mysql_error() expects parameter 1 to be resource, string given

Question

I'm working on code for a poll. The admin can create questions and delete them. The user can choose answers from a dropdown menu to answer the questions. This dropdown menu is filled dynamically with a query request.

Now I want the user's name to be inserted in the database when he completes a poll, so he can't do that poll twice. That's what the code here is for. This is the query I'm using:

$idfind = "SELECT `id` FROM `Umfragenteilnahme` WHERE `Username` ='$user'";
$id = mysql_query($idfind) or die (mysql_error());
if($id) {
    while($rpw = mysql_fetch_assoc($id)) {
        $id = $rpw['id']; // Using $rpw->id doesnt work
        echo $id; // it gives the right answer and everything works fine
                  //id= 4 for example
    }  
}
else {
    echo "Fehler"; // it does not report a fail
}

I know I should use the new mysqli_... tags. The code works absolutely fine and the query gives the right answer and works in MySql database as well.

But every time it gives this error

mysql_fetch_assoc() expects parameter 1 to be resource, string given in

or for the (mysql_error()) this one:

mysql_error() expects parameter 1 to be resource, string given in.

Everything works fine, but it shows these errors. Why are these errors happening? How can I avoid them?

Edit: Now i placed this in the beginning of the code after the first <?php :

set_error_handler ( function($errno, $errstr, $errfile, $errline) {
if (false!==stripos($errstr, 'mysql')) {
    $file = file($errfile);
    $code = '';
    for($i=max(0, $errline-8); $i<min($errline+8, count($file)); $i++) {
        $code .= sprintf('%4d | %s', $i+1, $file[$i]);
    }
    printf('<fieldset><legend>%s (%d), %s@%d</legend>
        <pre>%s</pre></fieldset>',
        htmlspecialchars($errstr), $errno,
        htmlspecialchars($errfile), $errline,
        htmlspecialchars( $code )
    );
}
else {
    return false;
}});

The output is this :

Benutzer id: 4
mysql_fetch_assoc() expects parameter 1 to be resource, string given (2), /censored/censored/www/Website2.0/btn_lehrer.php@36

29 | /* The $user is defined at start of code */
30 | /* With the next query i get the user-id from this specific user */
31 | 
32 | $idfind = "SELECT `id` FROM `Umfragenteilnahme` WHERE `Username`    =   '$user'";
33 | $id = mysql_query($idfind)or die (mysql_error());
34 |        if($id){
35 |            
36 |                    while($rpw = mysql_fetch_assoc($id)){
37 |                          $id = $rpw['id'];
38 |                           echo "Benutzer id: $id ";    //fehler       (Warning: mysql_fetch_assoc() expects parameter 1 to be   resource, string given line 15)                                    
39 |                        }  
40 |                      
41 |                }
42 |                else{
43 |                     echo "Fehler bei der Bestimmung der Benutzer id!";
44 |                    }

I hope that you dont mind that it took so long but as i said i completely restructured my code. Thanks for all the help so far maybe now someone has the answer to my problem. :)

Answer 1

You are assigning a string to $id within the loop, $id = $rpw['id'];. The next time mysql_fetch_assoc($id) is executed, $id is a string, hence the warning message.

while($rpw = mysql_fetch_assoc($id)) {
    echo "Benutzer id: ", $rpw['id'], "\r\n";
}