PHP 5.4.36 Released

mysqli::$insert_id

mysqli_insert_id

(PHP 5)

mysqli::$insert_id -- mysqli_insert_idReturns the auto generated id used in the last query

Descrierea

Stil obiect-orientat

Stil procedural

mixed mysqli_insert_id ( mysqli $link )

The mysqli_insert_id() function returns the ID generated by a query on a table with a column having the AUTO_INCREMENT attribute. If the last query wasn't an INSERT or UPDATE statement or if the modified table does not have a column with the AUTO_INCREMENT attribute, this function will return zero.

Notă:

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

Parametri

link

Numai stilul procedural: Un identificator al legăturii întors de mysqli_connect() sau mysqli_init()

Valorile întoarse

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.

Notă:

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

Exemple

Example #1 $mysqli->insert_id example

Stil obiect-orientat

<?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();
?>

Stil procedural

<?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);
?>

Exemplele de mai sus vor afișa:

New Record has id 1.
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User Contributed Notes 6 notes

up
15
will at phpfever dot com
8 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;
}
?>
up
5
bert at nospam thinc dot nl
6 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.]
up
1
owenzx at gmail dot com
1 year 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
up
2
Nick Baicoianu
7 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
?>
up
0
alan at commondream dot net
10 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.
up
-11
mail at nikha dot org
11 months ago
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??);
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