Hi guys i have code like this
include('db.php');
if (isset($_POST['save']) && $_POST['save'] == '') {
if(isset($_POST['misamarti'])){ $misamarti = $_POST['misamarti']; }
if(isset($_POST['teleponi'])) { $teleponi = $_POST['teleponi']; }
if(isset($_POST['posta'])) { $posta = $_POST['posta']; }
if(isset($_POST['paqsi'])) { $paqsi = $_POST['paqsi'];}
$query = "UPDATE contact SET `misamarti` = ".$misamarti.", `teleponi` = ".$teleponi.", `posta` = ".$posta.", `paqsi` = ".$paqsi." WHERE `contact`.`id` =1;";
$result = mysql_query($query);
}
<form method="post" action="" enctype="multipart/form-data">
<input type="hidden" name="about" value="template1" />
<input type="text" value="" name"misamarti" />
<input type="text" value="" name"teleponi" />
<input type="text" value="" name"posta" />
<input type="text" value="" name"paqsi" />
<input type="submit" value="" name="save" />
and result is
Notice: Undefined variable: misamarti in C:\xampp\htdocs\Template\admin_panel\contact.php on line 18
Notice: Undefined variable: teleponi in C:\xampp\htdocs\Template\admin_panel\contact.php on line 18
Notice: Undefined variable: posta in C:\xampp\htdocs\Template\admin_panel\contact.php on line 18
Notice: Undefined variable: paqsi in C:\xampp\htdocs\Template\admin_panel\contact.php on line 18
if anyone know why is this error pls comment.
You're using the variables even if they weren't submitted with the form, and update the fields in the db with those undefined varaibles, so you'll be destroying any data that was in the DB previously.
If nothing else, you need to at least set a default value for the fields, e.g:
$misamarti = isset($_POST['misamarti']) ? $_POST['misamarti'] : '';
^^--default empty string
And then realize that you are WIDE OPEN to an sql injection attack. You are BEGGING to get your server pwn3d.
As you can see in your HTML code, you have the names bad written, they have to be written this way: name="misamarti". You forgot the equal (=) between name and the name. If you see the rest of attributes, like text and value, you see that theres an equal, that's the correct way of setting attributes.
Related
I am working on a PHP project - I had one form post a date to another form
I made some changes (although none to the input in question)
Now all other inputs are updated with their Posted values, except the date
If I manually set the date in HTML it works:
<div><input type="date" class="form-control" id="DateCourse" name="DateCourse" value="2009-01-01"></div>
If I set it to the following, it doesn't:
<div><input type="date" class="form-control" id="DateCourse" name="DateCourse" value="<?php echo (isset($DateCourse))?$DateCourse:'';?>"></div>
The below:
$DateCourse = ($_POST["DateCourse"]);
var_dump($_POST["DateCourse"]);
var_dump($DateCourse);
Returns:
string(10) "2019-01-05" - means the post value is set
Notice: Undefined variable: DateCourse in /home/bitecons/bts.biteconsulting.co.za/v2/editccr.php on line 119 - how can it be undefined, I just defined it
NULL
What on earth am I doing wrong? Apart from using PHP :P
Flow as requested:
Records.php:
This is the function to prepopulate my posted fields:
function Prefill(x) {
TabletoEdit = x.closest('table').id;
SelectedRow = x.rowIndex;
document.getElementById("EntryEditing").value = x.cells[19].innerHTML;
document.getElementById("DateCourse").value = x.cells[0].innerHTML;
document.forms["records"].submit();
}
Then I also have:
<form action="editrec" method="post" id="records">
<input type='hidden' name='Period' id='Period' />
<input type='hidden' name='Month' id='Month' />
<input type='hidden' name='res' id='res' />
<input type='hidden' name='CustName' id='CustName' />
<input type='hidden' name='DateCourse' id='DateCourse' />
</form>
The Prefill gets called, then submits the form
I have tracked and DateCourse has data, but when getting to the other form, it "disappears":
if(!empty($_POST)) {
$DateCourse = ($_POST["DateCourse"]);
$CustName = ($_POST["CustName"]);
}
For example, CustName is filled in, but not DateCourse?
Side question:
Would this return true if another post var is not set (unrelated to this one):
if(!empty($_POST))
I think You use wrong Code
you Must Submit First Form And Then Use $DateCourse this In another Form in POSTBACK
One of the best way is to define $DateCourse too like
<?php
$DateCourse = "";
if(!empty($_POST["DateCourse"])) {
$DateCourse = ($_POST["DateCourse"]);
}
?>
<div><input type="date" class="form-control" id="DateCourse" name="DateCourse" value="<?php echo $DateCourse;?>"></div>
Okay, apologies folks, but it may help others in the future.
