PHP Echo results where input LIKE $_GET

Question

I'm trying to implement a hashtag system into my website. I have it set so user input that has a hashtag gets converted into a link to hashtag.php?q=%23$1 that echo's "Results for '.$_GET["q"].':"; which works fine, but it doesn't actually display any posts. For example, I have a post saying "This #website sucks" which is echoed out as

This <a href="hashtag.php?q=%23website">#website</a> sucks

But the following page only displays

Results for #website:

and the rest is blank. Here's my code for hashtag.php:

echo 'Results for '.$_GET["q"].':';

$connect = mysql_connect("localhost","root","");
mysql_select_db("database",$connect);
$mysql = "SELECT * FROM table WHERE input LIKE '".$_GET['q']."' ";
$myData = mysql_query($mysql, $connect);
while ($record = mysql_fetch_array($myData)){
echo $record['input'];
}

I'm working on using mysqli before I make the site public by the way.

I recommend to immediately stop using your code. It is completely open to SQL injection. What if I pass '; DROP TABLE table' for $_GET[q] ? — Kermit
– Kermit, Commented Oct 3, 2013 at 14:23
@FreshPrinceOfSO : +1 for effort, but your code will not do anything, as you can't run multiple commands :P — Nanne
– Nanne, Commented Oct 3, 2013 at 14:25
@FreshPrinceOfSO it wouldn't work because ext/mysql does not allow multiple queries to be executed with one function simultaneously. However, he should use properly parameterized queries to prevent injection. — Explosion Pills
– Explosion Pills, Commented Oct 3, 2013 at 14:25

Jonathon · Accepted Answer · 2013-10-03 14:25:31Z

1

Try putting % percentage signs on either side of the $_GET["p"] in your query. Don't forget to escape the $_GET["q"] as well. That'd be more important than using mysqli ;)

$mysql = "SELECT * FROM table WHERE input LIKE '%".mysql_real_escape_string($_GET['q'])."%' ";

answered Oct 3, 2013 at 14:25

Jonathon

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