mysqli_init
Note that if you Call a MySQL stored procedure to insert a new record and then reference $db->insert_id; you will get 0 back, not the last inserted ID.
It is therefore necessary to add a line to your MySQL Stored Procedure such as
select last_insert_id() as intRecordKey;
after the insert so that the query will return the new key value.
Then in your php code you can use the following
<?php
$objInsertResult = $db->query($strSQL);
$objInsertRow = $objInsertResult->fetch_object();
$intRecordKey = $objInsertRow->intRecordKey;
?>
It would however be nice if $db->insert_id did return the last insert ID following a stored procedure call.
mysqli_insert_id
mysqli->insert_id
(PHP 5)
mysqli_insert_id -- mysqli->insert_id — Retorna o id gerado automaticamente na última consulta
Descrição
Modo procedural:
Modo orientado a objeto (propriedade):
mysqli
int $insert_id;
A função mysqli_insert_id() retorna o ID gerado pela consulta em uma tabela com uma coluna com o atributo AUTO_INCREMENT. Se a última consulta não foi um INSERT ou UPDATE ou se a tabela modificada não tem uma coluna com o atributo AUTO_INCREMENT, esta função irá retornar zero.
Nota: Executar uma consulta INSERT ou UPDATE usando a função LAST_INSERT_ID() irá também modificar o valor retornado pela função mysqli_insert_id().
Parâmetros
- link
-
Apenas para estilo de procedimento: Um identificador de conexão retornado por mysqli_connect() or mysqli_init()
Valor Retornado
O valor do campo AUTO_INCREMENT que foi atualizado pela consulta anterior. Retorna zero se não houve uma consulta anterior na conexão ou se a consulta não atualizou um valor AUTO_INCREMENT.
Nota: Se o número for maior do que o maior valor inteiro possível, mysqli_insert_id() irá retornar uma string.
Exemplos
Exemplo #1 Estilo orientado a objeto
<?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();
?>
Exemplo #2 Estilo de procedimento
<?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);
?>
O exemplo acima irá imprimir:
New Record has id 1.
add a note
User Contributed Notesmysqli_insert_id
fizzyorange at gmail dot com
01-Apr-2009 03:27
01-Apr-2009 03:27
bert at nospam thinc dot nl
22-Jul-2008 03:58
22-Jul-2008 03:58
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 08:10
04-May-2007 08:10
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 01:40
21-Apr-2006 01:40
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;
}
?>
matix.bg a gmail
14-Mar-2006 10:56
14-Mar-2006 10:56
Some people are wondering how to get the ids of the rows inserted with loop. Here is it:
<?php
// stmt_init ...
for (;;;) {
// do some things...
$stmt->execute();
$ids[] = $mysqli->insert_id;
}
?>
alan at commondream dot net
03-Nov-2004 07:44
03-Nov-2004 07:44
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.