undefined index error in update form [duplicate] - php

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 5 years ago.
this img contain my updation.php code
this img contain my updation.php code
when i insert data into textbox and then click submit but at that time
javascript alertbox sys "your data has been updated successfully" and than browser display errorss messeges
error messeges
<?php
$con = mysqli_connect('127.0.0.1','root','','');
mysqli_select_db($con,'brm_db');
$q = "SELECT * FROM book";
$result = mysqli_query($con,$q);
$num = mysqli_num_rows($result);
mysqli_close($con);
?>
<!DOCTYPE html>
<html>
<head>
<title> Update Record </title>
<link rel="stylesheet" href="./css/viewstyle.css" />
</head>
<body>
<h1 align="center"> Book Record Management</h1>
<center>
<form action="updation.php" method="post">
<table id="view_table">
<tr>
<th>Book Id</th>
<th>Title</th>
<th>Price</th>
<th>Author</th>
</tr>
<?php
for($i=1;$i<=$num;$i++)
{
$row = mysqli_fetch_array($result);
?>
<tr>
<td> <?php echo $row['bookid'];?>
<input type="hidden" name="bookid <?php echo $i ;?>" value="<?php echo $row['bookid'];?>" /> </td>
<td><input type="text" name="title <?php echo $i ;?> "value="<?php echo $row['title'];?>" /></td>
<td><input type="text" name="price <?php echo $i ;?> "value="<?php echo $row['price'];?>" /></td>
<td><input type="text" name="author <?php echo $i ;?> "value="<?php echo $row['author'];?>" /></td>
</tr>
<?php
}
?>
</table>
<input type="submit" value="Update" style="background-color:lightgreen;width:100px;" />
</form>
</center>
</body>
</html>

Declare array assigned to those before entering into the for loop like the following
$bookid = array();
$title = array();
$price = array();
$author = array();
//for loop goes here
Instead of this type I would like to suggest you to have a single array as follows and then store in it
$books = array();
for($i = 0; $i <= $records; $i++){
$books[$i]['bookid'] = $_POST[$index1];
$books[$i]['title'] = $_POST[$index1];
$books[$i]['price'] = $_POST[$index1];
$books[$i]['author'] = $_POST[$index1];
}
Now loop the above $books array as follows, which will be very convenient to achieve
foreach($books as $book){
}

Related

ERROR :- Undefined variable: raw While Transfering data between 2 pages [duplicate]

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 4 years ago.
manageleavetype.php
Below code is my first file page I want to redirect this page form content into next file code edit.php. I want want to update the data from the next page.
<form action="" method="POST">
<tr>
<td><?php echo $type; ?></td>
<td><?php echo $des; ?></td>
<td><?php echo $time; ?></td>
<input type="hidden" name="hid" value="<?php echo $id; ?>">
<input type="hidden" name="htype" value="<?php echo $type; ?>">
<input type="hidden" name="hdes" value="<?php echo $des; ?>">
<td><input type="submit" name="del" value="DELETE">
<input type="submit" name="update" value="UPDATE">
</td></form></tr>
edit.php
This file except the code from above file and update the data that is received into MySQL database
<?php
require 'config.php';
if (isset($_POST['update'])) {
$eid = $_POST['hid'];
$que = "SELECT * FROM leavetype WHERE id='$eid'";
$exe = mysqli_query($conn, $que);
$raw = mysqli_fetch_array($exe);
}
if (isset($_POST['update'])) {
$nwtype = $_POST['ltype'];
$nwdes = $_POST['ldes'];
$que = "UPDATE leavetype SET leavetype='$nwtype',description='$nwdes' WHERE id='$eid'";
$exe = mysqli_query($conn, $que);
if ($exe) {
header("location:manageleavetype.php");
}
}
?>
<form method="POST">
<label>Leave Type</label>
<input type="text" name="ltype" value="<?php echo $raw['leavetype']; ?>">
<label>Description</label>
<input type="text" name="ldes" value="<?php echo $raw['description']; ?>">
<input type="submit" name="update" value="Update">
</form>
When I run this program following error is coming:
Notice: Undefined variable: raw in C:\wamp64\www\eLeaveSystem\edit.php on line 31
When you only define variables within an if block, you risk getting that error. In your case you define $raw only when update is posted:
if (isset($_POST['update'])) {
$eid = $_POST['hid'];
$que = "SELECT * FROM leavetype WHERE id='$eid'";
$exe = mysqli_query($conn, $que);
$raw = mysqli_fetch_array($exe);
}
But you access that variable unconditionally in the HTML generation part, like here:
<input type="text" name="ltype" value="<?php echo $raw['leavetype']; ?>">
So you have a few options:
Put the HTML generation part also in such an if block. Then you need to have another part in your code where you generate output for when no update is posted.
Define default values for those variables. For example $raw = ['leavetype' => 'something', 'description' => 'default description'].
You can simply solve this issue by using the isset() function to avoid undefined variable issue.
Try below:
<form method="POST">
<label>Leave Type</label>
<input type="text" name="ltype" value="<?php echo (isset($raw['leavetype']) ? $raw['leavetype'] : '');?>">
<label>Description</label>
<input type="text" name="ldes" value="<?php echo (isset($raw['description']) ? $raw['description'] : ''); ?>">
<input type="submit" name="update" value="Update">
</form>
I assume that the form block should appear for both Update and Insert.

