mysqli_stmt::$insert_id

mysqli_stmt_insert_id

(PHP 5, PHP 7, PHP 8)

mysqli_stmt::$insert_id -- mysqli_stmt_insert_id — Returns the value generated for an AUTO_INCREMENT column by the last statement executed

说明

面向对象风格

int|string $mysqli_stmt->insert_id;

过程化风格

function mysqli_stmt_insert_id(mysqli_stmt $statement): int|string

Returns the ID generated by an INSERT or UPDATE statement on a table with a column having the AUTO_INCREMENT attribute. In the case of a multiple-row INSERT statement, it returns the first automatically generated value that was successfully inserted. If LAST_INSERT_ID(expr) is used to set the AUTO_INCREMENT value, or in an UPDATE, the value of expr is returned instead.

Returns 0 if the statement did not change an AUTO_INCREMENT value.

参数

statement

仅以过程化样式：由 mysqli_stmt_init() 返回的 mysqli_stmt 对象。

返回值

The value of the AUTO_INCREMENT field that was updated by the statement. Returns zero if the statement has not been executed yet or if it did not update an AUTO_INCREMENT value.

The value is specific to the statement object. It is not affected by other statements executed on the same connection after this statement, nor by statements issued using other connections or clients.

注意:

If the number is greater than the maximum int value, it will be returned as a string.

示例

示例 #1 mysqli_stmt::$insert_id example

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 TEMPORARY TABLE myCity LIKE City");

$stmt = $mysqli->prepare("INSERT INTO myCity (Name, CountryCode, District, Population) VALUES (?, ?, ?, ?)");

$name = 'Stuttgart';
$countryCode = 'DEU';
$district = 'Stuttgart';
$population = 617000;
$stmt->bind_param('sssi', $name, $countryCode, $district, $population);
$stmt->execute();

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

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 TEMPORARY TABLE myCity LIKE City");

$stmt = mysqli_prepare($link, "INSERT INTO myCity (Name, CountryCode, District, Population) VALUES (?, ?, ?, ?)");

$name = 'Stuttgart';
$countryCode = 'DEU';
$district = 'Stuttgart';
$population = 617000;
mysqli_stmt_bind_param($stmt, 'sssi', $name, $countryCode, $district, $population);
mysqli_stmt_execute($stmt);

printf("New record has ID %d.\n", mysqli_stmt_insert_id($stmt));

以上示例会输出：

New record has ID 1.

参见

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用户贡献的备注 2 notes

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hgernhardt at tcnewengland dot org
15 years ago
It should be noted that using mysqli_stmt->insert_id will not result in a unique ID being returned for each execution of a prepared insert statement.  In practice, it appears that the first insertion ID is returned.  If you are performing multiple inserts with the same prepared statement (one invocation of mysqli_stmt::prepare and multiple invocations of mysqli_stmt::execute() for a given statement), and need to keep the unique ID for each insert, use mysqli_connection->insert_id.
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3
Aurelien Marchand
11 years ago
for those with zerofilled IDs beware as mysqli_stmt::$insert_id will return an integer. In other words, if the ID returned from your query was #000099, the value produced by mysqli_stmt::$insert_id is now #99.

I have not found a way to fix this beside using sprintf('%06d',$id) but this assumes you know in advance the size of the ID beforehand.
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