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mysqli::$insert_id

mysqli_insert_id

(PHP 5, PHP 7, PHP 8)

mysqli::$insert_id -- mysqli_insert_id返回最后一条插入语句产生的自增 ID

说明

面向对象风格

过程化风格

mysqli_insert_id(mysqli $link): mixed

mysqli_insert_id() 函数返回最后一个 SQL 语句(通常是 INSERT 语句) 所操作的表中设置为 AUTO_INCREMENT 的列的值。 如果最后一个 SQL 语句不是 INSERT 或者 UPDATE 语句, 或者所操作的表中没有设置为 AUTO_INCREMENT 的列, 返回值为 0。

注意:

如果在所执行的 INSERT 或者 UPDATE 语句中使用了数据库函数 LAST_INSERT_ID()。 有可能会影响 mysqli_insert_id() 函数的返回值。

参数

mysql

仅以过程化样式:由mysqli_connect()mysqli_init() 返回的 mysqli 对象。

返回值

最后一条 SQL(INSERT 或者 UPDATE)所操作的表中设置为 AUTO_INCREMENT 属性的列的值。 如果指定的连接上尚未执行 SQL 语句,或者最后一条 SQL 语句所操作的表中没有设为 AUTO_INCREMENT 的列,返回 0。

注意:

如果返回值超出了 php 允许的最大整数值, mysqli_insert_id() 函数会以字符串形式返回这个值。

范例

示例 #1 $mysqli->insert_id 例程

面向对象风格

<?php
$mysqli 
= new mysqli("localhost""my_user""my_password""world");

/* 检查连接 */
if (mysqli_connect_errno()) {
    
printf("Connect failed: %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 ("New Record has id %d.\n"$mysqli->insert_id);

/* 删除表 */
$mysqli->query("DROP TABLE myCity");

/* 关闭连接 */
$mysqli->close();
?>

过程化风格

<?php
$link 
mysqli_connect("localhost""my_user""my_password""world");

/* 检查连接 */
if (mysqli_connect_errno()) {
    
printf("Connect failed: %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 ("New Record has id %d.\n"mysqli_insert_id($link));

/* 删除表 */
mysqli_query($link"DROP TABLE myCity");

/* 关闭连接 */
mysqli_close($link);
?>

以上例程会输出:

New Record has id 1.
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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
14 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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