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convert 01, 02, 03, 04 .. 09 to 1,2,3,4 ... 9

Answer 1

+12 A:

I assume the input is a string?

$str = "01";
$anInt = intval($str);

You may think that the leading 0 would mean this is interpreted as octal, as in many other languages/APIs. However the second argument to intval is a base. The default value for this is 10. This means 09->9. See the first comment at the intval page, which states that the base deduction you might expect only happens if you pass 0 in as the base.

Doug T. 2009-12-04 15:09:06

If input is treated as an int, it will be handled as octal. (so 09->0)

erenon 2009-12-04 15:13:12

@erenon, see my update.

Doug T. 2009-12-04 15:17:59

Answer 2

A:

$str = "05";
$last = $str[1];

Fortega 2009-12-04 15:09:33

the intval() solution is more reliable in this case, as it can catch "009" as well.

Pekka 2009-12-04 15:10:57

true. But my solution is faster, and the TS didn't say anything about the 009 :-)

Fortega 2009-12-04 15:12:45

Answer 3

+1 A:

$old_nums = array("01","02","03");

foreach($old_nums as $k => $v) {

ltrim($v,"0");

}

David Thomas 2009-12-04 15:10:05

that's overkill

Gregory Pakosz 2009-12-04 15:11:31

That depends entirely on where the numbers are coming from, and how they're stored, doesn't it? In this instance I offered this only as an alternative to Doug T.'s -already submitted and clearly accurate and efficient- answer.

David Thomas 2009-12-04 16:09:08

Answer 4

A:

$i = substr($input, 1, 1);

erenon 2009-12-04 15:11:36

Answer 5

+1 A:

$x="01";
$x=+$x;

$x="02";
$x=+$x;

...

or

$x=+"01";

should work for both int, and string

S.Mark 2009-12-04 15:24:14

Answer 6

A:

If you want to use the generic regular expression solution: 's/[0]([0-9])/\1/' (add in anchors as appropriate)

Vitali 2009-12-04 16:33:49

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convert 01, 02, 03, 04 .. 09 to 1,2,3,4 ... 9

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