Updating mySQL database using an HTML FORM and PhP

Question

I am here again. I have lots of trouble but this is my main problem now.

I can successfully add dynamic rows to my html form then add it to my database. What I am doing now is, I retrieved the data, auto-populate my EDIT form. My problem is, I can't even add now to my table a row that already has 2 or 3 rows in it. What should I do? I just copy paste my code from my newform.php to editform.php. PLease help me. :(

Provide some code then only we can help you.

Shaikh Farooque
– Shaikh Farooque

2012-05-03 05:37:31 +00:00
Commented May 3, 2012 at 5:37 — Shaikh Farooque
– Shaikh Farooque, Commented May 3, 2012 at 5:37
provide your code here, then only people can help you

arun
– arun

2012-05-03 05:40:12 +00:00
Commented May 3, 2012 at 5:40 — arun
– arun, Commented May 3, 2012 at 5:40
provide your code so we can see whats the error

user1249679
– user1249679

2012-05-03 06:22:31 +00:00
Commented May 3, 2012 at 6:22 — user1249679
– user1249679, Commented May 3, 2012 at 6:22

KillABug · Accepted Answer · 2012-05-03 06:05:22Z

1

I think you are trying to do a CRUD(Create,Read,Update or EDIT,Delete) example using HTML/PHP.

If that is the case you can find many tutorials on googling.

Here is one which is quite good to start with. http://speckyboy.com/2011/02/17/getting-started-with-crud-in-php/

Hope it helps.Happy Coding

answered May 3, 2012 at 6:05

KillABug

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Comments

Er. Anurag Jain · Accepted Answer · 2012-05-03 06:19:23Z

I don't know what is your code but One way to achieve it.

1.create fields having name with unique id of rows from table. and a hidden field combine with all unique ids associate with a character or string like =||=.

<input type ="text" name = "naame_1" id = "name_1" value = "test" />
<input type ="text" name = "naame_5" id = "name_5" value = "test1" />
<input type="hidden" name = "allids" id = "allids" value = "1=||=5" />

And now in PHP file

 <?php
      if(isset($_POST)){
         $ids = explode("=||=" , $_POST['allids']);
          foreach($ids As $id){
             $query = "update tblename set fieldname = '".$_POST['name_'.$id]."' where id = ".$id;    
             mysql_query($query);
          }
       }
 ?>

Please try this , If this may helpful for you.

thanks

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