I had a function call to an old function - this failed, causing my variable to never get defined... I knew it was something stupid, but sometimes one needs a sound board...
My code and form:
<?php
include("includes/connect.php");
$content = "SELECT * FROM content where content_page='home'";
$result = mysql_query($content);
$row = mysql_fetch_array($result);
?>
<form method="post" action="admin_home.php">
Headline:</br>
<input type="text" name="content_title" value="<?php echo "$row[content_title]" ?>"></br> </br>
Main content:</br>
<textarea type="text" class="txtinput" cols="55" rows="20" name="content_text"><?php echo "$row[content_text]" ?></textarea></br></br>
<input type="submit" name="submit" value="Save changes">
</form>
Code for what I want to happen when the 'submit' button is pressed:
<?php
include("includes/connect.php");
if(isset($_POST['submit'])){
$order = "UPDATE content
SET content_title='$content_title', content_text='$content_text' WHERE content_page='home'";
mysql_query($order);
}
?>
The error I get:
Notice: Undefined variable: content_title in /Applications/XAMPP/xamppfiles/htdocs/templacreations/admin/admin_home.php on line 63
Notice: Undefined variable: content_text in /Applications/XAMPP/xamppfiles/htdocs/templacreations/admin/admin_home.php on line 63
Line 63 is the following line from my submit button code:
SET content_title='$content_title', content_text='$content_text' WHERE
is this not declaring them?
It looks like you are looking for the POST values of your form.
They are not contained in the variables $content_title and $content_text. In fact your error is telling you that you haven't assigned anything to those variables.
They are contained in the $_POST array just like the value of the submit.
I.e
<?php
include("includes/connect.php");
if(isset($_POST['submit'])){
//Sanitize data and assign value to variable
$content_title = mysql_real_escape_string($_POST['content_title']);
$content_text = mysql_real_escape_string($_POST['content_text']);
$order = "UPDATE content
SET content_title='$content_title', content_text='$content_text'
WHERE content_page='home'";
mysql_query($order);
}
?>
It should be a multiple upload form for pictures
I get the HTML Code for a Upload-Form:
<form action="upload.php" method="post" id="uploadform" name="uploadform" enctype="multipart/form-data">
<label id="filelabel" for="fileselect">Choose the Pictures</label>
<input type="file" id="fileselect" class="fileuplaod" name="uploads[]" multiple />
<span class="text">Exist Album</span><br />
<select id="existAlbum" name="existAlbum" size="1">
<option value="noAlbum">SELECT ALBUM</option>
</select>
<span class="text">OR</span>
<span class="text">New Album</span><br />
<input id="newAlbum" name="newAlbum" type="text" maxlength="20" placeholder="ALBUM NAME"/>
<input type="submit">
</form>
The form link to the uploaded.php. But there i get:
Notice: Undefined index: existAlbum in E:\xampp\htdocs\fotokurs\upload\upload.php on line 11
Notice: Undefined index: newAlbum in E:\xampp\htdocs\fotokurs\upload\upload.php on line 12
Here's the upload.php:
<?PHP
$allowedExtensions = array('png', 'jpg', 'jpeg');
$maxSize = 20971520;
$i = 0;
$first = 0;
$exist_album = $_POST['existAlbum'];
$new_album = $_POST['newAlbum'];
Where is my fault? I can't find it...
EDIT
Add following to my code:
if( isset( $_POST['existAlbum'] ) or isset( $_POST['newAlbum'] ) ){
$exist_album = $_POST['existAlbum'];
$new_album = $_POST['newAlbum'];
}else{
echo 'no album <br />';
}
print_r($_POST);
new output:
no album
Array ( )
Notice: Undefined variable: new_album in E:\xampp\htdocs\fotokurs\upload\upload.php on line 20
Notice: Undefined variable: exist_album in E:\xampp\htdocs\fotokurs\upload\upload.php on line 21
Notice: Undefined variable: new_album in E:\xampp\htdocs\fotokurs\upload\upload.php on line 22
Notice: Undefined variable: exist_album in E:\xampp\htdocs\fotokurs\upload\upload.php on line 23
One of your issues is that existAlbum has no actual values associated with it.
