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Closed 10 years ago.
Possible Duplicate:
PHP: “Notice: Undefined variable” and “Notice: Undefined index”
When do we get this error?
Notice: Undefined index: submit in C:\wamp\www\sample.php on line 25
What is the exact meaning of this perticular error?
This means exactly what it says: you're addressing to an undefined index in an array
$arr = array();
echo $arr['foo'];
In the example above the array is empty but I tried to output 'foo' item value, which doesn't exist.
It means you are trying to access a part of an arraythat isn't there.
If you have an array with 5 elements, you ca get to them via:
$array[0] through to $array[4]
But if you try $array[76] which doesn't exist, you will get an undefined Index error.
You've probably got an array that you're accessing like $_POST['submit']. That error message is saying is the element 'submit' of the array doesn't exist, and it's throwing a warning.
You should check that array elements exist before using them isset() before you access them to avoid avoid the warning.
Edit: possible duplicate of this: Undefined index in PHP
have you correctly mentioned the method in your form GET or POST ?? I think you are accessing/testing it without declaring it. Let me know if this is the case.
Related
This question already has answers here:
php array access on boolean
(3 answers)
Closed 5 years ago.
Why does PHP throw a "Notice: Undefined offset" error here:
<?php
$MyVar = array();
echo $MyVar[0]; //Notice: Undefined offset: 0
?>
But not here:
<?php
$MyVar = false;
echo $MyVar[0]; //No error
?>
It's ultimately because in your 2nd example $MyVar[0] is null which isn't an error. You could probably reference $MyVar[0][1][2][3] and get the same result.
The first example isn't null it's a missing index in an array so it warns you.
Undefined offset is provided when there are no values in an array and you're trying to reference an uninitialised index.
In case of the 2nd code, you've assigned(initialised) a value to the variable which is not different from variable[0].
If you assign values for 1st two indexes in 1st example, and refer to the 3rd index, you should get the undefined offset error.
I dont think it is about null or not, but a case of assigning values to indexes.
If you take a C parlance, array is an equivalent of malloc(upto a certain extent). Thankfully PHP does not crash, just throws an undefined index[offset] error
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Best way to test for a variable's existence in PHP; isset() is clearly broken
(17 answers)
Closed 7 years ago.
How do the following two function calls compare to check if a variable exists for a GET call or a POST call ?
if(isset($_GET['var']))
if($_GET['var'])
Both do their job but sometimes without isset() i get a warning :
PHP Notice: Undefined index: var on at Line XX
Update
It was a $_GET not $GET. Also i know it has nothing to do with GET/POST. All i am discussing is variable check instead of HTTP method. Thanks.
$_GET[''] is usually retrieved from the browser, so if you had something like the following:
www.hello.com/index.php?var=hello
You could use:
if(isset($_GET['var'])) {
echo $_GET['var']; // would print out hello
}
Using isset() will stop the undefined index error, so if you just had www.hello.com/index.php for example then you would not see the error even though the var is not set.
Posts are usually when one page posts information to another, the method is the same but using $_POST[''] instead of $_GET['']. Example you have a form posting information:
<form method="post" action="anotherpage.php">
<label></label>
<input type="text" name="text">
</form>
In anotherpage.php to get the information would be something like:
$text = isset($_POST['text']);
echo $text; // would echo what ever you input into the text field on the other page
In a nut shell, just putting $_GET['name'] if name is not set you will get an error.
Using isset($_GET['name']) will check is the var name has any value before continuing.
