Undefined index

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PHP: “Notice: Undefined variable” and “Notice: Undefined index”

hello everyone i am getting a problem while updating my record..when i click the edit button it doesnot update the record but give following errors to all post variables.. i would really appreciate any kind of comments or help...

this is courses-edit.php

<?php include("../includes/config.php"); ?>
<?php

if ($_SESSION["isteacher"])
{

$con=mysql_connect($dbserver,$dbusername,$dbpassword);
if (!$con) { die('Could not connect: ' . mysql_error()); }

mysql_select_db($dbname, $con);
$courseid=$_GET["id"];
$result = mysql_query("SELECT * FROM courses WHERE (id='".$courseid."')");
while($row = mysql_fetch_array($result))
{
   $id=$row['id'];
   $title = $row['title'];
   $des = $row['description'];
   $subjectid = $row['subjects-id'];
}
mysql_close($con);
?>


<script type="text/javascript">
$(function() {
            $("form").validity(function() {
                $("#title")
                    .require("This Field Must Be Filled!!")
            });
        });
    </script>
<?php include("../includes/header.php"); ?>
<?php include("includes/nav.php"); ?>
<div id="maincontent">
<div class="span-24 last">
<div id="breadcrumbs">
    <a href="index.php">Home</a> >
     <a href="manage-courses.php">Manage Courses</a> >
     <a href="courses-list.php">List Courses</a> >
     Edit Course
</div>
</div>
<?php include("../teacher/includes/manage-courses-aside.php"); ?>
<div class="span-18 last">
<h2 class="alt">Edit Course</h2>

<form id="form" method="post" action="courses-edit-action.php">
<input type="hidden" value="<?php echo $courseid; ?>" name="id" />
 <label>Course Name:</label><input type="text" name="title" id="title" class="text"   value="<?php echo $title; ?>" /><br /><br />
<label>Choose Subject:</label>
<?php

$con=mysql_connect($dbserver,$dbusername,$dbpassword);
if (!$con) { die('Could not connect: ' . mysql_error()); }

mysql_select_db($dbname, $con);
$result = mysql_query("SELECT * FROM subjects");

echo "<select name='subjects-id'>\n";
while($row = mysql_fetch_array($result))
{
echo "<option value='".$row['id'] . "'";
if ($subjectid==$row['id'])
echo 'selected="selected"';
echo " >" . $row['subjectname'] . "</option>\n";
}
echo "</select>\n";
mysql_close($con);
?>
<label>Description:</label><textarea name="description" id="description"><?php echo    $des; ?></textarea><br /> <br />
    <input type="submit" value="Edit Course" class="button" />
</form>

<?php include("../includes/footer.php"); ?>
<?php
}
else
{
    header("Location: ".$fullpath."login/unauthorized.php");

}
?>

and this is courses-edit-action.php

<?php include("../includes/config.php");?>
<?php

$id=$_POST["id"];
$title=$_POST["title"];
$des=$_POST["description"];
$subjectid=$_POST["subjects-id"];

$con=mysql_connect($dbserver,$dbusername,$dbpassword);
if (!$con) { die('Could not connect: ' . mysql_error()); }
mysql_select_db("ombts", $con);
$query=("UPDATE courses SET title='".$title."', description='".$des."', 'subjects-  id'='".$subjectid."' WHERE (id='".$id."')");
$query=mysql_query($query);
 if($result){
echo header("Location:manage-courses.php?status=2");
}
mysql_close($con);
?>

and the warnings are

Notice: Undefined index: id in C:\xampp\htdocs\project\teacher\courses-edit-action.php on line 4

Notice: Undefined index: title in C:\xampp\htdocs\project\teacher\courses-edit-action.php on line 5

Notice: Undefined index: description in C:\xampp\htdocs\project\teacher\courses-edit-action.php on line 6

Notice: Undefined index: subjects-id in C:\xampp\htdocs\project\teacher\courses-edit-action.php on line 7

Are you able to get any result if you add the following line into courses-edit-action.php `print_r($_POST);` ? — Steven, Sep 26 '12 at 08:18
also, use mysql_real_escape_string to stop the SQL injection — Menztrual, Sep 26 '12 at 08:20
You're variables arent being passed along from the form. try `print_r($_POST);` as suggested above. to see what you're getting. — Jacob Tomlinson, Sep 26 '12 at 08:21
@shakabra whts the right way then? please tell me where are the wrong quotes? thanx :) — trouble creator, Sep 26 '12 at 09:14
@Riu2009 Then your `POST` does not have any content. To be entirely sure, do a `print_r($_REQUEST)`. If it's still blank, then no data is sent from previous page. Personally I've started to use `$_REQUEST` instead of `POST` and `GET`. It prevents me from doing mistakes of trying to get POST data when I'm sending GET data. — Steven, Sep 26 '12 at 11:33

Gogol · Answer 1 · 2012-09-26T11:18:27.820

2

Modify your script like this....

    $id= isset($_POST["id"]) ?$_POST["id"] : 0 ;
    $title= isset($_POST["title"]) ? $_POST["title"] : '' ;
    $des=isset($_POST["description"]) ? $_POST["description"] : '';
    $subjectid= isset($_POST["subjects-id"]) ? $_POST["subjects-id"] : '';

Edited Also, i see an error here...

$query=mysql_query($query);
 if($result){

should be

$result=mysql_query($query);
 if($result){

If you still can't update the table, print the $query variable before you do mysql_query. Tell me what you see and i will help you.....

edited Sep 26 '12 at 11:18

answered Sep 26 '12 at 08:18

Gogol

3,033
4
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errors are vanished by this but still its not updating the record... :( i never met this tyype of sticky problem ever... :( – trouble creator Sep 26 '12 at 09:22
yup i just got tht error solved but still it wasnt working as u said now.. i printed the $query and its updaing.. this is what i see UPDATE courses SET title='jnjk', description='hgszhn', 'subjects-id'='5' WHERE (id='4') here the values right.. but i dont know where the problem is :( – trouble creator Sep 26 '12 at 14:20

score 2 · Answer 2 · answered Sep 26 '12 at 08:20

You try to access

$id=$_POST["id"];
$title=$_POST["title"];
$des=$_POST["description"];
$subjectid=$_POST["subjects-id"];

this POST variables. But when you access the script for the first time this values are not set in the POST variable.

Chck the values with isset.

if(!isset($_POST["id"]) {
    $_POST["id"] = '';
}

or safe the checked value in a new variable

$id = isset($_POST["id"]) ? $_POST["id"] : 0;

for example.

score 2 · Answer 3 · answered Sep 26 '12 at 11:42

@Stony's answer is good solution to check if you are sending any POST data - and if not, do not try to fetch non-existing data.

Now, if you think $_POST["id"]; should be there, then there is something missing. Are you using GET instead of POST?

To see what data is submitted do a print_r($_POST). If this is blank, do a print_r($_REQUEST). If this is still blank, then no data is being submitted from the previous page.

You can also check by adding this in your URL &id=75 (or ?id=75 if you have no other parameters in the url). This will submit data and you will see one less error.

Undefined index

3 Answers3