PHP Variable in Select Statement

Answer 1

• Please post in English. Everyone else does.
• Try using a different fetch method - fetch an associative array, then use the dynamic parameter to retrieve whatever column it is you need.
• Have you considered using PDO?

Answer 2

Wouldn't this work?

$result = mysql_fetch_array($query);

echo trim($result['name']);

Answer 3

Is it this you're looking for? Even your question in German isn't that clear to me :

$field = 'name';
$query = mysql_query("SELECT $field FROM contacts WHERE contact_id='". mysql_real_escape_string( $id ) ."' and user_id='1';");
$retval = mysql_fetch_object($query)->$field;

Answer 4

You should never put a variable into field list.
If want a variable field name, select * and then use your variable to fetch particular field

<?php
require_once 'config.php';

$id = mysql_real_escape_string($_GET["id"]); //ID DES DERZEITIGEN KONTAKTES
$user = $_GET["user"];  //ID DES DERZEITIGEN USERS

$query  = "SELECT * FROM contacts WHERE contact_id='$id' and user_id='1'";
$result = mysql_query($query) or trigger_error(mysql_error().$query);

$row = mysql_fetch_array($result);

//and finally
$fieldname = "name";
$retval = $row[$fieldname];

echo $retval;
?>

Answer 5

I believe you are confusing matters (unintentionally) due to your use of the word 'row'. Judging by your example you mean field/column. It sounds like you wish to specify the fields to select using a variable which can be done by any of these methods...

$fields = "name, age";

$sql = "SELECT $fields FROM table";
$sql = "SELECT {$fields} FROM table";
$sql = "SELECT ".$fields." FROM table";

NB it is important that you have secure date in the $fields element, I would suggest using a whitelist of allowed values i.e.

// assuming $_POST['fields'] looks something like array('name','age','hack');
$allowed = array('name', 'age');
$fields = array();

foreach ($_POST['fields'] as $field) {
   if (in_array($field, $allowed)) {
      $fields[] = $field;
   }
$fields = implode(', ', $fields);