Question

I’m using AJAX and MySQL combination to search a database and display the results within the table. Unfortunately it won’t echo the result from the database yet the code has found it.

$in=$_GET['txt'];
$msg="";
if(strlen($in)>0 and strlen($in) <20 ) {
    $t=mysql_query("SELECT user FROM users WHERE user LIKE '$in%'");
    while($nt=mysql_fetch_array($t)) {
        $msg.=$nt[name]."->$nt[id]<br>";
    }
}
echo $msg;

The issue seems to be in this part of my code.

Answer 1

You have selected user but you want to display name which the query is not returning so:

$in  = mysql_real_escape_string($_GET['txt']);
$msg = '';
if(strlen($in) > 0 && strlen($in) < 20){
  $t = mysql_query("SELECT id, name 
                    FROM users 
                    WHERE user 
                    LIKE '%$in%'");
  while($nt = mysql_fetch_array($t)){
    $msg .= $nt['name'] . "->" . $nt['id'] . "<br />";
  }
}

echo $msg;

Note that I’ve added mysql_real_escape_string() function for preventing mysql injection