Not able to edit a title...please help!

Question

totalnoob 0 Light Poster

15 Years Ago

I have a file with a table with movie titles. When I wish to edit I click on a link which is represented in this following code:

<a href="edit.php?autoid='.$row['autoid'].'">
           edit</a>

This redirects me to an edit file where I can change the name, year and category. After updating it says title updated but with the following error:
Notice: Undefined index: autoid in C:\wamp\www\edit.php on line 37
When I go back to the main page the title is never updated?
Can someone help me please?
I'll attach my file, so that anyone can see what I'm talking about...
Thanks in advance!

php

edit.php (2.09 KB)

<html>
<body style="background-color:powderblue;">
      <?php
      include("c:\wamp\www\conf.php");
       if(isset( $_GET['autoid']))
                    {

                  $connection=@mysql_connect($host, $user, $pass)or die("Unable to connect!");
                  mysql_select_db($db, $connection) or die ('Unable to select database!');

                   $id=$_GET['autoid'] ;
                   $query4= "select title, category, year from movie where autoid='$id'";
                $result2=mysql_query($query4) or die ("query failed!");


                     $row = mysql_fetch_assoc($result2);
         ?>

          <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST" style="text-align:center">
               Your title:
               </br> </br> </br>
               Title:<input type="text" name="title" value=<?php echo $row['title']; ?>>

               Category:<input type="text" name="category" value=<?php echo $row['category']; ?>>

               Year:<input type="text" name="year" value=<?php echo $row['year']; ?>>
               
               <input type="hidden" value="$autoid">
               <input type="submit" name="submit" value="change"  style=" height:35;width:70">



           <?php
                    }
                   if(isset($_POST['submit']))
                   {
                       $id=$_POST['autoid'] ;
                     $titel =  $_POST['title'];
                  $kategori =  $_POST['category'];
                  $ar =  $_POST['year'];

                  $connection=@mysql_connect($host, $user, $pass)or die("Unable to connect!");
                  mysql_select_db($db, $connection) or die ('Unable to select database!');

                    $query5= "update movie set title='$titel',category='$kategori', year='$ar'  where autoid='$id'";
                $result3=mysql_query($query5) or die ("query failed!");

                echo "Title updated!";

                   }
             ?>
                </br> </br> </br>
              <a href="members.php">Back to main page</a>
</body>
 </html>

3 Contributors
11 Replies
99 Views
1 Day Discussion Span
Latest Post 15 Years Ago Latest Post by totalnoob

darkagn 315 Veteran Poster

15 Years Ago

<input type="hidden" value="$autoid">

The error is in this line, you have forgotten to name this field, so it can't find it in the $_POST array. Try changing to this:

<input type="hidden" name="autoid" value="$autoid">

darkagn 315 Veteran Poster

15 Years Ago

Should it update the TITLE if you change the YEAR?

darkagn 315 Veteran Poster

15 Years Ago

Try echo'ing your update sql statement before you run the query. This way you can check the sql that is being run for the correct variables. Another good hint for debugging is to copy and paste that sql statement into an mysql command line and seeing what errors or results it gives you there (as opposed to php errors).

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totalnoob 0 Light Poster · Answer 1 · 2009-04-22T06:18:13+00:00

<input type="hidden" value="$autoid">
The error is in this line, you have forgotten to name this field, so it can't find it in the $_POST array. Try changing to this:
<input type="hidden" name="autoid" value="$autoid">

Hi! Thanks for your reply!
I've tried this out and the error dissapeared, but it didn't update the title, when I changed the year of the movie.

totalnoob 0 Light Poster · Answer 2 · 2009-04-22T06:30:33+00:00

Should it update the TITLE if you change the YEAR?

I can choose to update either the year, title or the category. So if I update the year it only updates the year. I'm sorry for not being clear. Instead of "editing a title" I should have written "edit movie info" as topic.

darkagn 315 Veteran Poster Featured Poster · Answer 3 · 2009-04-22T06:45:02+00:00

Also, you will need to check that each field has been set. Use the isset method on each of your $_POST retrievals, like so:

if( isset( $_POST[ 'title' ]))
  $titel = $_POST[ 'title' ];

Then you will need to qualify your SQL to only update the fields that are not null.

totalnoob 0 Light Poster · Answer 4 · 2009-04-22T07:01:48+00:00

Try echo'ing your update sql statement before you run the query. This way you can check the sql that is being run for the correct variables. Another good hint for debugging is to copy and paste that sql statement into an mysql command line and seeing what errors or results it gives you there (as opposed to php errors).

Good idea!
I've tried both things. In mySQL the query worked fine. However when I tried echoing the update statement everything was echoed (year,title, category) except autoid. It echoed $id, which means that autoid for some reason never is passed..

totalnoob 0 Light Poster · Answer 5 · 2009-04-22T07:11:12+00:00

Also, you will need to check that each field has been set. Use the isset method on each of your $_POST retrievals, like so:
if( isset( $_POST[ 'title' ]))
  $titel = $_POST[ 'title' ];
Then you will need to qualify your SQL to only update the fields that are not null.

I've tried:

if( isset( $_POST[ 'autoid' ]))
  $id = $_POST[ 'autoid' ];

but it acted normal without any errors and without updating.

darkagn 315 Veteran Poster Featured Poster · Answer 6 · 2009-04-22T08:05:50+00:00

Good idea!
I've tried both things. In mySQL the query worked fine. However when I tried echoing the update statement everything was echoed (year,title, category) except autoid. It echoed $id, which means that autoid for some reason never is passed..

If autoid is a numeric field, you don't need the ' marks around it in the SQL statement. If you remove them, the $id should be resolved.

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 7 · 2009-04-22T14:22:04+00:00

I checked your script.

<input type="hidden" value="$autoid">

is wrong. If you want to have a hidden element with php variable's value, you should do,

<input type="hidden" name="autoid" value="<?php echo $autoid; ?>">

Secondly, Everytime there is a database operation, you have mysql_connect and mysql_select_db in your script. You can put it on top of your script, or have it in a separate file and include it on top of this script.
And lastly, You are not sanitizing your input. Use mysql_real_escape_string to protect your application from sql injections.

totalnoob 0 Light Poster · Answer 8 · 2009-04-23T19:42:01+00:00

I checked your script.
is wrong. If you want to have a hidden element with php variable's value, you should do,
<input type="hidden" name="autoid" value="<?php echo $autoid; ?>">
Secondly, Everytime there is a database operation, you have mysql_connect and mysql_select_db in your script. You can put it on top of your script, or have it in a separate file and include it on top of this script.
And lastly, You are not sanitizing your input. Use mysql_real_escape_string to protect your application from sql injections.

Thank you!!
It worked as a charm! I would never have guessed that it required an echo. Now I learned something new. :icon_cheesygrin:
Once again thank you!