mysql_insert_id

(PHP 4, PHP 5)

mysql_insert_id Liefert die ID, die in der vorherigen Abfrage erzeugt wurde

Warnung

Diese Erweiterung ist seit PHP 5.5.0 als veraltet markiert und wurde in PHP 7.0.0 entfernt. Verwenden Sie stattdessen die Erweiterungen MySQLi oder PDO_MySQL. Weitere Informationen bietet der Ratgeber MySQL: Auswahl einer API. Alternativen zu dieser Funktion umfassen:

Beschreibung

mysql_insert_id(resource $link_identifier = NULL): int

Gibt die ID zurück, die für eine AUTO_INCREMENT-Spalte durch die vorherige Abfrage (meist INSERT) erzeugt wurde.

Parameter-Liste

link_identifier

Die MySQL-Verbindung. Wird die Verbindungskennung nicht angegeben, wird die letzte durch mysql_connect() geöffnete Verbindung angenommen. Falls keine solche Verbindung gefunden wird, wird versucht, eine Verbindung aufzubauen, wie es beim Aufruf von mysql_connect() ohne Angabe von Argumenten der Fall wäre. Falls zufällig keine Verbindung gefunden oder aufgebaut werden kann, wird eine Warnung der Stufe E_WARNING erzeugt.

Rückgabewerte

Gibt bei Erfolg die ID zurück, die durch die vorherige Abfrage für eine AUTO_INCREMENT-Spalte erzeugt wurde, 0, falls die vorherige Abfrage keinen AUTO_INCREMENT Wert erzeugt hatte, oder false, falls keine MySQL-Verbindung existierte.

Beispiele

Beispiel #1 mysql_insert_id()-Beispiel

<?php
$link
= mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!
$link) {
die(
'Keine Verbindung möglich: ' . mysql_error());
}
mysql_select_db('mydb');

mysql_query("INSERT INTO mytable (product) values ('kossu')");
printf("Der letze eingefügte Datensatz hat die ID %d\n", mysql_insert_id());
?>

Anmerkungen

Achtung

mysql_insert_id() konvertiert den Typ der Rückgabe der nativen MySQL C API-Funktion mysql_insert_id() in den Typ long (in PHP als int bezeichnet). Falls Ihre AUTO_INCREMENT-Spalte vom Typ BIGINT (64 Bit) ist, ist der Wert, den mysql_insert_id() zurückgibt, nicht korrekt. Verwenden Sie in einer SQL-Abfrage in diesem Fall stattdessen die MySQL-interne SQL-Funktion LAST_INSERT_ID(). Für weitergehende Informationen bezüglich PHPs maximaler Integer-Werte, lesen Sie bitte die Integer-Dokumenation.

Hinweis:

Da mysql_insert_id() mit der zuletzt durchgeführten Abfrage arbeitet, sollten Sie mysql_insert_id() unmittelbar nach der Abfrage aufrufen, die den Wert erzeugt.

Hinweis:

Der Wert der MySQL-SQL-Funktion LAST_INSERT_ID() gibt immer den zuletzt erzeugten AUTO_INCREMENT-Wert zurück. Dieser wird zwischen Abfragen nicht zurückgesetzt.

Siehe auch

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User Contributed Notes 12 notes

up
5
Alfred Nony Mouse
16 years ago
There's nothing inherently wrong with using auto-increment fields. There's also nothing wrong with the main competetive idea, which is for the database to supply a primitive sequence of non-repeating identifiers, typically integers. This is rather like which side of the road you drive on.

The bigger problem is when people don't understand what they are doing with database access. It's like driving a car without really knowing the rules of the road. Such people wind up making bad decisions without realizing it, and then, eventually, something breaks.

Databases are complex beasts, and worth taking the time to really understand. Learn about the implications and limitations of different approaches to solving problems. Then, you will be prepared to pick a solution based on what has to work.
up
4
bargainbatman at gmail dot com
14 years ago
I thought this would be relevant to all the people using mysqli and looking for the ID after INSERT command :

<?php
function insert_join($catid, $disc_id) {
// insert a new item into the database
  
$conn = db_connect();
  
// insert new item
  
$demande = "insert into categories_disc values ('', '".$catid."', '".$disc_id."')";
  
$resultat = $conn->query($demande);
   if (!
$resultat) {
     return
false;
   } else {
    return
$conn->insert_id; // function will now return the ID instead of true.
}

}
?>

Then, on the other side, let us call this function as follows :

<?php
$cat_id
= insert_join($catid, $disc_id);
if(
$cat_id !== false) {
   
        echo
"<p>Category stuff was added to the database as follows : <br>";
        echo
"<hr>ID de la category : ".$cat_id."</p><hr>";

        }
?>
up
0
Anonymous
18 years ago
Take care of setting an empty value for the AUTO_INCREMENT Field else you never get a value except zero returned from mysq_insert_id() ....

Ciao Ephraim
up
-1
elinor dot hurst at REMOVETHIS dot gmail dot com
15 years ago
I don't get all the fuss around this.

I read:
"The value of mysql_insert_id() is affected only by statements issued within the current client connection. It is not affected by statements issued by other clients."

See: http://dev.mysql.com/doc/refman/5.0/es/mysql-insert-id.html

I can't really see what's inaccurate about that.

