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;
}
?>
mysqli::$insert_id
mysqli_insert_id
(PHP 5, PHP 7, PHP 8)
mysqli::$insert_id -- mysqli_insert_id — Возвращает значение, созданное для столбца AUTO_INCREMENT последним запросом
Описание
Объектно-ориентированный стиль
Процедурный стиль
Возвращает идентификатор, сгенерированный запросом INSERT или
UPDATE для таблицы со столбцом,
имеющим атрибут AUTO_INCREMENT.
В случае многострочного оператора INSERT он возвращает первое автоматически сгенерированное значение,
которое было успешно добавлено.
Выполнение запроса INSERT или UPDATE с использованием MySQL-функции LAST_INSERT_ID() также изменит значение,
возвращаемое mysqli_insert_id().
Если LAST_INSERT_ID(expr) использовался для генерации значения AUTO_INCREMENT,
он возвращает значение последнего expr вместо сгенерированного значения AUTO_INCREMENT.
Возвращает 0, если предыдущий оператор не изменил значение AUTO_INCREMENT.
mysqli_insert_id() должен вызываться сразу после запроса, сгенерировавшего значение.
Список параметров
-
mysql -
Только для процедурного стиля: объект mysqli, который вернула функция mysqli_connect() или функция mysqli_init().
Возвращаемые значения
Значение поля AUTO_INCREMENT, которое было
затронуто предыдущим запросом. Возвращает ноль, если предыдущий
запрос не затронул таблицы, содержащие поле AUTO_INCREMENT.
Только запросы, выданные с использованием текущего соединения, влияют на возвращаемое значение. На значение не влияют запросы, выданные с использованием других подключений или клиентов.
Замечание:
Если число больше максимального значения целого числа, функция вернёт строку.
Примеры
Пример #1 Пример функции $mysqli->insert_id
Объектно-ориентированный стиль
<?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("ID новой записи: %d.\n", $mysqli->insert_id);
/* удаление таблицы */
$mysqli->query("DROP TABLE myCity");Процедурный стиль
<?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, $query);
printf("ID новой записи: %d.\n", mysqli_insert_id($link));
/* удаление таблицы */
mysqli_query($link, "DROP TABLE myCity");Результат выполнения приведённых примеров:
ID новой записи: 1.
add a note
User Contributed Notes 10 notes
will at phpfever dot com ¶
18 years ago
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.
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.
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.]
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
?>
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!
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.
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.
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
drburnett at mail dot com ¶
7 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.
