Php String to Integer conversion gives 0 as result

Question

When I try to convert a number as a string to an integer, The value changes to 0.

I did a preg_match and picked out a number value which as an output of preg_match is in a string datatype.

That is, When i tried to extract the number 7123 in the following xml data, <Count> 7123 </Count> using preg_match, I got the value 7123 as a string. [Used gettype function]. When I am checking if its 0 or anyother value, I need to convert it to integer datatype just to compare the values farely. So i do intval("7123") or intval($count) to get the integer value. But when I echo the value, its 0. Read few blogs online and one said this type of a thing happens when the string value is non-printable. Can someone help me understand this and help me solve it?

Community · Accepted Answer · 2017-05-23 12:27:25Z

2

It's very simple once you know how to convert a string to integer:

$str = "7123";
$val = (int) $str;

echo $val + 1;

result: 7124

This is not the only way to convert strings to integers, but it's been proven to be the fastest.

edited May 23, 2017 at 12:27

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answered Jun 5, 2014 at 18:26

l'L'l

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16 Comments

Vignesh Over a year ago

Nope. It still gives me the same 0 as result

l'L'l Over a year ago

Change your regex pattern to this: preg_match("/(\d+).+\/Count>/i",$handle,$match1);

l'L'l Over a year ago

Okay, I see what is going on. Are you only trying to capture the first <count> value? also, post your code in your question next time.

l'L'l Over a year ago

Change your regex to this: preg_match_all("/<Count\D+(\d+)\D+\/Count>/i",$handle,$match1); — It will grab all values in an array, but you can use it just as you did before, with one slight change $totalcount=$match1[1][0];

l'L'l Over a year ago

You're missing a semi-colon after echo $match1[$i]; but since it's an array you need print_r($match1[$i]); you can also do print_r($match1[1]); without the loop, it gives the same result.

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