PHP $_Get and Undefined Index

php $_GET and undefined index

Error reporting will have not included notices on the previous server which is why you haven't seen the errors.

You should be checking whether the index s actually exists in the $_GET array before attempting to use it.

Something like this would be suffice:

if (isset($_GET['s'])) {
    if ($_GET['s'] == 'jwshxnsyllabus')
        echo "<body onload=\"loadSyllabi('syllabus', '../syllabi/jwshxnporsyllabus.xml',         '../bibliographies/jwshxnbibliography_')\">";
    else if ($_GET['s'] == 'aquinas')
        echo "<body onload=\"loadSyllabi('syllabus', '../syllabi/AquinasSyllabus.xml')\">"; 
    else if ($_GET['s'] == 'POP2')
        echo "<body onload=\"loadSyllabi('POP2')\">";
} else {
    echo "<body>";
}

It may be beneficial (if you plan on adding more cases) to use a switch statement to make your code more readable.

switch ((isset($_GET['s']) ? $_GET['s'] : '')) {
    case 'jwshxnsyllabus':
        echo "<body onload=\"loadSyllabi('syllabus', '../syllabi/jwshxnporsyllabus.xml',         '../bibliographies/jwshxnbibliography_')\">";
        break;
    case 'aquinas':
        echo "<body onload=\"loadSyllabi('syllabus', '../syllabi/AquinasSyllabus.xml')\">";
        break;
    case 'POP2':
        echo "<body onload=\"loadSyllabi('POP2')\">";
        break;
    default:
        echo "<body>";
        break;
}

EDIT: BTW, the first set of code I wrote mimics what yours is meant to do in it's entirety. Is the expected outcome of an unexpected value in ?s= meant to output no <body> tag or was this an oversight? Note that the switch will fix this by always defaulting to <body>.

PHP: $_GET returns undefined index

From your code syntax I think you are problem is in the how you are getting the city. Your error indicates that it cannot find the index of city in $_GET; so basically it has not been set. Add a isset check before running the code, that will prevent any errors:

if ( isset($_GET['city']) ) {
    $urlContents = file_get_contents("http://api.openweathermap.org/data/2.5/weather?q=".urlencode($_GET['city']).",uk&appid=Hidden");
    //...
}

This will prevent the code to run and avoid any errors; Now check your url to make sure that it contains: https://example.com?city=somecity. That should fix the error.

Undefined index for $_GET

The problem is that you don't propagate the $code variable to when the button submit is clicked. To fix it, just add ?code= to your action form:

<?php $code = isset($_GET['code']) ? $_GET['code'] : ""; ?>
<form method="post" id="contact_form" action="feedback_form_entry.php?code=<?php echo $code; ?>">

Undefined Index while retrieving variable from $_GET in URL

What is $sitenumber = $_GET['site']; //Undefined index here supposed to be when it's not setup yet? What si the default value?

$sitenumber = $_GET['site'] ?? 'DEFAULT VALUE';

same as

$sitenumber = isset($_GET['site']) ? $_GET['site'] : 'DEFAULT VALUE';

Looks like you might want:

$sitenumber = $_GET['site'] ?? $sitenumber;

Since you define that var at the top.

Notice: Undefined index with used isset and $_GET

Check both of variables for exists or change code

if( isset($_GET['delt']) ) {
    $delt = intval($_GET['delt']);
  }
  if ( isset($_GET['editt']) ) {
    $editt= intval($_GET['editt']);
  }

Notice: Undefined variable , Notice: Undefined index , Warning: Undefined array key , and Notice: Undefined offset using PHP

Notice / Warning: Undefined variable

From the vast wisdom of the PHP Manual:

Relying on the default value of an uninitialized variable is problematic in the case of including one file into another which uses the same variable name. It is also a major security risk with register_globals turned on. E_NOTICE level error is issued in case of working with uninitialized variables, however not in the case of appending elements to the uninitialized array. isset() language construct can be used to detect if a variable has been already initialized. Additionally and more ideal is the solution of empty() since it does not generate a warning or error message if the variable is not initialized.

From PHP documentation:

No warning is generated if the variable does not exist. That means
empty() is essentially the concise equivalent to !isset($var) || $var
== false.

This means that you could use only empty() to determine if the variable is set, and in addition it checks the variable against the following, 0, 0.0, "", "0", null, false or [].

Example:

$o = [];
@$var = ["",0,null,1,2,3,$foo,$o['myIndex']];
array_walk($var, function($v) {
    echo (!isset($v) || $v == false) ? 'true ' : 'false';
    echo ' ' . (empty($v) ? 'true' : 'false');
    echo "\n";
});

Test the above snippet in the 3v4l.org online PHP editor

Although PHP does not require a variable declaration, it does recommend it in order to avoid some security vulnerabilities or bugs where one would forget to give a value to a variable that will be used later in the script. What PHP does in the case of undeclared variables is issue a very low level error, E_NOTICE, one that is not even reported by default, but the Manual advises to allow during development.

Ways to deal with the issue:

Recommended: Declare your variables, for example when you try to append a string to an undefined variable. Or use isset() / !empty() to check if they are declared before referencing them, as in:

//Initializing variable
$value = ""; //Initialization value; Examples
             //"" When you want to append stuff later
             //0  When you want to add numbers later
//isset()
$value = isset($_POST['value']) ? $_POST['value'] : '';
//empty()
$value = !empty($_POST['value']) ? $_POST['value'] : '';

This has become much cleaner as of PHP 7.0, now you can use the null coalesce operator:

    // Null coalesce operator - No need to explicitly initialize the variable.
    $value = $_POST['value'] ?? '';

Set a custom error handler for E_NOTICE and redirect the messages away from the standard output (maybe to a log file):
```
set_error_handler('myHandlerForMinorErrors', E_NOTICE | E_STRICT)
```
Disable E_NOTICE from reporting. A quick way to exclude just E_NOTICE is:
```
error_reporting( error_reporting() & ~E_NOTICE )
```
Suppress the error with the @ operator.

Note: It's strongly recommended to implement just point 1.

Notice: Undefined index / Undefined offset / Warning: Undefined array key

This notice/warning appears when you (or PHP) try to access an undefined index of an array.

Ways to deal with the issue:

Check if the index exists before you access it. For this you can use isset() or array_key_exists():

//isset()
$value = isset($array['my_index']) ? $array['my_index'] : '';
//array_key_exists()
$value = array_key_exists('my_index', $array) ? $array['my_index'] : '';

The language construct list() may generate this when it attempts to access an array index that does not exist:
```
list($a, $b) = array(0 => 'a');
//or
list($one, $two) = explode(',', 'test string');
```

Two variables are used to access two array elements, however there is only one array element, index 0, so this will generate:

Notice: Undefined offset: 1

#$_POST / $_GET / $_SESSION variable

The notices above appear often when working with $_POST, $_GET or $_SESSION. For $_POST and $_GET you just have to check if the index exists or not before you use them. For $_SESSION you have to make sure you have the session started with session_start() and that the index also exists.

Also note that all 3 variables are superglobals and are uppercase.

Notice: Undefined variable
Notice: Undefined Index

PHP tutorial, issues with Undefined index and recommended isset fixes

You have to check isset before you assign it to variable. change your php script as below:

<?php

if(isset($_GET['name']) && isset($_GET['age']) && !empty($_GET['name']) && !empty($_GET['age'])) {
    $name = $_GET['name'];
    $age = $_GET['age'];
    echo 'I am '.$name.' and I am '.$age.' years of age.';
} else {
    echo 'Please type something.';
}

?>

PHP $_Get and Undefined Index