mysql_error

(PHP 4, PHP 5)

mysql_error — Retourne le texte associé avec l'erreur générée lors de la dernière requête

Avertissement

Cette extension était obsolète en PHP 5.5.0, et a été supprimée en PHP 7.0.0. À la place, vous pouvez utiliser l'extension MySQLi ou l'extension PDO_MySQL. Voir aussi MySQL : choisir une API du guide. Alternatives à cette fonction :

Description

mysql_error(resource $link_identifier = NULL): string

mysql_error() retourne le message d'erreur généré par la dernière commande MySQL. Les erreurs retournées par le serveur MySQL ne génèrent plus de message d'alerte. À la place, vous devez utiliser la fonction mysql_error() pour lire le contenu du message. Notez que cette fonction ne retourne que le texte de l'erreur la plus récente(n'incluant pas mysql_error() et mysql_errno()), ce qui fait que si vous souhaitez l'utiliser, vous devez vous assurer de sa valeur avant de lancer une autre requête.

Liste de paramètres

link_identifier: La connexion MySQL. S'il n'est pas spécifié, la dernière connexion ouverte avec la fonction mysql_connect() sera utilisée. Si une telle connexion n'est pas trouvée, la fonction tentera d'ouvrir une connexion, comme si la fonction mysql_connect() avait été appelée sans argument. Si aucune connexion n'est trouvée ou établie, une alerte de niveau E_WARNING sera générée.

Valeurs de retour

Retourne le texte de l'erreur de la dernière fonction MySQL, ou '' (chaîne vide) si aucune erreur survient.

Exemples

Exemple #1 Exemple avec mysql_error()

<?php
$link = mysql_connect("localhost", "mysql_user", "mysql_password");

mysql_select_db("nonexistentdb", $link);
echo mysql_errno($link) . ": " . mysql_error($link). "\n";

mysql_select_db("kossu", $link);
mysql_query("SELECT * FROM nonexistenttable", $link);
echo mysql_errno($link) . ": " . mysql_error($link) . "\n";
?>

Résultat de l'exemple ci-dessus est similaire à :

1049: Unknown database 'nonexistentdb'
1146: Table 'kossu.nonexistenttable' doesn't exist

Voir aussi

mysql_errno() - Retourne le numéro d'erreur de la dernière commande MySQL
» les codes erreurs MySQL

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User Contributed Notes 14 notes

down

aleczapka _at) gmx dot net ¶

20 years ago


If you want to display errors like "Access denied...", when mysql_error() returns "" and mysql_errno() returns 0, use  $php_errormsg. This Warning will be stored there.  You need to have track_errors set to true in your php.ini.

Note. There is a bug in either documentation about error_reporting() or in mysql_error() function cause manual for mysql_error(), says:  "Errors coming back from the MySQL database backend no longer issue warnings." Which is not true.

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Pendragon Castle ¶

15 years ago


Using a manipulation of josh ><>'s function, I created the following. It's purpose is to use the DB to store errors. It handles both original query, as well as the error log. Included Larry Ullman's escape_data() as well since I use it in q().



<?php

function escape_data($data){

 global $dbc;

 if(ini_get('magic_quotes_gpc')){

  $data=stripslashes($data);

 }

 return mysql_real_escape_string(trim($data),$dbc);

}



function q($page,$query){ 

// $page

 $result = mysql_query($query);

 if (mysql_errno()) {

  $error = "MySQL error ".mysql_errno().": ".mysql_error()."\n<br>When executing:<br>\n$query\n<br>";

  $log = mysql_query("INSERT INTO db_errors (error_page,error_text) VALUES ('$page','".escape_data($error)."')");

 }

}



// Run the query using q()

$query = "INSERT INTO names (first, last) VALUES ('myfirst', 'mylast'");

$result = q("Sample Page Title",$query);

?>

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Florian Sidler ¶

14 years ago


Be aware that if you are using multiple MySQL connections you MUST support the link identifier to the mysql_error() function. Otherwise your error message will be blank.

