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.
(PHP 5, PHP 7)
mysqli::$insert_id -- mysqli_insert_id — 返回最后一条插入语句产生的自增 ID
面向对象风格
过程化风格
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。
Note:
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。
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。
Note:
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。
Example #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.
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.
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.
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
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??);
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
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.]
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
?>
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;
}
?>
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.