Iterating over numbers with leading zeroes with a `for` loop

Question

I'm working on a calendar site and need to change the following code to show 0 before the value of $list_day if it's a single digit number.

for($list_day = 1; $list_day <= $days_in_month; $list_day++):

How should I go about doing this?

Leading "0"s mean nothing to a *number*: they are simply an artifact of a given *text representation* (it can mean octal or just be padding, as in the post). — , Aug 04 '11 at 18:16

score 7 · Answer 1 · answered Aug 04 '11 at 18:14

7

When you're printing it, use sprintf("%02d", $list_day) to pad it with a zero.

answered Aug 04 '11 at 18:14

hughes

Don't need to print the value but this is great to know for later! – Philip Kirkbride Aug 04 '11 at 18:19
Oh ok... this doesn't actually print it though! Whatever you need it for, running it through `sprintf(...)` appends the `0`. – hughes Aug 04 '11 at 18:26
@Philip - If you're not printing it somewhere, the leading zero makes no difference – Brad Mace Aug 04 '11 at 18:28
@bemace he could conceivably be passing it to some other function that expects a zero-padded string. – hughes Aug 04 '11 at 18:31
@PhilipK sprintf don't prints, it returns string – RiaD Aug 04 '11 at 18:40

Pepe · Accepted Answer · 2011-08-04T18:16:42.937

2

if ($list_day < 10)
  echo "0" . $list_day;
else
  echo $list_day;

edited Aug 04 '11 at 18:16

answered Aug 04 '11 at 18:13

Pepe

use string concatenation '.' instead adding '+' – Jacek Kaniuk Aug 04 '11 at 18:15
@RiaD Thanks for taking care of that – Pepe Aug 04 '11 at 18:16
@Juhana Maaaaaaan ... I've been writing in VB6 for 3 days and I'm already missing ; :( – Pepe Aug 04 '11 at 18:17
Thanks I can mode this for what I need. Seems so obvious now! – Philip Kirkbride Aug 04 '11 at 18:19

score 1 · Answer 3 · answered Aug 04 '11 at 18:14

1

printf("%02d",$listday);


printf("%02d",3);  //prints 03

answered Aug 04 '11 at 18:14

RiaD

score 1 · Answer 4 · answered Aug 04 '11 at 18:15

1

echo sprintf('%02d',$list_day);

answered Aug 04 '11 at 18:15

Timur

`echo sprintf()` is an "antipattern". There is absolutely no reason that anyone should ever write `echo sprintf()` in any code for any reason -- it should be `printf()` every time. – mickmackusa Apr 08 '22 at 04:23

Brad Mace · Answer 5 · 2011-08-04T18:35:11.233

1

You could do this as

for ($list_day = "01"; $list_day <= $days_in_month; $list_day = sprintf("%02d", $list_day + 1))

but it's usually done like

for ($list_day = 1; $list_day <= $days_in_month; $list_day++) {
    print "<div>" . sprintf("%02d", $list_day) . "...</div>";
}

edited Aug 04 '11 at 18:35

answered Aug 04 '11 at 18:18

Brad Mace

In the first method you'd be incrementing a string after the first loop. PHP might just interpret `"02"` as the integer `2` but I wouldn't count on it. – JJJ Aug 04 '11 at 18:21
@Juhana - String concatenation is done with `.` in PHP; when using `+` it autoconverts strings to integers correctly. – Brad Mace Aug 04 '11 at 18:24
That's what I thought :) You should start with $list_day = "01" then, though. – JJJ Aug 04 '11 at 18:33
Instead of calling `sprintf()` inside of `print`, just call `printf()` and write all of the literal strings inside the first parameter of `printf()`. – mickmackusa Apr 08 '22 at 04:24

Jacek Kaniuk · Answer 6 · 2011-08-04T18:22:21.210

0

if($list_day < 10)
    $new_day = '0'.$list_day;
else
    $new_day = $list_day;

edited Aug 04 '11 at 18:22

answered Aug 04 '11 at 18:14

Jacek Kaniuk

1

This will probably mess up the loop. It's better not to touch the original variable. – JJJ Aug 04 '11 at 18:16

6 Answers6