Why does mysql_insert_id() return 0? - php

Question

When I run this code
$sql_select = "INSERT INTO `database`.`table`(Columns) VALUES (Values)... ";
$mysqlid = mysql_insert_id($sql_select->db);
echo ($mysqlid);
I get the error message
mysql_insert_id(): supplied argument is not a valid MySQL-Link
I have tried this variation;
$mysqlid = mysql_insert_id();
echo ($mysqlid);
but that returns a 0 which, according to the documentation, means an auto_increment field was not found. The only thing I can think of is that I am not calling the auto_increment column in the $sql_select, but there is an auto_increment column in there; will that affect the behavior of mysql_insert_id?

Answer 1

you need to actually run the query first:
$sql= "INSERT INTO `database`.`table`(Columns) VALUES (Values)...";
$result = mysql_query($sql);
and then after that:
$id = mysql_insert_id();
echo $id
Let me know if you have still problems.

Answer 2

That is because $sql_select->db is not a valid MySQL Link
What you are looking for is something like this:
$sql_select = "INSERT INTO `database`.`table`(Columns) VALUES (Values)...
$result = mysql_query($sql_select);
$mysqlid = mysql_insert_id($result->db);
$result is a valid MySQL resource.
Also don't forget to have created a mysql connection
NOTE: While I used code for the original MySQL driver it's use is discouraged. Instead you want to use MySQLi or PDO_MySQL

Answer 3

You need to run your query:
if($result = mysql_query($sql_select)){
$mysqlid = mysql_insert_id();
echo $mysqlid;
}
Cheers

Answer 4

your connection of mysqli is an object...so try this....
$mysqli = new mysqli("localhost", "user", "password", "dbname");
after inserting try this..
echo $mysqli->insert_id;