"In the case of a multiple-row INSERT statement, mysql_insert_id() returns the first automatically generated AUTO_INCREMENT value; if no such value is generated, it returns the last last explicit value inserted into the AUTO_INCREMENT column."

I must be missing something here but why would you insert multiple rows and then only handle the last one with some favoured behaviour? You could just as well insert them one at a time and then handle each row separately with the latest id.

I can't see what's wrong with that.

However I can see what's wrong with simply using max(my_table.id_column) because of the concurrent access issues this would imply.
up
-1
athies at gmail dot com
18 years ago
Just a quick note. mysql_insert_id() does work with REPLACE.
up
-2
foros (_AT_) anthalia.com
16 years ago
Forget about using MAX to get the last inserted id. Race conditions like other users inserting between your SELECT MAX(.. and your INSERT may render your id unusable.

The WAY to get the id is by using mysql_insert_id() or the mysql SQL function LAST_INSERT_ID().

Take care, if using mysql_insert_id() you should provide the resource returned by the mysql_connect, not the resultset returned by mysql_query.
up
-3
hoangvu4000 at gmail dot com
10 years ago
How to get ID of the last updated row in MySQL?

75
down vote
I've found an answer to this problem :)

by Pomyk

SET @update_id := 0;
UPDATE some_table SET row = 'value', id = (SELECT @update_id := id)
WHERE some_other_row = 'blah' LIMIT 1;
SELECT @update_id;
EDIT by aefxx

This technique can be further expanded to retrieve the ID of every row affected by an update statement:

SET @uids := null;
UPDATE footable
   SET foo = 'bar'
WHERE fooid > 5
   AND ( SELECT @uids := CONCAT_WS(',', fooid, @uids) );
SELECT @uids;
This will return a string with all the IDs concatenated by a colon.

(questions: 1388025  form stackoverflow)
up
-3
vksgeneric at hotmail dot com
24 years ago
You can't do an INSERT DELAYED and expect to get anything but zero, for it runs in a separate thread, and mysql_insert_id() is tied to the current thread.
Vlad
up
-3
louis at intoplay dot com
16 years ago
If mysql_insert_id() returns 0 or null, check your auto increment field is not being set by your sql query, also if you have multiple db connections like I did, the solution is to create a seperate db connection for this query.
up
-4
dhiraj dot webdeveloper at gmail dot com
6 years ago
MySQLi Procedural
//---------------------------------------------
$last_id = mysqli_insert_id($conn);

//---------------------------------------------
<?php
$servername
= "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die(
"Connection failed: " . mysqli_connect_error());
}

$sql = "INSERT INTO MyGuests (firstname, lastname, email)
VALUES ('John', 'Doe', 'john@example.com')"
;

if (
mysqli_query($conn, $sql)) {
   
$last_id = mysqli_insert_id($conn);
    echo
"New record created successfully. Last inserted ID is: " . $last_id;
} else {
    echo
"Error: " . $sql . "<br>" . mysqli_error($conn);
}

mysqli_close($conn);
?>
up
-6
heiligkind at yahoo dot de
18 years ago
If you insert a data row by using the ON DUPLICATE KEY UPDATE clause in an INSERT-statement, the mysql_insert_id() function will return not the same results as if you directly use LAST_INSERT_ID() in MySQL.

See the following example:

<?
   // insert a datarow, primary key is auto_increment
   // value is a unique key
   $query = "INSERT INTO test (value) VALUES ('test')";
   mysql_query( $query );

   echo 'LAST_INSERT_ID: ',
          mysql_query( "SELECT LAST_INSERT_ID()" ),
          '<br>mysql_insert_id: ',
          mysql_insert_id();

?>

This will print:

LAST_INSERT_ID: 1
mysql_insert_id: 1

In this case the function returns the same as the MySQL-Statement.
But see the insert on an existing key:

<?
   $query = "INSERT INTO test (value)
                  VALUES ('test')
                  ON DUPLICATE KEY UPDATE value = 'test2'";
   mysql_query( $query );

   echo 'LAST_INSERT_ID: ',
          mysql_query( "SELECT LAST_INSERT_ID()" ),
          '<br>mysql_insert_id: ',
          mysql_insert_id();

?>

This will print:

LAST_INSERT_ID: 2
mysql_insert_id: 1

By using the ON DUPLICATE KEY UPDATE clause, only the old datarow will be modified, if the INSERT statement causes a duplicate entry, but the LAST_INSERT_ID() function returns the next auto_increment value for the primary key, which is by the way not set as the next auto_increment value in the database.

The mysql_insert_id() function returns the primary key of the old (and changed) data row. For me this is the right operation method, because the LAST_INSERT_ID() function returns a value which is not referenced to a data row at all.

Greets from Munich.

heiligkind
up
-4
Steve Bond
19 years ago
If you use this function after doing an INSERT ... SELECT to insert multiple rows at once, you get the autonumber ID of the *first* row added by the INSERT.

e.g. if there are 4 records in table 'init' that have column 'type' = 2
I want to add these 4 records to table 'game'
Table game has an autonumber column 'game_id' that is currently at 32.

If I do this query:

INSERT INTO game (type, players, rounds)
SELECT type, players, rounds FROM init
WHERE type = 2

Then mysql_insert_id() will return 33, not 36.
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