Just spent a good 30 minutes trying to figure out why i didn't see my SQL errors.

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l dot poot at twing dot nl ¶

18 years ago


When creating large applications it's quite handy to create a custom function for handling queries. Just include this function in every script. And use db_query(in this example) instead of mysql_query.

This example prompts an error in debugmode (variable $b_debugmode ). An e-mail with the error will be sent to the site operator otherwise.

The script writes a log file in directory ( in this case /log ) as well.

The system is vulnerable when database/query information is prompted to visitors. So be sure to hide this information for visitors anytime.

Regars,

Lennart Poot
http://www.twing.nl

<?php
$b_debugmode = 1; // 0 || 1

$system_operator_mail = 'developer@company.com';
$system_from_mail = 'info@mywebsite.com';

function db_query( $query ){
  global $b_debugmode;
  
  // Perform Query
  $result = mysql_query($query);

  // Check result
  // This shows the actual query sent to MySQL, and the error. Useful for debugging.
  if (!$result) {
    if($b_debugmode){
      $message  = '<b>Invalid query:</b><br>' . mysql_error() . '<br><br>';
      $message .= '<b>Whole query:</b><br>' . $query . '<br><br>';
      die($message);
    }

    raise_error('db_query_error: ' . $message);
  }
  return $result;
}

  function raise_error( $message ){
    global $system_operator_mail, $system_from_mail;

    $serror=
    "Env:       " . $_SERVER['SERVER_NAME'] . "\r\n" .
    "timestamp: " . Date('m/d/Y H:i:s') . "\r\n" .
    "script:    " . $_SERVER['PHP_SELF'] . "\r\n" .
    "error:     " . $message ."\r\n\r\n";

    // open a log file and write error
    $fhandle = fopen( '/logs/errors'.date('Ymd').'.txt', 'a' );
    if($fhandle){
      fwrite( $fhandle, $serror );
      fclose(( $fhandle ));
     }
  
    // e-mail error to system operator
    if(!$b_debugmode)
      mail($system_operator_mail, 'error: '.$message, $serror, 'From: ' . $system_from_mail );
  }

?>

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-2

Anonymous ¶

23 years ago


some error can't handle. Example: 



ERROR 1044: Access denied for user: 'ituser@mail.ramon.intranet' to database 'itcom'



This error ocurrs when a intent of a sql insert of no authorized user. The results: mysql_errno = 0 and the mysql_error = "" .

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-3

Anonymous ¶

19 years ago


My suggested implementation of mysql_error():



$result = mysql_query($query) or die("<b>A fatal MySQL error occured</b>.\n<br />Query: " . $query . "<br />\nError: (" . mysql_errno() . ") " . mysql_error());



This will print out something like...



A fatal MySQL error occured.

Query: SELECT * FROM table

Error: (err_no) Bla bla bla, you did everything wrong



It's very useful to see your query in order to detect problems with syntax. Most often, the output message from MySQL doesn't let you see enough of the query in the error message to let you see where your query went bad- it a missing quote, comma, or ( or ) could have occured well before the error was detected. I do -not- recomend using this procedure, however, for queries which execute on your site that are not user-specific as it has the potential to leak sensative data. Recomended use is just for debugging/building a script, and for general user-specific queries which would at the worst, leak the users own information to themself.



Good luck,



-Scott

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-3

olaf at amen-online dot de ¶

20 years ago


When dealing with user input, make sure that you use
<?php
echo htmlspecialchars (mysql_error ());
?>
instead of
<?php
echo mysql_error ();
?>

Otherwise it might be possible to crack into your system by submitting data that causes the SQL query to fail and that also contains javascript commands.

Would it make sense to change the examples in the documentation for mysql_query () and for mysql_error () accordingly?

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-3

josh ><> ¶

21 years ago


Oops, the code in my previous post only works for queries that don't return data (INSERT, UPDATE, DELETE, etc.), this updated function should work for all types of queries (using $result = myquery($query);):

    function myquery ($query) {
        $result = mysql_query($query);
        if (mysql_errno())
            echo "MySQL error ".mysql_errno().": ".mysql_error()."\n<br>When executing:<br>\n$query\n<br>";
        return $result;
    }

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-3

Gianluigi_Zanettini-MegaLab.it ¶

17 years ago


A friend of mine proposed a great solution.