Receiving empty form values using php and post [duplicate]

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 6 years ago.
I am making a insertion form to add a product to the mysql database. I am using echo command to print the query. The printed query is showing empty values like this:
Notice: Undefined index: product_cat in F:\Apache24\htdocs\Site1\admin\product_add.php on line 40
Notice: Undefined index: product_brand in
F:\Apache24\htdocs\Site1\admin\product_add.php on line 41
Notice: Undefined index: product_title in
F:\Apache24\htdocs\Site1\admin\product_add.php on line 42
Notice: Undefined index: product_price in
F:\Apache24\htdocs\Site1\admin\product_add.php on line 43
Notice: Undefined index: product_description in
F:\Apache24\htdocs\Site1\admin\product_add.php on line 44
Notice: Undefined index: product_keywords in
F:\Apache24\htdocs\Site1\admin\product_add.php on line 45
Notice: Undefined index: product_images in
F:\Apache24\htdocs\Site1\admin\product_add.php on line 48
Notice: Undefined index: product_images in
F:\Apache24\htdocs\Site1\admin\product_add.php on line 49
insert into products ('product_cat', 'product_brand', 'product_title',
'product_price', 'product_desc', 'product_image', 'product_keywords')
values( '', '', '', '', '', '', '')
This is the html form. It is also the action page:
<!DOCTYPE html>
<html>
<head>
<title>Admin Panel - Add Products</title>
<link rel="stylesheet" type="text/css" href="../styles/admin-style.css">
<script src="//cdn.tinymce.com/4/tinymce.min.js"></script>
<script>tinymce.init({ selector:'textarea' });</script>
<?php
include("../includes/connect.php");
function getcats_add_products(){
global $con;
$get_cats="select * from categories";
$run_cats=mysqli_query($con, $get_cats);
echo "<option>Select Category</option>";
while($row_cats = mysqli_fetch_array($run_cats)){
$cat_id = $row_cats['cat_id'];
$cat_title = $row_cats['cat_title'];
echo "<option value='$cat_id'>$cat_title</option>";
}
}
function getbrands_add_products(){
global $con;
$get_brands="select * from brands";
$run_brands=mysqli_query($con, $get_brands);
echo "<option>Select Brand</option>";
while($row_brands = mysqli_fetch_array($run_brands)){
$brand_id = $row_brands['brand_id'];
$brand_title = $row_brands['brand_title'];
echo "<option value='$brand_id'>$brand_title</option>";
}
}
if(isset($_POST['submit'])){
$product_cat = $_POST['product_cat'];
$product_brand = $_POST['product_brand'];
$product_title = $_POST['product_title'];
$product_price = $_POST['product_price'];
$product_desc = $_POST['product_description'];
$product_keywords = $_POST['product_keywords'];
$product_images = $_FILES['product_images']['name'];
$product_images_temp = $_FILES['product_images']['tmp_name'];
$product_query = "insert into products
('product_cat',
'product_brand',
'product_title',
'product_price',
'product_desc',
'product_image',
'product_keywords')
values(
'$product_cat',
'$product_brand',
'$product_title',
'$product_price',
'$product_desc',
'$product_images',
'$product_keywords') ";
echo $product_query;
}
?>
</head>
<body>
<div class="wrapper">
<header>
</header>
<div class="heading">Add New Product</div>
<div class="product-table-div">
<form method="POST" action="" enctype="multipart/form-data">
<table class="product-table" border="1">
<tr>
<td id="product-add-label">Product Category</td>
<td>
<select id="product-table-input" name="product-cat">
<?php getcats_add_products(); ?>
</select>
</td>
</tr>
<tr>
<td id="product-add-label">Product Brand</td>
<td>
<select id="product-table-input" name="product-brand">
<?php getbrands_add_products(); ?>
</select>
</td>
</tr>
<tr>
<td id="product-add-label">Product title</td>
<td>
<input type="text" name="product-title" id="product-table-input">
</td>
</tr>
<tr>
<td id="product-add-label">Product Price</td>
<td>
<input type="number" name="product-price" id="product-table-input">
</td>
</tr>
<tr>
<td id="product-add-label">Product description</td>
<td>
<textarea rows="10" cols="30" name="product-description"></textarea>
</td>
</tr>
<tr>
<td id="product-add-label">Product image</td>
<td>
<input type="file" name="product-images" id="product-table-input">
</td>
</tr>
<tr>
<td id="product-add-label">Product Keywords</td>
<td>
<input type="text" name="product-keywords" id="product-table-input">
</td>
</tr>
<tr>
<td colspan="2">
<div id="product-submit-div">
<input type="reset" name="submitreset" id="product-submit" value="Clear">
<input type="submit" name="submit" id="product-submit" value="Add">
</div>
</td>
</tr>
</table>
</form>
</div>
</div>
</body>
</html>
This location of this page is root>admin>product_add.php. What seems to be the problem?
Oh, I have also used tinymce text editor in the textarea.
You are using Hyphen - signs in the HTML form for the field names; but when you try to read in PHP, you are using underscores.
$product_cat = $_POST['product_cat'];
$product_brand = $_POST['product_brand'];
$product_title = $_POST['product_title'];
$product_price = $_POST['product_price'];
$product_desc = $_POST['product_description'];
$product_keywords = $_POST['product_keywords'];
Use below code instead of above
$product_cat = $_POST['product-cat'];
$product_brand = $_POST['product-brand'];
$product_title = $_POST['product-title'];
$product_price = $_POST['product-price'];
$product_desc = $_POST['product-description'];
$product_keywords = $_POST['product-keywords'];

