preg_replace and literal $

[Justin Koivisto]

I am replacing a string in a text block that has a literal $ in it, and
preg_replace is seeing it as a backreference. Here is what I am using:

foreach($price_ lists as $list)
$x=preg_replace ('/--PRICE-LIST--/',$list,$x,1);

OK, so what this does it is takes each array element and replaces only
the first occurrance of "--PRICE-LIST--" with it. I would have used
str_replace, but I didn't think it should be necessary to create an
array with the same number of elements as $price_list just for a simple
thing like this.

I did, hoever try:
$x=str_replace( '--PRICE-LIST--',$price_lists, $x);

Since the manual page didn't say anything about that case to no avail.

So right now I have the following:
foreach($price_ lists as $list)
$x=preg_replace ('/--PRICE-LIST--/',str_replace(' $','\$',$list), $x,1);

Which compensates for this. However, that is just one case, I cane
forsee others showing up. Does anyone know of a way to make preg_replace
behave as I intend it to? Or does someone have a function for regex-safe
string encoding?

TIA!

--
Justin Koivisto - spam@koivi.com
PHP POSTERS: Please use comp.lang.php for PHP related questions,
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[Steven Vasilogianis]

Re: preg_replace and literal $

Justin Koivisto <spam@koivi.com > writes:
[color=blue]
> So right now I have the following:
> foreach($price_ lists as $list)
> $x=preg_replace ('/--PRICE-LIST--/',str_replace(' $','\$',$list), $x,1);
>
> Which compensates for this. However, that is just one case, I cane forsee
> others showing up. Does anyone know of a way to make preg_replace behave as
> I intend it to? Or does someone have a function for regex-safe string
> encoding?[/color]

Change it to:
foreach($price_ lists as $list)
$x=preg_replace ('/--PRICE-LIST--/', preg_quote($lis t), $x, 1);

And of course, see <http://us2.php.net/preg_quote>

HTH,
--
steven vasilogianis

[Justin Koivisto]

Re: preg_replace and literal $

Steven Vasilogianis wrote:
[color=blue]
> Justin Koivisto <spam@koivi.com > writes:
>
>[color=green]
>>So right now I have the following:
>>foreach($pric e_lists as $list)
>> $x=preg_replace ('/--PRICE-LIST--/',str_replace(' $','\$',$list), $x,1);
>>
>>Which compensates for this. However, that is just one case, I cane forsee
>>others showing up. Does anyone know of a way to make preg_replace behave as
>>I intend it to? Or does someone have a function for regex-safe string
>>encoding?[/color]
>
>
> Change it to:
> foreach($price_ lists as $list)
> $x=preg_replace ('/--PRICE-LIST--/', preg_quote($lis t), $x, 1);[/color]

Thanks! I'm surprized that I didn't see that one.

--
Justin Koivisto - spam@koivi.com
PHP POSTERS: Please use comp.lang.php for PHP related questions,
alt.php* groups are not recommended.

[Justin Koivisto]

Re: preg_replace and literal $

Steven Vasilogianis wrote:
[color=blue]
> Justin Koivisto <spam@koivi.com > writes:
>
>[color=green]
>>So right now I have the following:
>>foreach($pric e_lists as $list)
>> $x=preg_replace ('/--PRICE-LIST--/',str_replace(' $','\$',$list), $x,1);
>>
>>Which compensates for this. However, that is just one case, I cane forsee
>>others showing up. Does anyone know of a way to make preg_replace behave as
>>I intend it to? Or does someone have a function for regex-safe string
>>encoding?[/color]
>
> Change it to:
> foreach($price_ lists as $list)
> $x=preg_replace ('/--PRICE-LIST--/', preg_quote($lis t), $x, 1);
>
> And of course, see <http://us2.php.net/preg_quote>[/color]

Unfortunately, that didn't work for my application because there is HTML
in it, and it escaped all the < and >. However, After reviewing the
manual page, it looks like I should be OK with only escaping the $ in
the strings.

--
Justin Koivisto - spam@koivi.com
PHP POSTERS: Please use comp.lang.php for PHP related questions,
alt.php* groups are not recommended.