mysqli::$insert_id

mysqli_insert_id

(PHP 5, PHP 7, PHP 8)

mysqli::$insert_id -- mysqli_insert_id — Returns the value generated for an AUTO_INCREMENT column by the last query

Опис

Об'єктно-орієнтований стиль

int|string $mysqli->insert_id;

Процедурний стиль

mysqli_insert_id(mysqli $mysql): int|string

Returns the ID generated by an INSERT or UPDATE query 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.

Performing an INSERT or UPDATE statement using the LAST_INSERT_ID() MySQL function will also modify the value returned by mysqli_insert_id(). If LAST_INSERT_ID(expr) was used to generate the value of AUTO_INCREMENT, it returns the value of the last expr instead of the generated AUTO_INCREMENT value.

Returns 0 if the previous statement did not change an AUTO_INCREMENT value. mysqli_insert_id() must be called immediately after the statement that generated the value.

Параметри

mysql: Тільки процедурний стиль: об'єкт mysqli, якого повертає функція mysqli_connect() або mysqli_init()

Значення, що повертаються

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.

Only statements issued using the current connection affect the return value. The value is not affected by statements issued using other connections or clients.

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

Приклади

Приклад #1 $mysqli->insert_id example

Об'єктно-орієнтований 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("New record has ID %d.\n", $mysqli->insert_id);

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

Процедурний 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("New record has ID %d.\n", mysqli_insert_id($link));

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

Подані вище приклади виведуть:

New record has ID 1.

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User Contributed Notes 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

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