How to fix undefined variable in php [duplicate]

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 6 years ago.
I'm trying to do the update function in php with a mysql database connected. I put the codes for update in a file called parcelEdit.php. Here's my code for parcelEdit.php
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>Updating Parcel Details</title>
<link rel="stylesheet" href="css/style.css" />
</head>
<?php
include('db.php');
if(isset($_POST['update']))
{
$parcelID = $_POST['parcelID'];
$owner = $_POST['owner'];
$rcv_date = $_POST['rcv_date'];
$pck_date = $_POST['pck_date'];
$status = $_POST['status'];
// checking empty fields
if (empty($parcelID) || empty($owner) || empty($rcv_date)||
empty($pck_date)|| empty($status)) {
if(empty($parcelID)) {
echo "<font color='red'>Parcel ID field is empty.</font><br/>";}
if(empty($owner)) {
echo "<font color='red'>Owner Name field is empty.</font><br/>";}
if(empty($rcv_date)) {
echo "<font color='red'>Received Date field is empty.</font><br/>";}
if(empty($pck_date)) {
echo "<font color='red'>Picked Up Date field is empty.</font><br/>";}
if(empty($status)) {
echo "<font color='red'>Parcel Status field is empty.</font><br/>";}
} else {
//updating the table
$result = mysql_query("UPDATE parcel SET parcelOwner = '$owner',
dateReceived = '$rcv_date', datePickup = '$pck_date', parcelStatus =
'$status' WHERE parcelID='$parcelID'");
//redirectig to the display page. In our case, it is index.php
header("Location: parcelView.php");
}
}
?>
<?php
//getting id from url
if(isset($_GET['parcelID'])){
$parcelID = $_GET['parcelID'];
}
//selecting data associated with this particular id
if(isset($parcelID)){
$result = mysql_query("SELECT * FROM parcel WHERE parcelID='$parcelID'");
while($res = mysql_fetch_array($result))
{
//$mem_id= $res['mem_id'];
$parcelID= $res['parcelID'];
$owner= $res['parcelOwner'];
$rcv_date= $res['dateReceived'];
$pck_date= $res['datePickup'];
$status= $res['parcelStatus'];
}}
?>
<body>
<body style='background: url(mailbox.jpg)'>
<div align="center">
<h1>Update Parcel Details</h1>
<form method="post" enctype="multipart/form-data">
<table>
<tr>
<Td> PARCEL ID : </td>
<td><input name="parcelID" type="text" id="parcelID" value=<?php
echo $parcelID;?>></td>
</tr>
<tr>
<Td> OWNER : </td>
<td><input name="owner" type="text" id="owner" value=<?php echo
$owner;?>></td>
</tr>
<tr>
<Td> DATE RECEIVED : </td>
<td><input name="rcv_date" type="text" id="rcv_date" value=<?php
echo $rcv_date;?>></td>
</tr>
<tr>
<Td> DATE PICKED UP : </td>
<td><input name="pck_date" type="text" id="pck_date" value=<?php
echo $pck_date;?>></td>
</tr>
<tr>
<Td> STATUS : </td>
<td><input name="status" type="text" id="status" value=<?php
echo $status;?>></td>
</tr>
<tr>
<Td colspan="2" align="center">
<input type="submit" value="Update Records" name="update"/>
</Td>
</tr>
</table>
</form>
</div>
</body>
</html>
And i got these errors
Notice: Undefined variable: parcelID in
C:\xampp\htdocs\psmtest1\parcelEdit.php on line 77
Notice: Undefined variable: owner in
C:\xampp\htdocs\psmtest1\parcelEdit.php on line 81
Notice: Undefined variable: rcv_date in
C:\xampp\htdocs\psmtest1\parcelEdit.php on line 85
Notice: Undefined variable: pck_date in
C:\xampp\htdocs\psmtest1\parcelEdit.php on line 89
Notice: Undefined variable: status in
C:\xampp\htdocs\psmtest1\parcelEdit.php on line 93
I honestly can't find ways to solve this even after referring to different code examples.
You can check that whether the variable is coming as you thinks or not . You can dump all variables sent using POST method with the help of var_dump() or print_r() like this -
<?php
include('db.php');
if(isset($_POST['update']))
{
echo '<pre>';
print_r($_POST);
echo '</pre>';
$parcelID = $_POST['parcelID'];
$owner = $_POST['owner'];
$rcv_date = $_POST['rcv_date'];
$pck_date = $_POST['pck_date'];
$status = $_POST['status'];
...
?>
and you can check the key in $_POST and do the required changes .
Try adding an else after your if(isset($parcelID)) condition
if(isset($_GET["parcelID"])){
$parcelID = mysql_real_escape_string($_GET["parcelID"]);
$result = mysql_query("SELECT * FROM parcel WHERE parcelID = '$parcelID'");
while($res = mysql_fetch_array($result))
{
//$mem_id = $res['mem_id'];
$parcelID = $res['parcelID'];
$owner = $res['parcelOwner'];
$rcv_date = $res['dateReceived'];
$pck_date = $res['datePickup'];
$status = $res['parcelStatus'];
}
} else {
$parcelID = '';
$owner = '';
$rcv_date = '';
$pck_date = '';
$status = '';
}
Too many isset() condition which I think could be lessen, so I removed the first if(isset($_GET["parcelID"])) condition and just go straight and replace the if(isset($parcelID)) with it.
Use mysqli_* extension instead of deprecated mysql_*.

