Let's learn about the solution to Undefined index (selected)

After studying PHP for so long, today we will learn about the solution to PHP Notice: Undefined index. Maybe you don’t understand what it means, but I believe that when you read this article, you will definitely have new ideas. reward!

php Notice: Undefined index problem, Undefined index: means that there is an error in your code: "The variable is not defined yet, and it is used after assignment." This is not a fatal error and will not let you The PHP code is forcibly terminated, but there is a potential risk of problems...

Appears when reading data:

Notice: Undefined index: name in...

Notice: Undefined index: key in…

The source code is as follows:

name = isset ( name = isset(name=isset(_POST[‘name’]) ? filter_input(INPUT_POST, ‘name’, FILTER_SANITIZE_SPECIAL_CHARS) :htmlspecialchars($_GET[‘name’]);
key = isset ( key = isset(key=isset(_POST[‘key’]) ? filter_input(INPUT_POST,‘key’,FILTER_SANITIZE_SPECIAL_CHARS) :htmlspecialchars($_GET[‘key’]);

Problem analysis:

Undefined index: refers to your code There is an error: "The variable is not defined yet, and it is used before assigning a value." This is not a fatal error and will not forcefully stop the running of your php code, but there is a potential risk of problems, so it is recommended to modify it~~~~

Solution:

Use error_reporting = E_ALL & ~E_NOTICE in php.ini to turn off the display of notice and shield such warnings. However, it is recommended to change the code. Better, the code should always be written in a more standardized way, and there will be fewer problems in the future.

Cause of the problem: It is because you only checked whether POST exists, but did not check whether _POST exists, but did not check whether POST existed, but did not check the existence of _GET.

Perfect solution: Just change it to the following:

n a m e = i s s e t ( name = isset(name=isset(_POST[‘name’]) ? filter_input(INPUT_POST,‘name’,FILTER_SANITIZE_SPECIAL_CHARS) : isset($_GET[‘name’]) ? filter_input(INPUT_POST,‘name’,FILTER_SANITIZE_SPECIAL_CHARS) : ‘’;
k e y = i s s e t ( key = isset(key=isset(_POST[‘key’]) ? filter_input(INPUT_POST,‘key’,FILTER_SANITIZE_SPECIAL_CHARS) : isset($_GET[‘key’]) ? filter_input(INPUT_POST,‘key’,FILTER_SANITIZE_SPECIAL_CHARS) : ‘’;

First of all, this is not an error, it is a warning. So if the server cannot be changed, each variable should be defined before use.

Method 1: Modify the server configuration

修改php.ini配置文件，error_reporting = E_ALL & ~E_NOTICE

Method 2: Initialize the variables and write them in a standardized way (it is more cumbersome because there are a large number of variables). But I haven’t found a good definition method yet, I hope you can give me some advice
Method 3:

每个文件头部加上：error_reporting(0); 如果不行，只有打开php.ini,找到display_errors，设置为display_errors = Off

No errors will be prompted in the future.
Method 4:

做判断：isset($_GET["page"])      if-else判断

Or add ''@'' to indicate that if there is an error or warning in this line, do not output it
For example: @$page=$_GET["page" ]
Method 5:

file1.php文件把$xx变量付一个值，用post传递给file2.php，

If file2.php does not have the definition of $xx and uses $yy=$xx directly; the system will report an error: "undifined variaable $xx", if file2.php The file starts with $xx="";, then the $xx value of file1.php cannot be passed!

file2.php里可以这样
if(!isset($xx)) $xx="";

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