Home/ Questions/Q 9070611
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-06-16T17:43:31+00:00

I need to insert a new row in to a table, then grab the

  • 0

I need to insert a new row in to a table, then grab the ID of that row and update another table.
This is what I have:

$leadSQL="INSERT INTO $leadsTable (leadName, leadStatus, leadDescription, leadOpportunity, leadSource, leadSourceDescription, id, leadSince, contactID)
                     VALUES ('$_POST[leadName]', '$_POST[leadStatus]', '$_POST[leadDescription]', '$_POST[leadOpportunity]', '$_POST[leadSource]', '$_POST[leadSourceDescription]','$_POST[id]','$leadSince','$_POST[contactID]')";
$leadQuery = mysql_query($leadSQL);
$lastLeadID = mysql_insert_id();
$updateContactSQL = "UPDATE $contactsTable SET leadID = $lastLeadID WHERE contactID = $_POST[contactID]";
$updateContactQuery = mysql_query($updateContactSQL);

Everything works fine.. except that it inserts duplicate rows into the leads table. I have tried putting the update query into an if statement and it did the samething(this was just to try “something”). If I remove $lastID = mysql_insert_id(); it inserts just one row but obviously does not update the contacts table. So I am pretty sure it has to to with mysql_insert_id(). I need it to update the contacts table with the new id of the row inserted into the leads table. Any ideas would be greatly appreciated.

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    for all intents and purposes what you have should work.

    you might try trimming the fat a little bit though. Since you don’t need a resource for your query (it’s an INSERT), you can get rid of that variable. And I’d just put the mysql_insert_id() in your update statement.

    like this: 

    $leadSQL="INSERT INTO $leadsTable (leadName, leadStatus, leadDescription, leadOpportunity, leadSource, leadSourceDescription, id, leadSince, contactID)
                     VALUES ('$_POST[leadName]', '$_POST[leadStatus]', '$_POST[leadDescription]', '$_POST[leadOpportunity]', '$_POST[leadSource]', '$_POST[leadSourceDescription]','$_POST[id]','$leadSince','$_POST[contactID]')";
mysql_query($leadSQL);
$updateContactSQL = "UPDATE $contactsTable SET leadID = '".mysql_insert_id()."' WHERE contactID = $_POST[contactID]";
mysql_query($updateContactSQL);
    • 0