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
Related
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.
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 $book){
}
This question already has an answer here:
Post form and update multiple rows with mysql
(1 answer)
Closed 3 days ago.
Given this form, shown as a table:
<form action="multi.php" name="multi[]" method="POST" class="forms">
<table class="table-hovered">
<tr>
<th class="text-left highlight">attività svolta</th>
<th class="text-left highlight">categoria</th>
</tr>
<?php
foreach ($_POST['item'] as $k => $v)
{
$q_item = "SELECT * FROM eventi WHERE id = '".$v."'";
$qu_item = mysql_query($q_item);
while($res = mysql_fetch_array($qu_item))
{
?>
<tr>
<td><?php echo $res['descrizione'];?></td>
<td>
<select name="categoria">
<option value="<?php echo $res['categoria'];?>" selected><?php echo $res['categoria'];?>
<option value="80"> 80
<option value="40"> 40
<option value="70"> 70
</select>
</td>
<input type="hidden" name="idd" value="<?php echo $res['id'];?>">
</tr>
<?php
}
}
?>
</table>
<input type="submit" name="submit" value="modify" />
</form>
I am trying to edit multiple entries, using the code below:
<?php
$utente = $_SESSION["username"];
$id = $_POST["idd"];
$categoria = $_POST["categoria"];
if (!$id or !$categoria){
echo "Error";
}
else
if ($categoria!=='80' && $categoria!=='40' && $categoria!=='70'){
echo "Error";
}
else{
$sql="UPDATE eventi SET categoria='$categoria' WHERE id='$id'";
$update=mysql_query($sql);
echo "Entry modified correctly";
}
?>
As you see, this code changes just one item. I have tried making it recursive. Maybe using a "foreach" is the way to go.
Any hints are appreciated. And sorry for using an old version of PHP (I haven't switched to version 7 yet).
As you have same names for both input and select the last value of each of them overwrites previous values. For passing multiple values in inputs with same names - use [] notation:
<select name="categoria[]">
<option value="<?php echo $res['categoria'];?>" selected><?php echo $res['categoria'];?>
<option value="80"> 80
<option value="40"> 40
<option value="70"> 70
</select>
<input type="hidden" name="idd[]" value="<?php echo $res['id'];?>">
After that - check your $_POST values with print_r - you will see that
$_POST[categoria] and $_POST[idd] are arrays and you can iterate over them with for or foreach.
Btw, inserting an <input> right after </td> produces invalid html.
There's no need to create any hidden input element in the first place, you just have to change the name attribute of <select> element in the following way,
name="categoria[<?php echo $res['id'] ?>]"
So your code should be like this,
<form action="multi.php" name="multi[]" method="POST" class="forms">
<table class="table-hovered">
<tr>
<th class="text-left highlight">attività svolta</th>
<th class="text-left highlight">categoria</th>
</tr>
<?php
foreach ($_POST['item'] as $k => $v){
$q_item = "SELECT * FROM eventi WHERE id = '".$v."'";
$qu_item = mysql_query($q_item);
while($res = mysql_fetch_array($qu_item)){
?>
<tr>
<td><?php echo $res['descrizione'];?></td>
<td>
<select name="categoria[<?php echo $res['id'] ?>]">
<option value="<?php echo $res['categoria'];?>" selected><?php echo $res['categoria'];?>
<option value="80"> 80
<option value="40"> 40
<option value="70"> 70
</select>
</td>
</tr>
<?php
}
}
?>
</table>
<input type="submit" name="submit" value="modify" />
</form>
And this is how you can process your form to perform UPDATE operation,
foreach($_POST['categoria'] as $id => $categoria){
$sql="UPDATE eventi SET categoria='". $categoria . "' WHERE id='" . $id . "'";
// execute your query
}
Note: If you want to see the complete array structure, do var_dump($_POST);
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($parcelID)) with it.
Use mysqli_* extension instead of deprecated mysql_*.
I have made a script which saves a few lines of PHP to a .php file. All my scripts work perfectly fine, but just this page is starting to get annoying.
Explaination of what is happening in the GIF:
1: I change the settings - the last 2 settings need to appear when you enable the 2nd setting called "Custom style". That works fine and all.
2: So you enable it and for some reason it completely wipes the other 2 settings (the "primarycolor" and "adminbg").
How can this happen? What am I doing wrong? My script is below if you want to try it out yourself.
<?php
if (isset($_POST["submit"])) {
$string = '<?php
$customoptions = '. $_POST["customoptions"] .';
$primarycolor = "'. $_POST["primarycolor"] .'";
$adminbg = "'. $_POST["adminbg"] .'";
?>';
$fp = fopen("includes/userstyle.php", "w");
fwrite($fp, $string);
fclose($fp);
}
include("includes/userstyle.php");
?>
<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 $lang['chooselanguage']; ?></option>
<option value="dashboard.php?lang=en">English</option>
<option value="dashboard.php?lang=nl">Dutch</option>
</select>
</td>
</tr>
<tr>
<td>Custom Style</td>
<td><select name="customoptions" id="customoptions"><option value="true" <?php if($customoptions == true){ echo 'selected'; }; ?>><?php echo $lang['enabled']; ?></option><option value="false" <?php if($customoptions == false){ echo 'selected'; }; ?>><?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>
EDIT: userstyle.php
<?php
$customoptions = true;
$primarycolor = "555";
$adminbg = "fff";
?>
When you post your form for the second time, you post empty values, and they are saved to file, as expected.
You should add something like
if (isset($_POST["submit"]) && !empty($_POST['primarycolor']) && !empty($_POST['adminbg'])) {
// ...
But actually there can be another validation rules.
And as others noticed in comments — this is, even in educational purposes, very stupid idea to save user data into php file and then execute that file. The simplest alternative — save settings to a json file with json_encode, then decode, and don't forget to html-escape them with at least htmlspecialchars.