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How to insert data in db using jsp?

Get auto increment id for many to many relationship?

  • I'm attempting to create a form where the id will auto increment. I have 3 many to many tables in the form. I'm wondering 1. how to grab the auto increment the point being so that I can insert that id into the player_skill table. 2. How can I run the queries when the first query is going to one table and then the next 3 will be going to a many-to-many tables? The tables are setup as character (id,name) (skill_id, name, description) (advantage_id, name,cost) many-to-many ones player_skill (skillid,playerid) using an inner join by association player id and skill id to their respectively named ids. player_advantage (playerid, advantageid) Let me know if I didn't explain anything clear enough. //$mysqli is the database connection //findid should pull the last id made?? //this is my one attempt at making a query $findid = $mysqli_insert_id($mysqli); $action = isset($_POST['action']) ? $_POST['action'] : ""; //adding skills if($action=='create'){ //write query $query = "insert into player_skill set skillid = '".$mysqli->real_escape_string($_POST['skill_id'])."' WHERE charid='".$findid."' " }else{ echo mysqli_error($mysqli); } Googling this has brought about mostly msyql db schema's and how to approach many-to-many. They assume that data is already populated. I do not as these are characters and will be made as they go.

  • Answer:

    Use the insert_id method to get the id as a variable. Both Procedural and OO approaches are documented in the examples at php dot net Documentation - http://php.net/manual/en/mysqli.insert-id.php

Srihari Sankar Sahu at Quora Visit the source

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Well, the first should you do is inserting a character in characters table. Something like $mysqli->query('INSERT INTO `players` (`name`) VALUES (' . $name . ')'); Then you should save a last inserted id. You can do that like this: $id = $mysqli->insert_id; After that you can insert a new row in player_skill table: $mysqli->query('INSERT INTO `player_skill` (`playerid`, `skillid`) VALUES(' . $id . ', ' . $skill_id . ')'); Did I understand your question in a right way?

Nikita Slimov

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