You have <option>Select Album</option> which has no value associated with the option element. If there is no value associated, the select element is not posted to the server. You should change it to be:
<option value="">Select Album</option>
EDIT
Since the user only has to supply one or the other, you should use the following to set your variables:
$existsAlbum = (isset($_POST['existAlbum']) && !empty($_POST['existAlbum'])) ? $_POST['existAlbum'] : 'defaultValue';
$newAlbum = (isset($_POST['newAlbum']) && !empty($_POST['newAlbum'])) ? $_POST['newAlbum'] : 'defaultValue';
One important thing to note is that Internet Explorer does not support the placeholder attribute.
EDIT 2
Here is my quick test page that worked test.php:
<form action="upload.php" method="post" id="uploadform" name="uploadform" enctype="multipart/form-data">
<label id="filelabel" for="fileselect">Choose the Pictures</label>
<input type="file" id="fileselect" class="fileuplaod" name="uploads[]" multiple />
<span class="text">Exist Album</span><br />
<select id="existAlbum" name="existAlbum" size="1">
<option value="noAlbum">SELECT ALBUM</option>
</select>
<span class="text">OR</span>
<span class="text">New Album</span><br />
<input id="newAlbum" name="newAlbum" type="text" maxlength="20" placeholder="ALBUM NAME"/>
<input type="submit" value="Submit">
</form>
upload.php
<pre>
<?php print_r($_POST); ?>
<?php print_r($_FILES); ?>
</pre>
results
Array
(
[existAlbum] => noAlbum
[newAlbum] =>
)
Array
(
[uploads] => Array
(
//Contents here
)
)
Try if the value existAlbum get set, because it won't return any value if you there is nothing picked. You could give the existAlbum picker a default='1' or something:
if isset($_POST['existAlbum']){
echo 'yes';
}
else{
echo 'no';
}
I think that there is something wrong with the rule enctype="multipart/form-data". Try to just remove this, it should be set automatically by your browser.
You have no value for the option select album, even if you don't intend that option to be used give it a value such as 0 so that it will always be set in the POST variables.
<option value="0">SELECT ALBUM</option>
<option value="some album">Some Album</option>
...
If select is not picked you will not get it at all (you expect it to be empty, which is not true). You have to check first
$exist_album = isset($_POST['existAlbum']) ? $_POST['existAlbum'] : '<DEFAULT VALUE>';
and same for checkbox.
The newAlbum thing should work as text inputs are always there. See
print_r($_POST);
to see what's really in there, and in my case it is - on "empty" submit I get:
Array
(
[existAlbum] => SELECT ALBUM
[newAlbum] =>
)
BTW: you should use <?php rather than <?PHP.
print $_POST Array using print_r($_POST); Make sure your form action is correct
<form action="upload.php" method="post" id="uploadform" name="uploadform" enctype="multipart/form-data">
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
strange php error
Submit an HTML form with empty checkboxes
PHP form checkbox and undefined index
Some reason I am getting undefined index errors in the following code. Any help would be appreciated. Here are the errors I am Recieving
Notice: Undefined index: username in C:\xampp\htdocs\register.php on line 5
Notice: Undefined index: password in C:\xampp\htdocs\register.php on line 6
Notice: Undefined index: firstname in C:\xampp\htdocs\register.php on line 7
Notice: Undefined index: surname in C:\xampp\htdocs\register.php on line 8
Code:
<?php
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("clubresults") or die(mysql_error());
$username = $_POST['username'];
$password = md5($_POST['password']);
$firstname = $_POST['firstname'];
$surname = $_POST['surname'];
$sql = "INSERT INTO members (Username, Password, Firstname, Surname) VALUES ($username, $password, $firstname, $surname);";
mysql_query($sql);
?>
<html>
<div class="content">
<center>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<table border="0">
<tr><td colspan=2><h1>Register</h1></td></tr>
<tr><td>Username:</td><td>
<input type="text" name="username" maxlength="60">
</td></tr>
<tr><td>Password:</td><td>
<input type="password" name="password" maxlength="10">
</td></tr>
<tr><td>Firstname:</td><td>
<input type="text" name="firstname" maxlength=210">
</td></tr>
<tr><td>Surname:</td><td>
<input type="text" name="surname" maxlength=210">
</td></tr>
<tr><th colspan=2><input type="submit" name="submit"
value="Register"> </th></tr> </table>
</form>
</center>
</div>
</html>
When your page loads for the first time (unless it was the action of another page), the REQUEST_METHOD is 'GET'; as a result, no POST variables exist (although the $_POST superglobal exists).
When you POST, either from another page, or from the same page, all the data set in the form will be present in the $_POST superglobal, and can be accessed the way you accessed them earlier; unfortunately, if a data is not present, and you try to access it, an error will be thrown, similar to what you have.