PHP - Page needs Twice a Submit to show Changes [duplicate]

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 7 years ago.
I have the most annoying PHP ever encountered. I need to submit the page TWICE before the changes are being made to the page. For example; if you submit the page when you enable a tool, it will show the tools' settings: for these settings to show up you need to submit the page twice. What is wrong with this code?
Another example:
CODE:
<?php
if (isset($_POST["submit"]))
{
$string = '<?php
$customoptions = ' . $_POST["customoptions"] . ';
$primarycolor = "' . $_POST["primarycolor"] . '";
$adminbg = "' . $_POST["adminbg"] . '";
?>';
$fp = fopen("includes/userstyle.php", "w");
fwrite($fp, $string);
fclose($fp);
}
?>
<form action="" name="customopt" method="post">
<table>
<tr>
<td>Panel language</td>
<td>
<select onchange="this.options[this.selectedIndex].value && (window.location = this.options[this.selectedIndex].value);">
<option><?php echo $lang['chooselanguage']; ?></option>
<option value="dashboard.php?lang=en">English</option>
<option value="dashboard.php?lang=nl">Dutch</option>
</select>
</td>
</tr>
<tr>
<td>Custom Style</td>
<td>
<select name="customoptions" id="customoptions">
<option <?php echo ($customoptions == true) ? 'selected' : '' ?> value="true">
<?php echo $lang['enabled']; ?>
</option>
<option <?php echo ($customoptions == true) ? 'selected' : '' ?> value="false">
<?php echo $lang['disabled']; ?>
</option>
</select>
</td>
</tr>
<?php
if ($customoptions)
{
?>
<tr>
<td>Primary Color</td>
<td><input name="primarycolor" type="text" id="primarycolor" value="<?php echo $primarycolor; ?>"></td>
</tr>
<tr>
<td>Background Color</td>
<td><input name="adminbg" type="text" id="adminbg" value="<?php echo $adminbg; ?>"></td>
</tr>
<?php
}
?>
</table>
<input type="submit" name="submit" value="<?php echo $lang['ok']; ?>">
</form>
PS: I know this isn't a good way to save settings in a php file but this is just a test, it will never go live.
OK, try to do it has to do with the top that says if (isset($_POST['submit']))
Try and give it a default value if it isn't set. if (!(isset($_POST['submit'])))
I'm referrin to the $customoptions variable $primarycolor and $adminbg

Undefined index php error [duplicate]

