PHP/mySQL Like function does not work

Question

I have a simple query in PHP but I can't get Like to work.

Here is the code:

$var = $_GET['q'];
$trimmed = trim($var);
$query = "SELECT * FROM vm_regiony WHERE nazev LIKE "%$trimmed%" order by id     LIMIT 10";
$result = mysql_query($query);
if(mysql_num_rows($result)==0){
  echo "nothing";
  echo "<br />";
  echo $trimmed;
}else{
  while($rene=mysql_fetch_array($result)){
    $jmeno = $rene['nazev'];
    echo '<a id="hled" onclick="javascript:vybrat()">'.$jmeno.'</a>';

i was just about to fix it too lol couldn't understand a word.. — Andres
– Andres, Commented Dec 7, 2011 at 21:14
Please note that you have written a script that is vulnerable to SQL Injection vulnerabilities because you have not sanitized any of the user-supplied variables in your SQL queries. Please use PHP Prepared Statements to prevent these vulnerabilities. Thanks. — sarnold
– sarnold, Commented Jan 21, 2012 at 1:41

Kai Qing · Accepted Answer · 2011-12-07 21:12:26Z

5

For one you need to use single quotes there

$query = "SELECT * FROM vm_regiony WHERE nazev LIKE '%$trimmed%' order by id LIMIT 10";

answered Dec 7, 2011 at 21:12

Kai Qing

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4 Comments

René Beneš Over a year ago

Yeah, thanks you really much man, really quick answer THANKS!

Adrian Cornish Over a year ago

Or escape the double quotes - but single quotes is simpler :-)

Kai Qing Over a year ago

no problem - to ensure fast response in the future you should check this answer as accepted.

René Beneš Over a year ago

Yeah, i was waiting, bacause there was 10 miutes remaining, and Ive forgot about it :D sorry

Cyclonecode · Accepted Answer · 2011-12-07 22:00:27Z

0

$query = "SELECT * FROM vm_regiony 
          WHERE nazev LIKE '%' . $trimmed . '%' 
          ORDER BY id LIMIT 10";

edited Dec 7, 2011 at 22:00

Cyclonecode

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answered Dec 7, 2011 at 21:15

Andres

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