Hello JS Forum (again!)

I'm having some trouble working out a problem dealing with string manipulation. I have an input box, and it replaces whatever the user types with x's and 1's. Letters are turned to x's and numbers to 1's. Spaces are kept.

But I want to extend this script, and offer users to keep certain parts of the string intact. So if they type 'h4llo' it will be replaced to 'x1xxx', but if they type 'h[4]llo' it will be replaced to 'x[4]xxx'. I just need to know how to get the script to not replace characters in between the [ and ]'s. Here is my JS (one line)

return val.replace(/[a-zA-Z]/g,'x').replace(/[0-9]/g,'1').replace(/[^a-zA-Z 0-9 \[\]]+/g,'');

Currently I have the non-alphanumeric replacement excluding [ and ], so they aren't removed like any other characters like !@#$%^&*(). The PHP is what actually does something with the value, but I know PHP enough to tell it to work with just x's and leave everything else as it is.

Thanks!!

-PG

Let me first say that I like regex. However, if you write a Regex that is insanely hard to understand it will be insanely hard to modify later as well.

I think you should do what you've done but include other string manipulation techniques without the use of Regex. So basically mix Regex and standard string manipulation in order to create something that works and is easy to read.

I'm not sure what you mean, stbuchok. You want me to use a combination of regex and other kinds of string manipulation?

I suppose that could work... By splitting by [ and ] I could grab what is between those two array keys and not replace that... But I was hoping more of a regex-based answer; something I could literally 'pop' on the end of my already written regex.

Like this:

/[^!(\[|\])$]/g

Meaning 'Not [ or ]', but that doesn't work, or I'm just putting it in the wrong place (in a separate replace() after the others)

The ^ can be used as a not-operator, as in

[^0-9]

which will match everythign except digits 0-9. In your case, i believe this should do the trick.

[^\[.+\]]

which will match everything except any charcter(s) enclosed in brackets.
I havn't looked at your code at all, but I'm sure you can work it in there.

Thanks for the reply, lizciz! Here is my code:

JS:

function replace_chars(val) {
	return val.replace(/[a-zA-Z]/g,'x').replace(/[0-9]/g,'1').replace(/[^a-zA-Z 0-9 \[\]]+/g,'');
}

<input type="text" onchange="this.value=replace_chars(this.value)" />

So if you were to input 'hello' into this currently, it would replace it to 'xxxxx'. I tried putting your regex into it, but it ended up replacing the square brackets instead. Basically, I need it to make whatever that is inside the square brackets stay intact when it replaces.

I'm no regex expert myself, but I think you could use a replace with a callback function. So, first use my piece of code to find any sub string that is not in brackets, that part is sent to your callback function where you replace letters with x and numbers with 1.

Something like this (not tested):

function replace_chars(val) {
   return val.replace(/[^\[.+\]]/, your_callback_function);
}

function your_callback_function(match) {
   return match.replace(/[a-zA-Z]/g,'x').replace(/[0-9]/g,'1');
}

Also, I dont know about javascript, but php's *_replace functions can be called with array parameters instead of calling them twice. Perhaps that is also possible in javascript(?).

I'm sorry, lizciz, but that didn't work .

I'm determined to get this working. Fortunately I know PHP a little better than JS, so I wrote what I want to do in PHP:

PHP Code:

$str="[no] yes";
$parts=array_filter(preg_split('//',$str));
$first=strpos($str,"[")+1;
$last=strpos($str,"]")+1;
for($p=0;$p<=count($parts);$p++){
    if ($p>$first&&$p<$last){$output.=$parts[$p];}
    else {
        if(ctype_alnum($parts[$p])){$output.='x';}
        elseif(ctype_digit($parts[$p])){$output.='1';}
        else{$output.=$parts[$p];}
    }
}
echo $output; 


Is there a way to translate that to JS? I know about php.js, but I'd like to use JS-only functions. If there's no way to do it, I'll just stick an AJAX call to do it for me xD

Well, if you're gonna do it without regex any way, perhaps this will work?
Again, not tested

function replace_chars(val) {
   for (i = 0; i < val.length; i++) {
      if (val.charAt(i) == '[') {
         while (val.charAt(++i) != ']');
      } elseif (val.charAt(i) == parseInt(val.charAt(i))) {
         setCharAt(val, i, '1');
      } else {
         setCharAt(val, i, 'x');
      }
   }
}

function setCharAt(str,index,chr) {
   if(index > str.length-1) return str;
   return str.substr(0,index) + chr + str.substr(index+1);
}

Oh, and the setCharAt() function is not my code, I copied it from here.

Thanks lizciz! Unfortunately, that doens't seem to work, either (I changed elseif to else if, too). It returns 'undefined'.

I'll whip up an AJAX query, and that'll be the end of my problem Thanks for your help, though!

Well, i forgot to actually return anything
You can try to add "return val", or stick with your ajax option

Are the chars you want to keep between the [ and ] numeric only or alphanumeric?

They are alphanumeric (a-zA-Z 0-9)

Just for the record, I've now fixed my previous errors and actually tested it

function replace_chars(val) {
   for (i = 0; i < val.length; i++) {
      if (val.charAt(i) == '[') {
         while (val.charAt(++i) != ']');
      } else if (val.charAt(i) == parseInt(val.charAt(i))) {
         val = setCharAt(val, i, '1');
      } else {
         val = setCharAt(val, i, 'x');
      }
   }
   return val;
}

function setCharAt(str,index,chr) {
   if(index > str.length-1) return str;
   return str.substr(0,index) + chr + str.substr(index+1);
}

It works!!!! Fantastic! Thank you lizciz