like this:
$digit = 122;
$digits = sprintf ("%08d", $digit);
print $digits;
Greets,
Edward
----- Original Message -----
From: "Jeff Sheltren" <[EMAIL PROTECTED]>
To: "Andy" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Saturday, February 02, 2002 6:41 PM
Subject: Re: [PHP] Adding 6 digits to a str?
> Why is it that you want to represent an int with leading 0's? Is it just
> to print it out that way? If that is the case, then you can use the
printf
> function for formatted printing.
>
>
> Jeff
>
> At 10:53 AM 2/2/2002 +0100, Andy wrote:
> >Hi guys,
> >
> >I am trying to force a int to be 8 digits. If it is only 3 dig filling
the
> >first ones with 0.
> >
> >Anyhow I tryed it with type casting, but it does not matter what I try, I
> >always get an addition.
> >
> >E.g:
> > $member_id is: 136
> > should be: 00000136
> >
> >Here is the code, which is still returning 136:
> >
> > for($i=0;$i<count($member_id);$i++){
> > $length = strlen($member_id[$i]);
> >
> > settype ($member_id[$i], "string");
> > $zero = '0';
> > settype ($zero, "string");
> >
> > $member_id[$i] = $zero + $member_id[$i];
> > echo $member_id[$i];
> > }
> >
> >Does anybody know how to solve this thing??
> >
> >Thanx Andy
>
>
>
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