PHP coding question.....???

Question

$imgres = mysql_db_query("database", "select ID, URL1 from contacts where ID=$idno;");
$imgdata = mysql_fetch_row($imgres);
echo mysql_error();
?>

This code doesn't seem to work....I've tried everything....here's the error messages I'm getting when trying to view it in the browser....

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/nethinga/public_html/Unt... on line 8
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 '' at line 1

What I really don't understand, is that it worked earlier, and I messed with it and now it doesn't. I'm SOOOO FRUSTRATED with this....please help!@!@!@

Answer 1

First off I'm going to suggest you look here http://uk2.php.net/manual/en/function.mysql-db-query.php
says that the function is depreciated.

It also looks like you have a semi colon that shouldn't be there

"select ID, URL1 from contacts where ID=$idno;<- (not sure that should be there)");

hope it helps

for eg

$sql = "INSERT
INTO users (reg_username,
reg_password,
reg_email)
VALUES ('bad_guy', 'mypass', ''), ('good_guy',
'1234',
'[e-mail]')"; ?>

taken from http://phpsec.org/projects/guide/3.html

Answer 2

this isn't a problem with the depreciated function, especially if you are still using PHP 4.x

the problem is in your sql statement. I know the ";" should not be after idno, also, to optimise your code, you should include $idno in quotes (single ' work). Doing this last will make sure that the index is used, often if the quotes are omitted the index is ignored.

Answer 3

Well, mysql_db_query is deprecated (going out of use). Use mysql_connect, and then mysql_query. Make sure the variable $idno is correct, you can hardcode it yourself and see if it works that way. (I.e, try "select ID, URL1 from contacts where ID=1" if you know you have a row where ID=1 in your table.)