mysqli::$insert_id

mysqli_insert_id

(PHP 5, PHP 7, PHP 8)

mysqli::$insert_id -- mysqli_insert_id — Devuelve el id autogenerado que se utilizó en la última consulta

Descripción

Estilo orientado a objetos

mixed $mysqli->insert_id;

Estilo por procedimientos

mysqli_insert_id(mysqli $link): mixed

La función mysqli_insert_id() devuelve el ID generado por una query (normalmente INSERT) en una tabla con una columna que tenga el atributo AUTO_INCREMENT. Si no se enviaron declaraciones INSERT o UPDATE a través de esta conexión, o si la tabla modificada no tiene una columna con el atributo AUTO_INCREMENT, esta función devolverá cero.

Nota:
Realizar una sentencia INSERT o UPDATE usando la función LAST_INSERT_ID() modificará el valor retornado por la función mysqli_insert_id().

Parámetros

link: Sólo estilo por procediminetos: Un identificador de enlace devuelto por mysqli_connect() o mysqli_init()

Valores devueltos

El valor de el campo AUTO_INCREMENT que fué actualizado por la consulta anterior. Devuelve cero si no hubo una consulta previa en la conexión o si la consulta no actualiza un valor AUTO_INCREMENT.

Nota:
Si el número es mayor que el valor máximo int, mysqli_insert_id() devolverá un string.

Ejemplos

Ejemplo #1 Ejemplo de $mysqli->insert_id

Estilo orientado a objetos

<?php
$mysqli = new mysqli("localhost", "mi_usuario", "mi_password", "world");

/* comprobar la conexión. */
if (mysqli_connect_errno()) {
    printf("Error de conexión: %s\n", mysqli_connect_error());
    exit();
}

$mysqli->query("CREATE TABLE myCity LIKE City");

$query = "INSERT INTO myCity VALUES (NULL, 'Stuttgart', 'DEU', 'Stuttgart', 617000)";
$mysqli->query($query);

printf ("Nuevo registro con el id %d.\n", $mysqli->insert_id);

/* drop table */
$mysqli->query("DROP TABLE myCity");

/* close connection */
$mysqli->close();
?>

Estilo por procedimientos

<?php
$link = mysqli_connect("localhost", "mi_usuario", "mi_password", "world");

/* comprobar la conexión. */
if (mysqli_connect_errno()) {
    printf("Error de conexión: %s\n", mysqli_connect_error());
    exit();
}

mysqli_query($link, "CREATE TABLE myCity LIKE City");

$query = "INSERT INTO myCity VALUES (NULL, 'Stuttgart', 'DEU', 'Stuttgart', 617000)";
mysqli_query($link, $query);

printf ("Nuevo registro con el id %d.\n", mysqli_insert_id($link));

/* drop table */
mysqli_query($link, "DROP TABLE myCity");

/* close connection */
mysqli_close($link);
?>

El resultado de los ejemplos sería:

Nuevo registro con el id 1.

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User Contributed Notes 9 notes

down

will at phpfever dot com ¶

18 years ago

I have received many statements that the insert_id property has a bug because it "works sometimes".  Keep in mind that when using the OOP approach, the actual instantiation of the mysqli class will hold the insert_id.  

The following code will return nothing.
<?php
$mysqli = new mysqli('host','user','pass','db');
if ($result = $mysqli->query("INSERT INTO t (field) VALUES ('value');")) {
   echo 'The ID is: '.$result->insert_id;
}
?>

This is because the insert_id property doesn't belong to the result, but rather the actual mysqli class.  This would work:

<?php
$mysqli = new mysqli('host','user','pass','db');
if ($result = $mysqli->query("INSERT INTO t (field) VALUES ('value');")) {
   echo 'The ID is: '.$mysqli->insert_id;
}
?>

down

mmulej at gmail dot com ¶

3 years ago

There has been no examples with prepared statements yet.

```php
$u_name = "John Doe";
$u_email = "johndoe@example.com";

$stmt = $connection->prepare(
    "INSERT INTO users (name, email) VALUES (?, ?)"
);
$stmt->bind_param('ss', $u_name, $u_email);
$stmt->execute();

echo $stmt->insert_id;
```

For UPDATE you simply change query string and binding parameters accordingly, the rest stays the same.