Not sure if this is what you were after, but from the question this is my best guess at an answer
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 just started coding so i dont know whats wrong with my code
A PHP Error was encountered
Severity: Notice
Message: Undefined variable: kd_user
Filename: views/dokter.php
Line Number: 1
Controller
function read_user()
{
$this->load->model('dokter_model');
$data['datauser']=$this->dokter_model->read_user();
$this->load->view('dokter', $data);
}
Model
function read_user()
{
$q="SELECT a.*, b.*
FROM users a
LEFT JOIN dokter b ON a.kd_user=b.kd_user
WHERE a.kd_user='".$this->session->userdata('kd_user')."'";
$query=$this->db->query($q);
return $query->result();
}
Views
<p align="center">Selamat Datang <?php echo $kd_user;?></p>
Sorry for my poor english
Change :
$data['datauser']=$this->dokter_model->read_user();
to:
$data['kd_user']=$this->dokter_model->read_user();
In general, the error
Message: Undefined variable: kd_user
Means that you are referring to a variable (in this case, $kd_user), and it was never defined (you never initialized it and/or assigned a default value to it.)
Looking at your code, you are referring to kd_user in two places -- once in the associative array user_data in your model, and once in your view as kd_user.
To begin, you may wish to debug in each of those files (most probably the view) if that variable actually exists.
Without seeing all of your app's code, its hard to say for sure, but I'd verify that your view has a variable available called $kd_user for you to access. If it isn't there, you either need to find a way to get it into there, or reference another variable to get to the value you need.
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Closed 10 years ago.
Possible Duplicate:
PHP: “Notice: Undefined variable” and “Notice: Undefined index”
I've skimmed through docs and this site but still unsure. I'm working with Drupal 7.
Printing the variable with this code:
<?php print $info['comments'] ?>
If there are zero comments I get
Notice: Undefined index as a message.
If it exists, it prints the correct #.
Any help would be appreciated. How do I set it so if there are no comments it displays a zero?
Thanks!
To print an integer no matter what:
<?php print #intval($info['comments']) ?>
Or to irrevocably suppress the notice:
<?php print isset($info['comments']) ? $info['comments'] : 0 ?>
I don't know drupal, and don't know your context exactly, but if you try to access an index that doesn't exists from an array, this kind of notice is displayed.
To avoid such a thing, you can do something like :
if(isset($info['comments'])) {
print $info['comments'];
}
PHP arrays are associative by nature, thus you can have an array composed of all types of keys/values:
arry['stringkey'] = 'somestring';
arry[1] = 'some other string';
etc..
If you were to try and reference an index in the array that is not present without doing any error checking, like so
if(arry[2] == 'some third string')
you'd get an 'undefined index' error. Look into the isset() function for a solution.
You could test if the array has that key:
if (array_key_exists('comments', $info)) {
...
}
But I am not sure if that is what you really want to do.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
PHP: “Notice: Undefined variable” and “Notice: Undefined index”
first time asking questions here. I am trying to create a small, primitive forum. I am basing it on this: while making changes to it where it is needed.
So far, I have setup the database and all that. Now I am trying to fix the view_topic.php and it is returning "Notice: Undefined index: id" on line 30 something.
This is line 30: $id=$_GET['id'];
You should make sure id is a valid index before accessing it.
$id = isset($_GET['id']) ? $_GET['id'] : null;
Notice: Undefined index:
This means that a piece of code attempts to access an element of an array that does not exist.
So, for example:
$myArray = Array(); // an empty array
echo $myArray['id']; // print 'id' element
// ^^ Oops! No such element yet!
The $_GET array contains the arguments from the querystring, which is the ?id=1 part of your request URL.
You will need to find out why this does not contain the element with key 'id' when id was provided in the URL, and then make your code decide what to do in the case that this value is missing, using a function such as isset or array_key_exists and an if statement.
Most likely you will present a more useful error message and terminate the script, if your code cannot continue without a valid id value.
Are you sure id= exists in the URL? Try adding the following near the top of your code.
var_dump($_GET);
var_dump dumps information about a variable to your screen. In this case, it will display all the GET data sent from your form. This is a debugging technique so you can see if your code is receiving what you expect. If the form uses method="post" and id is a field in the form, then you will need to use $_POST['id'] or $_REQUEST['id'] to access it.