mysqli::$insert_id

mysqli_insert_id

(PHP 5, PHP 7, PHP 8)

mysqli::$insert_id -- mysqli_insert_id — 返回上次查询为 AUTO_INCREMENT 列生成的值

说明

面向对象风格

int|string $mysqli->insert_id;

过程化风格

mysqli_insert_id(mysqli $mysql): int|string

返回有 AUTO_INCREMENT 属性的表执行 INSERT 或 UPDATE 查询生成的 ID。对于多行 INSERT 语句，返回成功插入的第一个自动生成的值。

使用 LAST_INSERT_ID() MySQL 函数执行 INSERT 或 UPDATE 语句也会修改 mysqli_insert_id() 返回的值。如果使用 LAST_INSERT_ID(expr) 生成 AUTO_INCREMENT 的值，将返回最后 expr 的值而不是生成的 AUTO_INCREMENT 值。

如果之前的语句没有修改 AUTO_INCREMENT 的值，则返回 0。mysqli_insert_id() 必须在生成值的语句之后立即调用。

参数

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

返回值

上一个查询更新的 AUTO_INCREMENT 字段的值。如果连接上没有之前的查询或者查询没有更新 AUTO_INCREMENT 值。则返回 0。

只有使用当前连接发出的语句才会影响返回值。该值不受使用其它连接或客户端发出的语句的影响。

注意:
如果数字大于最大 int 值，将以字符串返回该值。

示例

示例 #1 $mysqli->insert_id 示例

面向对象风格

<?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);

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

过程化风格

<?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));

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

以上示例会输出：

New record has ID 1.

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用户贡献的备注 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