Of course the table needs to have AUTOINCREMENT PRIMARY KEY.

down

adrian dot nesse dot wiik at gmail dot com ¶

1 year ago

If you try to INSERT a row using ON DUPLICATE KEY UPDATE, be aware that insert_id will not update if the ON DUPLICATE KEY UPDATE clause was triggered.

When you think about it, it's actually very logical since ON DUPLICATE KEY UPDATE is an UPDATE statement, and not an INSERT.

In a worst case scenario, if you're iterating over something and doing INSERTs while relying on insert_id in later code, you could be pointing at the wrong row on iterations where ON DUPLICATE KEY UPDATE is triggered!

down

bert at nospam thinc dot nl ¶

16 years ago

Watch out for the oo-style use of $db->insert_id. When the insert_id exceeds 2^31 (2147483648) fetching the insert id renders a wrong, too large number. You better use the procedural mysqli_insert_id( $db ) instead.



[EDIT by danbrown AT php DOT net: This is another prime example of the limits of 32-bit signed integers.]

down

www dot wesley at gmail dot com ¶

5 years ago

When using "INSERT ... ON DUPLICATE KEY UPDATE `id` = LAST_INSERT_ID(`id`)", the AUTO_INCREMENT will increase in an InnoDB table, but not in a MyISAM table.

down

Nick Baicoianu ¶

17 years ago

When running extended inserts on a table with an AUTO_INCREMENT field, the value of mysqli_insert_id() will equal the value of the *first* row inserted, not the last, as you might expect.

<?
//mytable has an auto_increment field
$db->query("INSERT INTO mytable (field1,field2,field3) VALUES ('val1','val2','val3'),
('val1','val2','val3'),
('val1','val2','val3')");

echo $db->insert_id; //will echo the id of the FIRST row inserted
?>

down

jpage at chatterbox dot fyi ¶

1 year ago

What is unclear is how concurrency control affects this function.   When you make two successive calls to mysql where the result of the second depends on the first,  another user may have done an insert in the meantime.

The documentation is silent on this, so I always determine the value of an auto increment before and after an insert to guard against this.

down

alan at commondream dot net ¶

20 years ago

I was having problems with getting the inserted id, and did a bit of testing. It ended up that if you commit a transaction before getting the last inserted id, it returns 0 every time, but if you get the last inserted id before committing the transaction, you get the correct value.

down

-5

owenzx at gmail dot com ¶

11 years ago

The example is lack of insert_id in multi_query. Here is my example:
Assuming you have a new test_db in mysql like this:

create database if not exists test_db;
use test_db;
create table user_info (_id serial, name varchar(100) not null);
create table house_info (_id serial, address varchar(100) not null);

Then you run a php file like this:

<?php
define('SERVER', '127.0.01');
define('MYSQL_USER', 'your_user_name');
define('MYSQL_PASSWORD', 'your_password');

$db = new mysqli(SERVER, MYSQL_USER, MYSQL_PASSWORD, "test_db", 3306);
if ($db->connect_errno)
  echo "create db failed, error is ", $db->connect_error;
else {
  $sql = "insert into user_info "
    . "(name) values "
    . "('owen'), ('john'), ('lily')";
  if (!$result = $db->query($sql))
    echo "insert failed, error: ", $db->error;
  else
    echo "last insert id in query is ", $db->insert_id, "\n";
  $sql = "insert into user_info"
    . "(name) values "
    . "('jim');";
  $sql .= "insert into house_info "
    . "(address) values "
    . "('shenyang')";
  if (!$db->multi_query($sql))
    echo "insert failed in multi_query, error: ", $db->error;
  else {
    echo "last insert id in first multi_query is ", $db->insert_id, "\n";
    if ($db->more_results() && $db->next_result())
      echo "last insert id in second multi_query is ", $db->insert_id, "\n";
    else
      echo "insert failed in multi_query, second query error is ", $db->error;
  }
  $db->close();
}
?>

You will get output like this:

last insert id in query is 1
last insert id in first multi_query is 4
last insert id in second multi_query is 1

Conclusion:
1 insert_id works in multi_query
2 insert_id is the first id mysql has used if you have insert multi values

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