[carmelraj]

$sql="INSERT INTO order_details(co_id,company_name,bookingf_no,amount,discount,amount_payable,amount_collected,balance_payable_on,balance_payed,executive,receipt_no,receipt_date,no_of_iss,add_size,remarks)VALUES('$id','$_POST[company_name]','$_POST[bookingf_no]','$_POST[amount]','$_POST[discount]','$_POST[amount_payable]','$_POST[amount_collected]','$_POST[balance_payable_on]','$_POST[balance_payed]','$_SESSION[User]','$_POST[receipt_no]','$_POST[receipt_date]','$_POST[no_of_issues]','$_POST[size]','$_POST[remarks]') ON DUPLICATE KEY UPDATE order_id=LAST_INSERT_ID(order_id)";

I am trying to update the values which is already available in mysql databases.If not it will insert new values. Insertion is working but UPDATING THE EXISTING VALUES IS NOT WORKING.please help me

[Dormilich]

for updating already existing datasets use the UPDATE statement.

[carmelraj]

i don't how to update it... i try to use LAST_INSERT_ID() but its not working...

[japanir]

Perhaps you are looking for the 'ON DUPLICATE KEY' operation.
check here:
http://dev.mysql.com...-duplicate.html

[carmelraj]

id=LAST_INSERT_ID(id)

this is not working do u know why.....

[Dormilich]

it depends on the statement, but I guess you’re using it wrong.

about LAST_INSERT_ID(expr):

The MySQL Manual said:

If expr is given as an argument to LAST_INSERT_ID(), the value of the argument is returned by the function and is remembered as the next value to be returned by LAST_INSERT_ID().

see also LAST_INSERT_ID() @ MySQL

This post has been edited by Dormilich: 10 May 2011 - 07:09 AM

[JackOfAllTrades]

Why do people not want to spend any time LEARNING the language in which they're writing and READING the documentation? Geez.