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 8 years ago.
I'm sorry for taking a bit of your time i've been trying to figure out this error for hours
I have the following error:
Notice: Undefined index: Action in C:\xampp\htdocs\Earl.com\Earl.com\EditEverything.php on line 71
Notice: Undefined index: BUSSINESSID in C:\xampp\htdocs\Earl.com\Earl.com\EditEverything.php on line 129
Notice: Undefined index: ID in C:\xampp\htdocs\Earl.com\Earl.com\EditEverything.php on line 129
Notice: Undefined index: Action in C:\xampp\htdocs\Earl.com\Earl.com\EditEverything.php on line 129
Notice: Undefined index: BUSSINESSID in C:\xampp\htdocs\Earl.com\Earl.com\EditEverything.php on line 129
Notice: Undefined index: ID in C:\xampp\htdocs\Earl.com\Earl.com\EditEverything.php on line 129
Notice: Undefined index: Action in C:\xampp\htdocs\Earl.com\Earl.com\EditEverything.php on line 129
Can't seem to work out whats wrong Here's the code:
<html>
<head>
</head>
<body>
<?php
$user="root";
$pass="";
$database="epages";
$localhost="localhost";
$conn=mysql_connect($localhost,$user,$pass);
mysql_select_db("epages") or die(mysql_error());
//*** Add Condition ***//
if($_POST["hdnCmd"] == "Add") {
$strSQL = "INSERT INTO BUSINESS ";
$strSQL .="(BUSINESSID,BUSINESSNAME,SERVICES) ";
$strSQL .="VALUES ";
$strSQL .="('".$_POST["txtAddBUSINESSID"]."','".$_POST["txtAddName"]."' ";
$strSQL .=",'".$_POST["txtAddService"]."') ";
$objQuery = mysql_query($strSQL);
if(!$objQuery) {
echo "Error Save [".mysql_error()."]";
}
//header("location:$_SERVER[PHP_SELF]");
//exit();
}
//*** Update Condition ***//
if($_POST["hdnCmd"] == "Update") {
$strSQL = "UPDATE BUSINESS SET ";
$strSQL .="BUSINESSID = '".$_POST["txtEditBUSINESSID"]."' ";
$strSQL .=",BUSINESSNAME = '".$_POST["txtEditName"]."' ";
$strSQL .=",SERVICES = '".$_POST["txtEditService"]."' ";
$strSQL .="WHERE BUSINESSID = '".$_POST["hdnEditBUSINESSID"]."' ";
$objQuery = mysql_query($strSQL);
if(!$objQuery) {
echo "Error Update [".mysql_error()."]";
}
//header("location:$_SERVER[PHP_SELF]");
//exit();
}
//*** Delete Condition ***//
if($_GET["Action"] == "Del") {
$strSQL = "DELETE FROM BUSINESS WHERE BUSINESSID ='".$_GET["ID"]."' ";
$objQuery = mysql_query($strSQL);
if(!$objQuery) {
echo "Error Delete [".mysql_error()."]";
}
//header("location:$_SERVER[PHP_SELF]");
//exit();
}
$strSQL = "SELECT * FROM BUSINESS";
$objQuery = mysql_query($strSQL) or die ("Error Query [".$strSQL."]"); ?>
<form name="frmMain" method="post" action="<?php echo $_SERVER["PHP_SELF"];?>">
<input type="hidden" name="hdnCmd" value="">
<table width="600" border="1">
<tr>
<th width="91"> <div align="center">BUSINESSID</div></th>
