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php3-list | 200003

Re: [PHP3] Variable substitution in sprintf() format string ?! From: NickM (nick <email protected>)
Date: 03/31/00

When you use single quotes ( ' ) it means to take the string literally,
therefore
    $s = '$row["x"]/%s';
literally means that string, no variable parsed. Use double quotes ( " ) to
pass a variable. BUT, as you can see that won't work because of the
associative array, this is what you do instead:
    $s = $row["x"]."string";

However, on top of all that I don't see why you put the sprintf formatting
in the variable, this is unreadable to other programmers and senseless, why
not just do this in the first place:
    $row = Array("x" => "teststring: ");
    $dummy = "also a test";
    echo sprintf("%s blabla\n%s blabla",$row["x"], $dummy);

Much more compact and neater.

Nick

----- Original Message -----
From: "Andreas Jung" <ajung <email protected>>
To: <php3 <email protected>>
Sent: Friday, March 31, 2000 1:43 PM
Subject: [PHP3] Variable substitution in sprintf() format string ?!

> Take a look the following code. I am trying to use a variable
> as format string for sprintf(). The string itsself contains
> some other variable that should be replaced within the sprintf()
> call however I are not replaced. I tried a lot with escaping $
> and quoting but without result. Maybe someone can help.
>
> <?
> $row = Array();
> $row["x"] = "teststring: ";
> $dummy = "also a test";
>
> $s = '$row["x"]/%s';
> $s1 = '$dummy/%s';
>
>
> echo sprintf("$s\n",'blabla');
> echo sprintf("$s1\n",'blabla');
> ?>
>
>
> Thanks,
> Andreas
>
>
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