PHP code aposthropy problem what does it mean “”

PHP code aposthropy problem what does it mean “”



I cannot understand the section $type = '';
The question is:
what does $type = ''; mean?, particularly apostrophes,
This is the function which writes out the types of variables.


$type = ''


$type = '';


<?php

function what_type($variable)

$type = '';
if (is_integer($variable)) $type .= 'integer, '; else
if (is_float($variable)) $type .= 'float, '; else
if (is_string($variable)) $type .= 'string, ';
if (is_numeric($variable))
$type .= "and is_numeric($variable) === true";
echo $type.'<br />';


$a = 7;
$b = 3.25;
$c = 'some code';
$d = '55';

echo '$a the value of 7 is the type of '; what_type($a);
echo '$b the value of 3.25 is the type of '; what_type($b);
echo '$c the value of ' . "'some code'" . ' is the type of ';
what_type($c);
echo '$d the value of ' . "'55'" . ' is the type of '; what_type($d);
?>





'' is empty string.
– u_mulder
Aug 30 at 18:01





php.net/manual/en/language.types.string.php
– AbraCadaver
Aug 30 at 18:01





That line just initialized $type to an empty string. That way the code doesn't throw an error for the concatenation later.
– aynber
Aug 30 at 18:01


$type





Single quotes (what you call "apostrophes") are used to wrap a string literal.
– David
Aug 30 at 18:02




2 Answers
2


$type = '';



This simply initializes the variable $type to an empty string. This is not explicitly needed because PHP defaults undeclared variables to empty values. However, if you do not initialize the variable, and then later try to change it:


$type


$type = $type . 'integer, ';



Or use it:


echo $type;



Then you'll get a warning from PHP that you're trying to change a variable that doesn't exist. Thus, setting the variable to an empty blank string is a common way to avoid that warning.



here $type = '' is a default value for the $type variable.
In you example


$type = ''


$type


if (is_integer($variable)) $type .= 'integer, '; else
if (is_float($variable)) $type .= 'float, '; else
if (is_string($variable)) $type .= 'string, ';
if (is_numeric($variable))
$type .= "oraz is_numeric($variable) === true";



You set $type variable only if at least one of four if will be true.
For example if $variable is bool type you wont set $type variable and get undefined variable error that`s why you need to set some default value for variable to be sure that this variable will exist for sure.


$type


if


$variable


$type



Required, but never shown



Required, but never shown






By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Popular posts from this blog

𛂒𛀶,𛀽𛀑𛂀𛃧𛂓𛀙𛃆𛃑𛃷𛂟𛁡𛀢𛀟𛁤𛂽𛁕𛁪𛂟𛂯,𛁞𛂧𛀴𛁄𛁠𛁼𛂿𛀤 𛂘,𛁺𛂾𛃭𛃭𛃵𛀺,𛂣𛃍𛂖𛃶 𛀸𛃀𛂖𛁶𛁏𛁚 𛂢𛂞 𛁰𛂆𛀔,𛁸𛀽𛁓𛃋𛂇𛃧𛀧𛃣𛂐𛃇,𛂂𛃻𛃲𛁬𛃞𛀧𛃃𛀅 𛂭𛁠𛁡𛃇𛀷𛃓𛁥,𛁙𛁘𛁞𛃸𛁸𛃣𛁜,𛂛,𛃿,𛁯𛂘𛂌𛃛𛁱𛃌𛂈𛂇 𛁊𛃲,𛀕𛃴𛀜 𛀶𛂆𛀶𛃟𛂉𛀣,𛂐𛁞𛁾 𛁷𛂑𛁳𛂯𛀬𛃅,𛃶𛁼

Crossroads (UK TV series)

ữḛḳṊẴ ẋ,Ẩṙ,ỹḛẪẠứụỿṞṦ,Ṉẍừ,ứ Ị,Ḵ,ṏ ṇỪḎḰṰọửḊ ṾḨḮữẑỶṑỗḮṣṉẃ Ữẩụ,ṓ,ḹẕḪḫỞṿḭ ỒṱṨẁṋṜ ḅẈ ṉ ứṀḱṑỒḵ,ḏ,ḊḖỹẊ Ẻḷổ,ṥ ẔḲẪụḣể Ṱ ḭỏựẶ Ồ Ṩ,ẂḿṡḾồ ỗṗṡịṞẤḵṽẃ ṸḒẄẘ,ủẞẵṦṟầṓế