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I'm still learning PHP/mySQL and for the life of me I can't figure out what's wrong with my code. I'm trying to take several javascript variables from an html page and add them to a table in my database. I'm using ajax and passing the variables in the URL, then retrieving them with $_GET. I've tested this and the variables are passed correctly to the PHP script, but they don't appear in the database. All I know is that the script dies on this line: mysql_query($query) or die('data entry failed'); I used phpmyadmin to create the table and my host is 000webhost. Thanks in advance!

$database='a9293297_blog';
$con=mysql_connect('mysql2.000webhost.com','my_username','my_password');
mysql_select_db($database,$con) or die('failed to connect to database');
$username=$_GET['username'];
$password=$_GET['password'];
$charName=$_GET['charName'];
$sex=$_GET['sex'];
$class=$_GET['class'];
$race=$_GET['race'];
$str=$_GET['str'];
$sta=$_GET['sta'];
$dex=$_GET['dex'];
$int=$_GET['int'];
$cha=$_GET['cha'];
$query="INSERT INTO Players (username, password, charName, sex, class, race, str, sta, dex, int, cha)
VALUES ('".$username."', '".$password."', '".$charName."', '".$sex."', '".$class."', '".$race."', '".$str."', '".$sta."', '".$dex."', '".$int."', '".$cha."')";
mysql_query($query) or die('data entry failed');
mysql_close($con);
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Does it output anything? – Wikeno Apr 2 '11 at 19:08
Are you sure that the $query variable is being populated? – Jeff V Apr 2 '11 at 19:10
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2 Answers

up vote 3 down vote accepted

To know better what's wrong with your SQL query, use mysql_error():

mysql_query($query) or die(mysql_error());

Escape your string variables with mysql_real_escape_string(). Example:

$query = "INSERT INTO MYTABLE(MYFIELD) VALUES ('".mysql_real_escape_string($myVar)."');

EDIT

int seems to be a reserved MySQL keyword. Escape it with backquotes:

INSERT INTO Players (username, password, ..., str, sta, dex, `int`, cha) ...
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thanks for the tips, I didn't know about mysql_error() – Austen Apr 2 '11 at 19:18
now it gives me this error: 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 'int, cha) – Austen Apr 2 '11 at 19:19
edited answer, following your MySQL error message. – Frosty Z Apr 2 '11 at 19:32
Thank you, I never would have figured that out! Everything is working now. – Austen Apr 2 '11 at 19:37
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up vote 0 down vote

im not sure but try this

Your Code

$query="INSERT INTO Players (username, password, charName, sex, class, race, str, sta, dex, int, cha)VALUES ('".$username."', '".$password."', '".$charName."', '".$sex."', '".$class."', '".$race."', '".$str."', '".$sta."', '".$dex."', '".$int."', '".$cha."')";

In this code

$query="INSERT INTO Players (username, password, charName, sex, class, race, str, sta, dex, int, cha)VALUES ('$username', '$password', '$charName', '$sex', '$class', '$race', '$str', '$sta', '$dex', '$int', '$cha')";

Maybe this helps if you remove " and .

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thanks, that's what I tried originally. But putting the variables in quotes treats them as strings... – Austen Apr 2 '11 at 19:20
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