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

2019-09-17 21:58发布

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?


This is a General Reference question for people to link to as duplicate, instead of having to explain the issue over and over again. I feel this is necessary because most real-world answers on this issue are very specific.

Related Meta discussion:

28条回答
男人必须洒脱
2楼-- · 2019-09-17 22:21

In PHP you need fist to define the variable after that you can use it.
We can check variable is defined or not in very efficient way!.

//If you only want to check variable has value and value has true and false value.
//But variable must be defined first.

if($my_variable_name){

}

//If you want to check variable is define or undefine
//Isset() does not check that variable has true or false value
//But it check null value of variable
if(isset($my_variable_name)){

}

Simple Explanation

//It will work with :- true,false,NULL
$defineVarialbe = false;
if($defineVarialbe){
    echo "true";
}else{
    echo "false";
}

//It will check variable is define or not and variable has null value.
if(isset($unDefineVarialbe)){
    echo "true";
}else{
    echo "false";
}
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Root(大扎)
3楼-- · 2019-09-17 22:22

WHY IS THIS HAPPENING?

Over time, PHP has become a more security-focused language. Settings which used to be turned off by default are now turned on by default. A perfect example of this is E_STRICT, which became turned on by default as of PHP 5.4.0.

Furthermore, according to PHP documentation, by default, E_NOTICE is disabled in php.ini. PHP docs recommend turning it on for debugging purposes. However, when I download PHP from the Ubuntu repository–and from BitNami's Windows stack–I see something else.

; Common Values:
;   E_ALL (Show all errors, warnings and notices including coding standards.)
;   E_ALL & ~E_NOTICE  (Show all errors, except for notices)
;   E_ALL & ~E_NOTICE & ~E_STRICT  (Show all errors, except for notices and coding standards warnings.)
;   E_COMPILE_ERROR|E_RECOVERABLE_ERROR|E_ERROR|E_CORE_ERROR  (Show only errors)
; Default Value: E_ALL & ~E_NOTICE & ~E_STRICT & ~E_DEPRECATED
; Development Value: E_ALL
; Production Value: E_ALL & ~E_DEPRECATED & ~E_STRICT
; http://php.net/error-reporting
error_reporting = E_ALL & ~E_DEPRECATED & ~E_STRICT

Notice that error_reporting is actually set to the production value by default, not to the "default" value by default. This is somewhat confusing and is not documented outside of php.ini, so I have not validated this on other distributions.

To answer your question, however, this error pops up now when it did not pop up before because:

  1. You installed PHP and the new default settings are somewhat poorly documented but do not exclude E_NOTICE.

  2. E_NOTICE warnings like undefined variables and undefined indexes actually help to make your code cleaner and safer. I can tell you that, years ago, keeping E_NOTICE enabled forced me to declare my variables. It made it a LOT easier to learn C, were not declaring variables is much bigger of a nuisance.

WHAT CAN I DO ABOUT IT?

  1. Turn off E_NOTICE by copying the "Default value" E_ALL & ~E_NOTICE & ~E_STRICT & ~E_DEPRECATED and replacing it with what is currently uncommented after the equals sign in error_reporting =. Restart Apache, or PHP if using CGI or FPM. Make sure you are editing the "right" php.ini. The correct one will be Apache if you are running PHP with Apache, fpm or php-fpm if running PHP-FPM, cgi if running PHP-CGI, etc. This is not the recommended method, but if you have legacy code that's going to be exceedingly difficult to edit, then it might be your best bet.

  2. Turn off E_NOTICE on the file or folder level. This might be preferable if you have some legacy code but want to do things the "right" way otherwise. To do this, you should consult Apache2, Nginx, or whatever your server of choice is. In Apache, you would use php_value inside of <Directory>.

  3. Rewrite your code to be cleaner. If you need to do this while moving to a production environment or don't want someone to see your errors, make sure you are disabling any display of errors, and only logging your errors (see display_errors and log_errors in php.ini and your server settings).

To expand on option 3: This is the ideal. If you can go this route, you should. If you are not going this route initially, consider moving this route eventually by testing your code in a development environment. While you're at it, get rid of ~E_STRICT and ~E_DEPRECATED to see what might go wrong in the future. You're going to see a LOT of unfamiliar errors, but it's going to stop you from having any unpleasant problems when you need to upgrade PHP in the future.

WHAT DO THE ERRORS MEAN?

Undefined variable: my_variable_name - This occurs when a variable has not been defined before use. When the PHP script is executed, it internally just assumes a null value. However, in which scenario would you need to check a variable before it was defined? Ultimately, this is an argument for "sloppy code". As a developer, I can tell you that I love it when I see an open source project where variables are defined as high up in their scopes as they can be defined. It makes it easier to tell what variables are going to pop up in the future and makes it easier to read/learn the code.

function foo()
{
    $my_variable_name = '';

    //....

    if ($my_variable_name) {
        // perform some logic
    }
}

Undefined index: my_index - This occurs when you try to access a value in an array and it does not exist. To prevent this error, perform a conditional check.

// verbose way - generally better
if (isset($my_array['my_index'])) {
    echo "My index value is: " . $my_array['my_index'];
}

// non-verbose ternary example - I use this sometimes for small rules.
$my_index_val = isset($my_array['my_index'])?$my_array['my_index']:'(undefined)';
echo "My index value is: " . $my_index_val;   

Another option is to declare an empty array at the top of your function. This is not always possible.

$my_array = array(
    'my_index' => ''
);

//...

$my_array['my_index'] = 'new string';

(additional tip)

  • When I was encountering these and other issues, I used NetBeans IDE (free) and it gave me a host of warnings and notices. Some of them offer very helpful tips. This is not a requirement, and I don't use IDEs anymore except for large projects. I'm more of a vim person these days :).
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男人必须洒脱
4楼-- · 2019-09-17 22:22

One common cause of a variable not existing after an HTML form has been submitted is the form element is not contained within a <form> tag:

Example: Element not contained within the <form>

<form action="example.php" method="post">
    <p>
        <input type="text" name="name" />
        <input type="submit" value="Submit" />
    </p>
</form>

<select name="choice">
    <option value="choice1">choice 1</option>
    <option value="choice2">choice 2</option>
    <option value="choice3">choice 3</option>
    <option value="choice4">choice 4</option>
</select>

Example: Element now contained within the <form>

<form action="example.php" method="post">
    <select name="choice">
        <option value="choice1">choice 1</option>
        <option value="choice2">choice 2</option>
        <option value="choice3">choice 3</option>
        <option value="choice4">choice 4</option>
    </select>
    <p>
        <input type="text" name="name" />
        <input type="submit" value="Submit" />
    </p>
</form>
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Bombasti
5楼-- · 2019-09-17 22:24

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);
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ら.Afraid
6楼-- · 2019-09-17 22:25

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:

    browser developer tools / network tab

    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.
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倾城 Initia
7楼-- · 2019-09-17 22:26

In a very Simple Language.
The mistake is you are using a variable $user_location which is not defined by you earlier and it doesn't have any value So I recommend you to please declare this variable before using it, For Example:


$user_location = '';
Or
$user_location = 'Los Angles';
This is a very common error you can face.So don't worry just declare the variable and Enjoy Coding.

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