“Notice: Undefined variable”, “Notice: Undefined i

Question

I'm running a PHP script and continue to receive errors like:

Notice: Undefined variable: my_variable_name in C:\wamp\www\mypath\index.php on line 10

Notice: Undefined index: my_index C:\wamp\www\mypath\index.php on line 11

Line 10 and 11 looks like this:

echo "My variable value is: " . $my_variable_name;
echo "My index value is: " . $my_array["my_index"];

What is the meaning of these error messages?

Why do they appear all of a sudden? I used to use this script for years and I've never had any problem.

How do I fix them?

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Answer 1

Those notices are because you don't have the used variable defined and my_index key was not present into $my_array variable.

Those notices were triggered every time, because your code is not correct, but probably you didn't have the reporting of notices on.

Solve the bugs:

$my_variable_name = "Variable name"; // defining variable
echo "My variable value is: " . $my_variable_name;

if(isset($my_array["my_index"])){
    echo "My index value is: " . $my_array["my_index"]; // check if my_index is set 
}

Another way to get this out:

ini_set("error_reporting", false)

Answer 2

It means you are testing, evaluating, or printing a variable that you have not yet assigned anything to. It means you either have a typo, or you need to check that the variable was initialized to something first. Check your logic paths, it may be set in one path but not in another.

Answer 3

undefined index means in an array you requested for unavailable array index for example

<?php 

$newArray[] = {1,2,3,4,5};
print_r($newArray[5]);

?>

undefined variable means you have used completely not existing variable or which is not defined or initialized by that name for example

<?php print_r($myvar); ?>

undefined offset means in array you have asked for non existing key. And the solution for this is to check before use

php> echo array_key_exists(1, $myarray);

Answer 4

In PHP 7.0 it's now possible to use Null coalescing operator:

echo "My index value is: " . ($my_array["my_index"] ?? '');

Equals to:

echo "My index value is: " . (isset($my_array["my_index"]) ? $my_array["my_index"] : '');

PHP manual PHP 7.0

Answer 5

Error display @ operator

For undesired and redundant notices, one could use the dedicated @ operator to »hide« undefined variable/index messages.

$var = @($_GET["optional_param"]);

• This is usually discouraged. Newcomers tend to way overuse it.
• It's very inappropriate for code deep within the application logic (ignoring undeclared variables where you shouldn't), e.g. for function parameters, or in loops.
• There's one upside over the isset?: or ?? super-supression however. Notices still can get logged. And one may resurrect @-hidden notices with: set_error_handler("var_dump");
  • Additonally you shouldn't habitually use/recommend if (isset($_POST["shubmit"])) in your initial code.
  • Newcomers won't spot such typos. It just deprives you of PHPs Notices for those very cases. Add @ or isset only after verifying functionality.

  • Fix the cause first. Not the notices.

• @ is mainly acceptable for $_GET/$_POST input parameters, specifically if they're optional.

And since this covers the majority of such questions, let's expand on the most common causes:

$_GET / $_POST / $_REQUEST undefined input

• First thing you do when encountering an undefined index/offset, is check for typos:
  $count = $_GET["whatnow?"];

  • Is this an expected key name and present on each page request?
  • Variable names and array indicies are case-sensitive in PHP.

• Secondly, if the notice doesn't have an obvious cause, use var_dump or print_r to verify all input arrays for their curent content:

  var_dump($_GET);
  var_dump($_POST);
  //print_r($_REQUEST);
  

  Both will reveal if your script was invoked with the right or any parameters at all.

• Alternativey or additionally use your browser devtools (F12) and inspect the network tab for requests and parameters:

  

  POST parameters and GET input will be be shown separately.

• For $_GET parameters you can also peek at the QUERY_STRING in

  print_r($_SERVER);
  

  PHP has some rules to coalesce non-standard parameter names into the superglobals. Apache might do some rewriting as well. You can also look at supplied raw $_COOKIES and other HTTP request headers that way.

• More obviously look at your browser address bar for GET parameters:

  http://example.org/script.php?id=5&sort=desc

  The name=value pairs after the ? question mark are your query (GET) parameters. Thus this URL could only possibly yield $_GET["id"] and $_GET["sort"].

• Finally check your <form> and <input> declarations, if you expect a parameter but receive none.

  • Ensure each required input has an <input name=FOO>
  • The id= or title= attribute does not suffice.
  • A method=POST form ought to populate $_POST.
  • Whereas a method=GET (or leaving it out) would yield $_GET variables.
  • It's also possible for a form to supply action=script.php?get=param via $_GET and the remaining method=POST fields in $_POST alongside.
  • With modern PHP configurations (≥ 5.6) it has become feasible (not fashionable) to use $_REQUEST['vars'] again, which mashes GET and POST params.

• If you are employing mod_rewrite, then you should check both the access.log as well as enable the RewriteLog to figure out absent parameters.

$_FILES

• The same sanity checks apply to file uploads and $_FILES["formname"].
• Moreover check for enctype=multipart/form-data
• As well as method=POST in your <form> declaration.
• See also: PHP Undefined index error $_FILES?

$_COOKIE

• The $_COOKIE array is never populated right after setcookie(), but only on any followup HTTP request.
• Additionally their validity times out, they could be constraint to subdomains or individual paths, and user and browser can just reject or delete them.

Answer 6

Its because the variable '$user_location' is not getting defined. If you are using any if loop inside which you are declaring the '$user_location' variable then you must also have an else loop and define the same. For example:

$a=10;
if($a==5) { $user_location='Paris';} else { }
echo $user_location;

The above code will create error as The if loop is not satisfied and in the else loop '$user_location' was not defined. Still PHP was asked to echo out the variable. So to modify the code you must do the following:

$a=10;
if($a==5) { $user_location='Paris';} else { $user_location='SOMETHING OR BLANK'; }
echo $user_location;