mysqli::$insert_id

mysqli_insert_id

(PHP 5, PHP 7)

mysqli::$insert_id -- mysqli_insert_id — Returns the auto generated id used in the latest query

说明

面向对象风格

mixed $mysqli->insert_id;

过程化风格

mixed mysqli_insert_id ( mysqli $link )

The mysqli_insert_id() function returns the ID generated by a query (usually INSERT) on a table with a column having the AUTO_INCREMENT attribute. If no INSERT or UPDATE statements were sent via this connection, or if the modified table does not have a column with the AUTO_INCREMENT attribute, this function will return zero.

Note:

Performing an INSERT or UPDATE statement using the LAST_INSERT_ID() function will also modify the value returned by the mysqli_insert_id() function.

参数

link

仅以过程化样式：由mysqli_connect() 或 mysqli_init() 返回的链接标识。

返回值

The value of the AUTO_INCREMENT field that was updated by the previous query. Returns zero if there was no previous query on the connection or if the query did not update an AUTO_INCREMENT value.

Note:

If the number is greater than maximal int value, mysqli_insert_id() will return a string.

范例

Example #1 $mysqli->insert_id example

面向对象风格

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

/* check connection */
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);

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

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

过程化风格

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

/* check connection */
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));

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

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

以上例程会输出：

New Record has id 1.

User Contributed Notes

drburnett at mail dot com 19-Jan-2017 12:35
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.
Himadri himadri dot s dot roy at ansysoft dot com 18-Mar-2016 09:31
Hi  mail at nikha dot org,

I must say your way of getting the key is not correct. Consider a multi user situation where everyone is registering and you are returning their id. Let's say A, B and C are registering at the same time.

Regards
Himadri
mail at nikha dot org 20-Jan-2014 09:37
Hi Dears,

msqli_insert_id() simply does not ALWAYS return the correct value.

Use it only, if you performed some inserts just before. Then you get what you want.

In all other cases: may be, may be not.

I never found out why and why not.

I'm now performing a query like this:
SELECT MAX(`id`) FROM `table`

(by calling mysqli_query() in procedural style, for OO may be similar.)

It' s simple and  reliable - if you have set your id colum to "auto-increment". (if not: hm, why not??);
owenzx at gmail dot com 06-Sep-2013 11:54
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
bert at nospam thinc dot nl 22-Jul-2008 05:58
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.]
Nick Baicoianu 04-May-2007 10:10
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
?>
will at phpfever dot com 21-Apr-2006 03:40
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;
}
?>
alan at commondream dot net 03-Nov-2004 08:44
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.