<th width="98"> <div align="center">BusinessName</div></th>
<th width="198"> <div align="center">Service</div></th>
<th width="30"> <div align="center">Edit</div></th>
<th width="30"> <div align="center">Delete</div></th>
</tr>
<?php
while($objResult = mysql_fetch_array($objQuery)) {
if($objResult["BUSSINESSID"] == $_GET["ID"] and $_GET["Action"] == "Edit") { ?>
<tr>
<td><div align="center">
<input type="text" name="txtEditBUSINESSID" size="5" value="<?php echo $objResult["BUSINESSID"];?>">
<input type="hidden" name="hdnEditBUSINESSID" size="5" value="<?php echo $objResult["BUSINESSID"];?>">
</div></td>
<td><input type="text" name="txtEditName" size="20" value="<?php echo $objResult["BUSINESSNAME"];?>"></td>
<td><input type="text" name="txtEditService" size="20" value="<?php echo $objResult["SERVICES"];?>"></td>
<td colspan="2" align="right"><div align="center">
<input name="btnAdd" type="button" id="btnUpdate" value="Update" OnClick="frmMain.hdnCmd.value='Update';frmMain.submit();">
<input name="btnAdd" type="button" id="btnCancel" value="Cancel" OnClick="window.location='<?php echo $_SERVER["PHP_SELF"];?>';">
</div></td>
</tr>
<?php
} else { ?>
<tr>
<td><div align="center"><?php echo $objResult["BUSINESSID"];?></div></td>
<td><?php echo $objResult["BUSINESSNAME"];?></td>
<td><?php echo $objResult["SERVICES"];?></td>
<td align="center">Edit</td>
<td align="center">Delete</td>
</tr>
<?php
}
} ?>
<tr>
<td><div align="center"><input type="text" name="txtAddBUSSINESSID" size="5"></div></td>
<td><input type="text" name="txtAddName" size="20"></td>
<td><input type="text" name="txtAddService" size="20"></td>
<td colspan="2" align="right"><div align="center"><input name="btnAdd" type="button" id="btnAdd" value="Add" OnClick="frmMain.hdnCmd.value='Add';frmMain.submit();"></div></td>
</tr>
</table>
</form>
<?php mysql_close($conn); ?>
</body>
</html>
`
The lines that gives me the errors is this if($_GET["Action"] == "Del")
and this if($objResult["BUSSINESSID"] == $_GET["ID"] and $_GET["Action"] == "Edit").
Thanks in advance if you can help me out.
You need to take whole code using $_POST or $_GET arrays inside if block and check for any of the main $_GET/$_POST variable that is set by (isset() method) if it is set then do that stuff in the if block else don't.
Check whether the index is set in the arrays before using them, like so:
if (isset($_GET["Action"]) {
if($_GET["Action"] == "Del") {
// Do stuff
}
}
As #A.B mentioned in the other answer, you should always do this check.
The reason you're getting this error is because the GET/POST requests don't have a value set for this index on the client side.

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