Debugging in PHP or Debugging in gernal

Question

I'm experienced in Javascript, the way i debug now is to console.log(vars), so the way i debug is oberserving my vars in the program, if they contain sth.,an object etc. what sometime also helps it's the error log in the browser, but this is more a side effect, this approach works for me in javascript, but since a few months i'm trying to debug PHP with it, and made little to no progress, it seems quite difficult to set up PHP for this, and there are log-files in general ...

Question: Is watching the values of variables not the way how you debug PHP? and if not, what approach is the right one?

You can use var_dump($your_var) in php for oberserving variables values — jiali
– jiali, Commented Aug 17, 2021 at 23:36

Tangentially Perpendicular · Accepted Answer · 2021-08-17 23:44:06Z

2

There are tools like xDebug that can help debugging PHP, although I've never had a lot of luck with it beyond its profiling capabilities.

Place echo and var_dump() statements in appropriate places in the code and watch the output.

You can include error_log() statements to log details to a dedicated log file that you can examine.

And there's the server error log.

That's what I use - others may have more.

answered Aug 17, 2021 at 23:44

Tangentially Perpendicular

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2 Comments

Maksim Over a year ago

And the xDebug is only way to debug. All other ways such var_dump(), dd(), etc - is not debug.

Tangentially Perpendicular Over a year ago

@Maksim you must have a frightfully narrow definition of the term 'debug'

Gonzalo Rodriguez · Accepted Answer · 2021-08-17 23:45:29Z

If you want to debug variable content in php there are several ways to do it You could use var_dump () To fully show the content of the variable (object, class or whatever this variable is). If you want to debug for errors you can active the error display config on .htaccess, .user.ini or using php comands to active the error display options.

E.g. If you are using a local PHP server, in most cases the error display options are active for default.

So If you just want to debug a variable conten, like $_POST, you can use

 var_dump($_POST);

You cant prevent error using isset() for variables that you dont know if have any value.

if ( isset($_POST) ) var_dump($variable_name);

you can try this code

URL: http://localhost/vardump.php?test=this%20is%20a%20variable%20content%20test&value=123

echo 'This is a $_POST debug';
var_dump( $_POST );

echo 'This is a $_GET debug'; 
var_dump( $_GET );

echo 'This is a non setted variable';
var_dump( $variable_name );

result

    This is a $_POST debug

D:\wamp64\www\vardump.php:4:
array (size=0)
  empty

This is a $_GET debug

D:\wamp64\www\vardump.php:7:
array (size=2)
  'test' => string 'this is a variable content test' (length=31)
  'value' => string '123' (length=3)

This is a non setted variable
( ! ) Notice: Undefined variable: variable_name in D:\wamp64\www\vardump.php on line 10
Call Stack
#   Time    Memory  Function    Location
1   0.0002  407224  {main}( )   ...\vardump.php:0

D:\wamp64\www\vardump.php:10:null

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