What are the four common error types in PHP?

In PHP, errors may occur due to grammatical or logical problems, so PHP errors can be divided into several types.

There are various types of errors in PHP, but it can basically be divided into four main types of errors. Let’s take a look at these four main types of errors:

Parse errors or syntax errors: (Recommended learning: PHP video tutorial)

This is the type of error the programmer makes in the source code of the program. Syntax errors are caught by the compiler. After fixing the syntax errors, the compiler compiles the code and executes it. Parsing errors can result from undisclosed quotes, missing or extra parentheses, unclosed braces, missing semicolons, and more.

For example:

<?php
$x = "php中文网";
y = "www.php.cn";
echo $x;
echo $y;
?>

Explanation: In the above program, the $ symbol is missing in line 3, so an error message will be given.

Fatal Error:

This is the type of error where the PHP compiler understands the PHP code but it identifies an undeclared function. This means calling a function without a function definition.

For example:

<?php 
function add($x, $y) { 
    $sum = $x + $y; 
    echo "sum = " . $sum; 
} 
$x = 0; 
$y = 20; 
add($x, $y); 
   
diff($x, $y); 
?>

Explanation: In line 10, the function diff() is called, but the function diff() does not have a declaration definition, so it will give An error occurred.

Warning error:

The main reason why warning error occurs is that the missing file is included, which also means that the PHP function calls the missing file.

Such as:

<?php  
header("content-type:text/html;charset=utf-8");
$x = "PHP中文网"; 
include ("header.php"); 
echo $x . "，网址为：www.php.cn"; 
?>

Description: This program calls an undefined file header.php that is not available, so it will generate an error.

NOTE Error:

It is similar to a warning error, which means that the program contains an error, but it allows execution of the script.

Such as:

<?php  
header("content-type:text/html;charset=utf-8");
$x = "PHP中文网"; 
echo $x; 
echo $y; 
?>

Description: This program uses an undeclared variable $y, so it gives an error message.

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