PHP 8.2.0 Alpha 2 available for testing

mysqli::$insert_id

mysqli_insert_id

(PHP 5, PHP 7, PHP 8)

mysqli::$insert_id -- mysqli_insert_id直近のクエリの AUTO_INCREMENT カラムで生成した値を返す

説明

オブジェクト指向型

int|string $mysqli->insert_id;

手続き型

mysqli_insert_id(mysqli $mysql): int|string

AUTO_INCREMENT 属性を持つカラムがあるテーブル上での INSERTUPDATE クエリが生成したIDを返します。 INSERT 文が複数の行を変更する場合、 最初に自動生成された値を返します。

MySQL 関数 LAST_INSERT_ID() を使って INSERTUPDATE を実行すると、 mysqli_insert_id() が返す値も変更されます。 AUTO_INCREMENT の値を生成するために LAST_INSERT_ID(expr) を使った場合、 生成された AUTO_INCREMENT の値ではなく、 最後の expr の値を返します。

直前のクエリが AUTO_INCREMENT の値を変更しなかった場合は、 0 を返します。 クエリが値を生成した直後に、 mysqli_insert_id() をコールする必要があります。

パラメータ

link

手続き型のみ: mysqli_connect() あるいは mysqli_init() が返す mysqliオブジェクト。

戻り値

直前のクエリで更新された AUTO_INCREMENT フィールドの値を返します。接続での直前のクエリがない場合や クエリが AUTO_INCREMENT の値を更新しなかった場合は ゼロを返します。

現在の接続を使って発行されたクエリだけが、戻り値に影響します。 他の接続やクライアントが発行したクエリからは影響を受けません。

注意:

もし数値が int の最大値をこえた場合、 文字列で結果を返します。

例1 $mysqli->insert_id の例

オブジェクト指向型

<?php

mysqli_report
(MYSQLI_REPORT_ERROR MYSQLI_REPORT_STRICT);
$mysqli = new mysqli("localhost""my_user""my_password""world");

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

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

printf("New record has ID %d.\n"$mysqli->insert_id);

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

手続き型

<?php

mysqli_report
(MYSQLI_REPORT_ERROR MYSQLI_REPORT_STRICT);
$link mysqli_connect("localhost""my_user""my_password""world");

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

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

printf("New record has ID %d.\n"mysqli_insert_id($link));

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

上の例の出力は以下となります。

New record has ID 1.
add a note

User Contributed Notes 8 notes

up
9
mmulej at gmail dot com
1 year 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.
up
43
will at phpfever dot com
16 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;
}
?>
up
2
www dot wesley at gmail dot com
3 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.
up
4
bert at nospam thinc dot nl
13 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.]
up
2
Nick Baicoianu
15 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
?>
up
2
alan at commondream dot net
17 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.
up
-6
owenzx at gmail dot com
8 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
up
-9
drburnett at mail dot com
5 years ago
msqli_insert_id();
This seems to return that last id entered.
BUT,  if you have multiple users running the same code, depending on the server or processor I have seen it return the wrong id.

Test Case:
Two users added an item to their list.
I have had a few times where the id was the id from the other user.
This is very very rare and it only happens on my test server and not my main server.

I am guessing it is because of multicores (maybe hyperthreading) or how the operating system handles multi-threads.

It is rare, but it happens.
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