEFAULT ‘0’,
`cssfile` varchar(80) NOT NULL DEFAULT ”,
`icon_dir` varchar(80) DEFAULT NULL,
`disp_last_played` char(1) DEFAULT ‘1’,
`disp_times_played` char(1) DEFAULT ‘1’,
`disp_id_numbers` char(1) DEFAULT ‘1’,
`disp_jump_to` char(1) DEFAULT ‘1’,
`disp_duration` char(1) DEFAULT ‘1’,
`disp_totals` char(1) DEFAULT ‘1’,
`disp_related_performers` char(1) DEFAULT ‘1’,
`confirm_delete` char(1) DEFAULT ‘1’,
`can_download` char(1) DEFAULT ‘0’,
`can_upload` char(1) DEFAULT ‘0’,
`disp_download` char(1) DEFAULT ‘0’,
`disp_upload` char(1) DEFAULT ‘0’,
`disp_lyrics` char(1) DEFAULT ‘0’,
`hide_icon_text` char(1) NOT NULL DEFAULT ‘0’,
`disp_fav_shares` char(1) DEFAULT ‘0’,
`disp_small_images` char(1) DEFAULT ‘1’,
`browse_albums_by_covers` char(1) DEFAULT ‘0’,
`browse_performer_by_picture` char(1) DEFAULT ‘0’,
`autoplay` char(1) DEFAULT ‘0’,
`autoplay_num_tracks` int(11) NOT NULL DEFAULT ‘1’,
`autoplay_list` varchar(80) DEFAULT ‘Tracks’,
`autoplay_last` char(1) DEFAULT ‘0’,
`autoplay_last_list` varchar(80) DEFAULT NULL,
`ask4favoritelist` char(1) DEFAULT ‘0’,
`ask4favoritelist_disp_suggestion` char(1) NOT NULL DEFAULT ‘0’,
`disp_now_playing` char(1) DEFAULT ‘0’,
`disp_now_playing_add2favorite` char(1) NOT NULL DEFAULT ‘0’,
`avoid_duplicate_entries` char(1) DEFAULT ‘1’,
`auto_add2favorite` char(1) NOT NULL DEFAULT ‘0’,
`auto_add2favorite_create_new` char(1) NOT NULL DEFAULT ‘1’,
`auto_add2favorite_prefix` varchar(50) NOT NULL DEFAULT ‘AmpJuke_Automatically_added’,
`disp_help` char(1) DEFAULT ‘1’,
`welcome_num_items` smallint(6) NOT NULL DEFAULT ’10’,
`welcome_content_1` varchar(80) DEFAULT ‘Recently played tracks’,
`welcome_content_2` varchar(80) DEFAULT ‘Random albums’,
`welcome_content_3` varchar(80) DEFAULT ‘Random albums’,
`lame_local_enabled` char(1) DEFAULT ‘1’,
`lame_local_parameters` varchar(80) DEFAULT NULL,
`lastfm_active` char(1) DEFAULT ‘0’,
`lastfm_username` varchar(80) DEFAULT NULL,
`lastfm_password` varchar(80) DEFAULT NULL,
`xspf_active` char(1) NOT NULL DEFAULT ‘0’,
PRIMARY KEY (`id`),
KEY `name` (`name`),
KEY `password` (`password`)
) ENGINE=MyISAM;”;

?>
here are the errors :

Static analysis:

7 errors were found during analysis.

1. Unexpected character. (near “?” at position 1)
2. Unexpected character. (near “$” at position 7)
3. Unexpected beginning of statement. (near “?” at position 1)
4. Unexpected beginning of statement. (near “php” at position 2)
5. Unexpected beginning of statement. (near “$” at position 7)
6. Unexpected beginning of statement. (near “c_album” at position
7. Unexpected beginning of statement. (near “”CREATE TABLE IF NOT EXISTS album ( aid int(11) NOT NULL auto_increment, aperformer_id int(11) NOT NULL default ‘0’, aname varchar(200) NOT NULL, bio_short text, bio_long text, PRIMARY KEY (aid), KEY aperformer_id (aperformer_id), KEY aname (aname) ) ENGINE=MyISAM;”” at position 16)

SQL query:

<?php $c_album=”CREATE TABLE IF NOT EXISTS album ( aid int(11) NOT NULL auto_increment, aperformer_id int(11) NOT NULL default ‘0’, aname varchar(200) NOT NULL, bio_short text, bio_long text, PRIMARY KEY (aid), KEY aperformer_id (aperformer_id), KEY aname (aname) ) ENGINE=MyISAM;”