CODEWITHSUNDEEP
phpstring
Lea Cohen
Lea Cohen

Reputation: 8200

String concatenation of algebric term

I would like to add two integers and concatenate their result to a string, that is:

Add 1 to $i, and concatenate that result to the string 'icon'. I thought the following syntax would work:

$x = 'icon'.$i+1;

However it doesn't do what I want - it keeps returning the value 'icon1', disregarding the value of $i.

What's the right way to do what I want?

Upvotes: 2

Views: 46

Answers (3)

Prasanth Bendra
Prasanth Bendra

Reputation: 32800

try this :

$temp = $i+1;
$x = 'icon'.$temp;

You are getting wrong answer because of "Operator Precedence",

Ref this link : http://php.net/manual/en/language.operators.precedence.php

Here see the line : left + - . arithmetic and string

. has more Precedence than +, So your expression will be like : $x = ('icon'.$i)+1;

To solve it either use the method i mentioned above or hsz answer

ie : $x = 'icon'.($i+1);

Upvotes: 1

Bjoern
Bjoern

Reputation: 16304

Operator Precedence explains why this is happening.

You can use brackets:

$x = 'icon'.($i+1);

This should do the job.

My test:

$i = 18;
$x = 'icon'.($i+1);
var_dump($x);

--> string(6) "icon19"

Upvotes: 1

hsz
hsz

Reputation: 152284

Try with:

$i = 0; // init $i
$x = 'icon'.($i+1);

If you want to regularly increment $i variable:

$x = 'icon'.(++$i);

Upvotes: 2

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