Why does this MySQL query not work in PHP code?

Question

I am writing PHP code for a dynamic web page. The SQL query code below looks correct, but whenever the page runs, I keep getting the message within the die parameter.

$recipeid = $_GET['id'];
$query = "select title from recipes where recipeid = $recipeid";
$result = mysql_query($query) or die('Could not retrieve file');

Why does the query keep failing?

Try to run $result = mysql_query($query) or die(mysql_error()); — Nicola Cossu, Jun 03 '11 at 21:47
Do what @nick says above and tell use the error you get. Here's a tip, make you SQL easier to read by capitalising key words like so, SELECT title FROM recipes WHERE recipeid = $recipeid — Ash Burlaczenko, Jun 03 '11 at 21:51
How to work safely with mysql_* functions: http://stackoverflow.com/questions/6198104/reference-what-is-a-perfect-code-sample-using-the-mysql-extension/6198584#6198584 — OZ_, Jun 03 '11 at 21:52
indeed, you need to 1) do mysql_escape_strin(), 2) check that you did get an id (it's not empty), 3)check that if you do have $recipeid , it's an integer. — AR., Jun 03 '11 at 21:53
Link for learning how to stop sql-injections >>> http://stackoverflow.com/questions/60174/best-way-to-stop-sql-injection-in-php — Mr Meow, Jun 03 '11 at 21:53

score 2 · Accepted Answer · answered Jun 03 '11 at 21:52

2

Aside from the SQL Injection ( look into mysql_real_escape_string ) add this to your error statement to see the exact error:

 $result = mysql_query($query) or die('Could not retrieve file: ' . mysql_error());

And it will tell you what is wrong.

answered Jun 03 '11 at 21:52

Jim

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score 1 · Answer 2 · answered Jun 03 '11 at 21:50

Could be a handful of different things, since we're only seeing a little snippet of the code, it would just be speculation. Posting code or more details about what you know the error isn't would be helpful.

First try putting single quotes around $recipeid. like so:

$query = "select title from recipes where recipeid = '$recipeid'";

If that doesn't work, time for more debugging.

score 1 · Answer 3 · answered Jun 03 '11 at 21:57

1

If invoke the page without passing the id parameter in url (?id=22, for instance) you'll get a syntax error.

Try checking if $recipeid has a value prior to execute the query.

answered Jun 03 '11 at 21:57

Doug

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Why does this MySQL query not work in PHP code?

3 Answers3