<?php
$old_track = ini_set('track_errors', '1');

.....

        if ($this->db_handle!=FALSE && $db_selection_status!=FALSE)
            {
            $this->connected=1;
            ini_set('track_errors', $old_track);
            }
        else
            {
            $this->connected=-1;
            $mysql_warning=$php_errormsg;
            ini_set('track_errors', $old_track);
            throw new mysql_cns_exception(1, $mysql_warning . " " . mysql_error());
            }
?>

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-4

scott at rocketpack dot net ¶

21 years ago


My suggested implementation of mysql_error():

$result = mysql_query($query) or die("<b>A fatal MySQL error occured</b>.\n<br />Query: " . $query . "<br />\nError: (" . mysql_errno() . ") " . mysql_error());

This will print out something like...

<b>A fatal MySQL error occured</b>.
Query: SELECT * FROM table
Error: (err_no) Bla bla bla, you did everything wrong

It's very useful to see your query in order to detect problems with syntax. Most often, the output message from MySQL doesn't let you see enough of the query in the error message to let you see where your query went bad- it a missing quote, comma, or ( or ) could have occured well before the error was detected. I do -not- recomend using this procedure, however, for queries which execute on your site that are not user-specific as it has the potential to leak sensative data. Recomended use is just for debugging/building a script, and for general user-specific queries which would at the worst, leak the users own information to themself.

Good luck,

-Scott

down

-5

Gianluigi_Zanettini-MegaLab.it ¶

17 years ago


"Errors coming back from the MySQL database backend no longer issue warnings." Please note, you have an error/bug here. In fact, MySQL 5.1 with PHP 5.2:

Warning: mysql_connect() [function.mysql-connect]: Unknown MySQL server host 'locallllllhost' (11001)

That's a warning, which is not trapped by mysql_error()!

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-5

phpnet at robzazueta dot com ¶

17 years ago


This is a big one - As of MySQL 4.1 and above, apparently, the way passwords are hashed has changed. PHP 4.x is not compatible with this change, though PHP 5.0 is. I'm still using the 4.x series for various compatibility reasons, so when I set up MySQL 5.0.x on IIS 6.0 running PHP 4.4.4 I was surpised to get this error from mysql_error():

MYSQL: Client does not support authentication protocol requested by server; consider upgrading MySQL client 

According to the MySQL site (http://dev.mysql.com/doc/refman/5.0/en/old-client.html) the best fix for this is to use the OLD_PASSWORD() function for your mysql DB user. You can reset it by issuing to MySQL:

Set PASSWORD for 'user'@'host' = OLD_PASSWORD('password');

This saved my hide.

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-5

miko_il AT yahoo DOT com ¶

21 years ago


Following are error codes that may appear when you call MySQL from any host language:

http://www.mysql.com/doc/en/Error-returns.html

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-9

Gerrit ¶

10 years ago


The following code returns two times the same error, even though I would have expected only one: 

   $ conn = mysql_connect ('localhost', 'root', ''); 
   $ conn2 = mysql_connect ('localhost', 'root', ''); 
  
   mysql_select_db ('db1', $ conn); 
   mysql_select_db ('db2', $ conn2); 
  
   $ result = mysql_query ("select 1 from dual", $ conn); 
   $ result2 = mysql_query ("select 1 from luad", $ conn2); 
  
   echo mysql_error ($ conn) "<hr>". 
   echo mysql_error ($ conn2) "<hr>". 

The reason for this is that mysql_connect not working as expected a further connection returns. Since the parameters are equal, a further reference to the previous link is returned. So also changes the second mysql_select_db the selected DB of $conn to 'db2'. 

If you change the connection parameters of the second connection to 127.0.0.1, a new connection is returned. In addition to the parameters new_link the mysql_connect() function to be forced.

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