PHP converting variable string to int

Question

To start, I could not find this answer online because of the way my variable string is defined. Normally I should be able to add 0 to the variable, or use (int), but it does not work.

<?php

$casestringid = "'118'";
$caseid = $casestringid + 0;

echo $casestringid;

echo $caseid;

?>

Output: '118'0

As you can see, because of the way my first variable is declared, the standard methods of converting a string to an integer does not work. My $casestringid is written like that because it requests a number from another page. Rather than trying to change how to format that, I figure it will be easier for help on how to convert a string that looks like that, into an integer. I would like the output of caseid to be 118. Thanks for any help in advance.

Replace (remove) the single quotes first then type cast as integer? — ɴᴀᴛᴇ ᴅᴇᴠ
– ɴᴀᴛᴇ ᴅᴇᴠ, Commented Aug 6, 2013 at 18:38

federicot · Accepted Answer · 2013-08-06 18:40:03Z

3

The problem is that '118' is not an integer as far as the PHP parser is concerned, it's a string. It looks like an integer to us, of course, but it has slashes (') which make it "unconvertible".

Use str_replace for this:

intval(str_replace("'", '', $casestringid));

answered Aug 6, 2013 at 18:40

federicot

12.3k19 gold badges71 silver badges111 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

donald123 Over a year ago

typecasting (int) str_replace(...) is performanter then intval()

federicot Over a year ago

@donald123 That's true. In this case we are using intval with base 10, so it's better to use int casting. 0 + str_replace(...) would be even faster.

steven · Accepted Answer · 2013-08-06 18:38:54Z

1

i think you have no other chance like this:

intval(str_replace("'",'',$casestringid));

answered Aug 6, 2013 at 18:38

steven

4,8952 gold badges31 silver badges61 bronze badges

2 Comments

user2562568 Over a year ago

someone said the exact same thing a minute before you. Thanks though it is right!

user2562568 Over a year ago

wait nvm you said it before

Zaziki · Accepted Answer · 2013-08-06 18:39:34Z

1

Replace the '':

intval(str_replace("'",'',$casestringid));

answered Aug 6, 2013 at 18:39

Zaziki

4182 silver badges12 bronze badges

Comments

Nerius · Accepted Answer · 2013-08-06 18:43:20Z

1

Try intval ($casestringid) + 0.

EDIT:

How about this, then:

filter_var ($casestringid, FILTER_SANITIZE_NUMBER_INT);

edited Aug 6, 2013 at 18:43

answered Aug 6, 2013 at 18:36

Nerius

9806 silver badges17 bronze badges

5 Comments

MightyPork Over a year ago

I dont think intval will digest quotes.

steven Over a year ago

i think intval would result in zero.

Nerius Over a year ago

You are mistaken, one of the many curiosities of PHP is the way it handles string-to-int conversion. It will basically take the largest substring of digits starting at index 0 it can find, and discard the rest, then convert those to int. For instance, saying intval('51 pigs') will return 51. EDIT: Never mind, I just noticed the ' at index 0.

user2562568 Over a year ago

I'm still getting the same result. $caseid is still 0

robz228 Over a year ago

@Thanix using your example, it would be intval("'51 pigs'") which will return 0

theftprevention · Accepted Answer · 2013-08-06 18:38:49Z

0

You have to remove the single quotes and use intval().

<?php

$casestringid = "'118'";
$parseid = str_replace("'", "", $casestringid);
$caseid = intval($parseid);

echo $casestringid;

echo $caseid;

?>

answered Aug 6, 2013 at 18:38

theftprevention

5,2433 gold badges21 silver badges32 bronze badges

Comments

robz228 · Accepted Answer · 2013-08-06 18:39:47Z

0

$casestringid = "'118'";
$int = str_replace("'", "", $casestringid);
echo intval($int);

answered Aug 6, 2013 at 18:39

robz228

6401 gold badge4 silver badges11 bronze badges

1 Comment

Cfreak Over a year ago

should still use intval() even after the string is replaced

Shawn G. · Accepted Answer · 2013-08-06 18:51:52Z

0

If it is just an integer you are looking for, this could work. it will remove any non digit characters then return it as an int

function parseInt( $s ){
    return (int)(preg_replace( '~\D+~' , '' , $s ));
}

answered Aug 6, 2013 at 18:51

Shawn G.

6224 silver badges16 bronze badges

Collectives™ on Stack Overflow

PHP converting variable string to int

7 Answers 7

2 Comments

2 Comments

Comments

5 Comments

Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

7 Answers 7

2 Comments

2 Comments

Comments

5 Comments

Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related