To solve the issue, especially if you want to post to the same page as the form, use something like
if (isset($_POST["username"]))
{
// do whatever
}
if (isset($_POST["password"]))
{
// do whatever
}
if (isset($_POST["firstname"]))
{
// do whatever
}
if (isset($_POST["surname"]))
{
// do whatever
}
With that, your code won't throw an error if those data aren't present.
Hope that helps.
On an unrelated, but important note, do not use mysql_* functions anymore. They have been deprecated; use mysqli or PDO functions for data access.
Happy coding!
I'm new in PHP and I'm getting this error:
Notice: Undefined index: productid in /var/www/test/modifyform.php on
line 32
Notice: Undefined index: name in /var/www/test/modifyform.php on line
33
Notice: Undefined index: price in /var/www/test/modifyform.php on line
34
Notice: Undefined index: description in /var/www/test/modifyform.php
on line 35
I couldn't find any solution online, so maybe someone can help me.
Here is the code:
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="POST">
<input type="hidden" name="rowID" value="<?php echo $rowID;?>">
<p>
Product ID:<br />
<input type="text" name="productid" size="8" maxlength="8" value="<?php echo $productid;?>" />
</p>
<p>
Name:<br />
<input type="text" name="name" size="25" maxlength="25" value="<?php echo $name;?>" />
</p>
<p>
Price:<br />
<input type="text" name="price" size="6" maxlength="6" value="<?php echo $price;?>" />
</p>
<p>
Description:<br />
<textarea name="description" rows="5" cols="30">
<?php echo $description;?></textarea>
</p>
<p>
<input type="submit" name="submit" value="Submit!" />
</p>
</form>
<?php
if (isset($_POST['submit'])) {
$rowID = $_POST['rowID'];
$productid = $_POST['productid']; //this is line 32 and so on...
$name = $_POST['name'];
$price = $_POST['price'];
$description = $_POST['description'];
}
What I do after that (or at least I'm trying) is to update a table in MySQL.
I really can't understand why $rowID is defined while the other variables aren't.
Thank you for taking your time to answer me.
Cheers!
Try:
<?php
if (isset($_POST['name'])) {
$name = $_POST['name'];
}
if (isset($_POST['price'])) {
$price = $_POST['price'];
}
if (isset($_POST['description'])) {
$description = $_POST['description'];
}
?>
Apparently the index 'productid' is missing from your html form.
Inspect your html inputs first. eg <input type="text" name="productid" value="">
But this will handle the current error PHP is raising.
$rowID = isset($_POST['rowID']) ? $_POST['rowID'] : '';
$productid = isset($_POST['productid']) ? $_POST['productid'] : '';
$name = isset($_POST['name']) ? $_POST['name'] : '';
$price = isset($_POST['price']) ? $_POST['price'] : '';
$description = isset($_POST['description']) ? $_POST['description'] : '';
This is happening because your PHP code is getting executed before the form gets posted.
To avoid this wrap your PHP code in following if statement and it will handle the rest no need to set if statements for each variables
if(isset($_POST) && array_key_exists('name_of_your_submit_input',$_POST))
{
//process PHP Code
}
else
{
//do nothing
}
TRY
<?php
$rowID=$productid=$name=$price=$description="";
if (isset($_POST['submit'])) {
$rowID = $_POST['rowID'];
$productid = $_POST['productid']; //this is line 32 and so on...
$name = $_POST['name'];
$price = $_POST['price'];
$description = $_POST['description'];
}
There should be the problem, when you generate the <form>. I bet the variables $name, $price are NULL or empty string when you echo them into the value of the <input> field. Empty input fields are not sent by the browser, so $_POST will not have their keys.
Anyway, you can check that with isset().
Test variables with the following:
if(isset($_POST['key'])) ? $variable=$_POST['key'] : $variable=NULL
You better set it to NULL, because
NULL value represents a variable with no value.
Hey this is happening because u r trying to display value before assignnig it
U just fill in the values and submit form it will display correct output
Or u can write ur php code below form tags
It ll run without any errors
If you are using wamp server , then i recommend you to use xampp server .
you . i get this error in less than i minute but i resolved this by using (isset) function . and i get no error .
and after that i remove (isset) function and i don,t see any error.
by the way i am using xampp server
this error occurred sometime method attribute ( valid passing method )
Error option :
method="get" but called by $Fname = $_POST["name"];
or
method="post" but called by $Fname = $_GET["name"];
More info visit http://www.doordie.co.in/index.php
To remove this error, in your html form you should do the following in enctype:
<form enctype="multipart/form-data">
The following down is the cause of that error i.e if you start with form-data in enctype, so you should start with multipart:
<form enctype="form-data/multipart">