BigCroyd Posted March 12, 2009 Share Posted March 12, 2009 Hi, Can anyone help me im trying to use the mysql_insert_id() to give me the value of last auto-incremented value created in my database but the value always returns as 0. The information is entering into the database fine but i just cant seem to get the ID back out: $sql3="INSERT INTO `kf`.`order` (`order_id` ,`cust_id`,`order_date`,`order_status`,`order_total`) VALUES (0,'".$cust_id."','".$today."','".$status."','".$order_tot."')"; $order_ID = mysql_insert_id(); echo "$order_ID"; Result off the echo is always = 0 even though the query runs fine. Any help would be appreciated Thanks Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/ Share on other sites More sharing options...
Maq Posted March 12, 2009 Share Posted March 12, 2009 I don't see where you actually insert anything, you create the string and that's it, unless you didn't paste all the code... You have to use mysql_query() and mysql_insert_id() gets the resource id from the last inserted record. Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/#findComment-782877 Share on other sites More sharing options...
timmah1 Posted March 12, 2009 Share Posted March 12, 2009 Is order_id set to auto increment? If not, it will always be defined as 0 Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/#findComment-782879 Share on other sites More sharing options...
BigCroyd Posted March 12, 2009 Author Share Posted March 12, 2009 Yes, Sorry the mysql_query is in the code i just forgot to paste it in : $sql3="INSERT INTO `kf`.`order` (`order_id` ,`cust_id`,`order_date`,`order_status`,`order_total`) VALUES (0,'".$cust_id."','".$today."','".$status."','".$order_tot."')"; $result = mysql_query($sql3, $con) $order_ID = mysql_insert_id(); or die ("Unable to perform query"); echo "$order_ID"; } And yes, the order_ID feild is set to auto-increment in the database Thanks Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/#findComment-782944 Share on other sites More sharing options...
premiso Posted March 12, 2009 Share Posted March 12, 2009 $sql3="INSERT INTO `kf`.`order` (`order_id` ,`cust_id`,`order_date`,`order_status`,`order_total`) VALUES ('','".$cust_id."','".$today."','".$status."','".$order_tot."')"; You have it set to "0" in your sql query. Change it to '' (as seen above) and it should work. Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/#findComment-782948 Share on other sites More sharing options...
BigCroyd Posted March 12, 2009 Author Share Posted March 12, 2009 Hi premiso I changed the query to what you suggested but i am still getting the same problem. The value is still 0. The database order_ID field is auto-incrementing properly Thanks for the help so far. Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/#findComment-782966 Share on other sites More sharing options...
Maq Posted March 12, 2009 Share Posted March 12, 2009 Can you post exactly what you have? Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/#findComment-782967 Share on other sites More sharing options...
BigCroyd Posted March 12, 2009 Author Share Posted March 12, 2009 require "connect.php"; $sql3="INSERT INTO `kf`.`order` (`order_id` ,`cust_id`,`order_date`,`order_status`,`order_total`) VALUES (0,'".$cust_id."','".$today."','".$status."','".$order_tot."')"; $result = mysql_query($sql3, $con) $order_ID = mysql_insert_id(); or die ("Unable to perform query"); echo "$order_ID"; } The result from the echo gives 0. Just to make it clear - the query runs ok and inserts the correct data into the database and the order_id field in the database is auto-incrementing ok e.g 1,2,3,4 etc. Hope this is clearer Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/#findComment-783044 Share on other sites More sharing options...
Maq Posted March 12, 2009 Share Posted March 12, 2009 You're still inserting order_id as 0. Please read premiso's post. Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/#findComment-783047 Share on other sites More sharing options...
BigCroyd Posted March 12, 2009 Author Share Posted March 12, 2009 it doesnt seem to make a difference what i put in the insert statement for the auto-incrementing field i have tried both the following insert queries with the same result: $sql3="INSERT INTO `kf`.`order` (`order_id` ,`cust_id`,`order_date`,`order_status`,`order_total`) VALUES (0,'".$cust_id."','".$today."','".$status."','".$order_tot."')"; $sql3="INSERT INTO `kf`.`order` (`order_id` ,`cust_id`,`order_date`,`order_status`,`order_total`) VALUES ('','".$cust_id."','".$today."','".$status."','".$order_tot."')"; $sql3="INSERT INTO `kf`.`order` (`order_id` ,`cust_id`,`order_date`,`order_status`,`order_total`) VALUES ("NULL",'".$cust_id."','".$today."','".$status."','".$order_tot."')"; Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/#findComment-783075 Share on other sites More sharing options...
premiso Posted March 12, 2009 Share Posted March 12, 2009 Then your table is setup wrong. Post your table structure and note: mysql_insert_id() converts the return type of the native MySQL C API function mysql_insert_id() to a type of long (named int in PHP). If your AUTO_INCREMENT column has a column type of BIGINT, the value returned by mysql_insert_id() will be incorrect. Instead, use the internal MySQL SQL function LAST_INSERT_ID() in an SQL query. Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/#findComment-783081 Share on other sites More sharing options...
Psycho Posted March 12, 2009 Share Posted March 12, 2009 You shouldn't even include the ORDER_ID in the insert statement at all. Also, that is the oddest use of an "or die()" [on line all by itself, previous line was ended with a semi-colon] I have ever seen. I don't even know how that would process. But it could be the fact that you have your "$order_ID = mysql_insert_id()" sandwiched in-between the query and the "or die()" that is causing the problem. Try this: require "connect.php"; $sql3="INSERT INTO `kf`.`order` (`cust_id`,`order_date`,`order_status`,`order_total`) VALUES ('{$cust_id}','{$today}','{$status}','{$order_tot}')"; $result = mysql_query($sql3, $con) or die ("Unable to perform query"); $order_ID = mysql_insert_id(); echo "$order_ID"; Also, pleas use CODE or PHP tags when posting code. Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/#findComment-783093 Share on other sites More sharing options...
BigCroyd Posted March 12, 2009 Author Share Posted March 12, 2009 hi premiso. My table structure is shown below: CREATE TABLE IF NOT EXISTS `order` ( `order_id` bigint(20) unsigned NOT NULL AUTO_INCREMENT, `cust_id` int(5) NOT NULL, `order_date` varchar(40) NOT NULL, `order_status` varchar(30) NOT NULL, `order_total` varchar(30) NOT NULL, PRIMARY KEY (`order_id`), UNIQUE KEY `order_id` (`order_id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ; I have noticed that my auto-increment is a Bigint as suggested in your post. Can you point me in the right direction about how i could get the value into a variable useing LAST_INSERT_ID()? Thank you Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/#findComment-783110 Share on other sites More sharing options...
PFMaBiSmAd Posted March 12, 2009 Share Posted March 12, 2009 From the mysql manual - LAST_INSERT_ID(), LAST_INSERT_ID(expr) LAST_INSERT_ID() (with no argument) returns the first automatically generated value that was set for an AUTO_INCREMENT column by the most recently executed INSERT statement to affect such a column. For example, after inserting a row that generates an AUTO_INCREMENT value, you can get the value like this: mysql> SELECT LAST_INSERT_ID(); -> 195 Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/#findComment-783116 Share on other sites More sharing options...
Psycho Posted March 12, 2009 Share Posted March 12, 2009 Did you see my response? The code you're posting can't be working. I tested it and there is a parse error because there is no semicolon after $result = mysql_query($sql3, $con) When I ran the same code (modified for a test DB) I get this error Parse error: parse error, unexpected T_VARIABLE in C:\xampp\htdocs\test\test.php on line 14 (14 being the line after the line with the missing semi-colon, as expected) So, I don't see how what you posted is the actual code you are using. Did you try the code I posted (which works for me)? Quote Link to comment https://forums.phpfreaks.com/topic/149098-mysql_insert_id/#findComment-783124 Share on other sites More sharing options...
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