#21. Kevin on 18 April 2008

@ Jonas Raoni: That's astonishing :D The non-overlapping approach was convincing enough for me not to take the effort to test if this was the actual mechanism of PHP. It would make sense though right? Anyway, I will reinstate your version Jonas Raoni, and adjust the test case. Thanks for all of your work! You too Peter, some functions come easy, some tend to be a little of a pain. But that's the challenge & learning experience, right.

#20. Philip on 18 April 2008

That is incredibly weird... I was certain that PHP replaced all the stuff at once so as not to create overlap... ah well, I guess it's not like that after all. It was a cool function though :P

#19. Jonas Raoni on 18 April 2008

I didn't looked the comments before, I just saw the code was long and made a version :)~

But it's strange... What php version are you using? The following code, on php 5, gave me as output "hemmo, mars":

echo str_replace(array('{name}', 'l'), array('hello', 'm'), '{name}, lars');

#18. Philip Peterson on 18 April 2008

Woah, I didn't see comments #16 or #15... hmm, I'm surprised, it definitely has merit, though...

#17. Philip Peterson on 18 April 2008

Wow... bravo is all I can say... that's more than 20x faster than my code, too, lol...

#16. Kevin on 17 April 2008

@ Jonas Raoni: I've updated, committed, generated, deployed, tested, reverted, commited, generated & deployed ;) And srry dude but your str_replace did not survive Peter's testcase:

// should return: hello, mars
str_replace(['{name}', 'l'], ['hello', 'm'], '{name}, lars');

More info in the comments here. Guess I'll be needing those medals back ;) nah just kidding :)

#15. Kevin on 17 April 2008

@ Jonas Raoni: That looks very promising :) I still have to test it but if you have reduced our function to this, preserving all of it's functionality, then you deserve 3 medals ;)

#14. Jonas Raoni on 17 April 2008

I just made this code, it's using the already present is_array function.

function str_replace(f, r, s){
var ra = is_array(r), sa = is_array(s), f = [].concat(f), r = [].concat(r), i = (s = [].concat(s)).length;
while(j = 0, i--)
... [more] while(s[i] = s[i].split(f[j]).join(ra ? r[j] || "" : r[0]), ++j in f);
return sa ? s : s[0];
}

Example:

alert(str_replace(["a", "fb", "c"], ["f", "g"], ["abcd"]));

#13. Kevin on 15 April 2008

@ Philip Peterson: If you like you can now do this in SVN ;)

#12. Philip Peterson on 15 April 2008

Hmm, I actually can see where that could come in useful though...

nsub = nsub.replace(new RegExp(__regexp_escape(search[x]), "g"), "[cod"+fincods[x]+"]");
 
could work with
 
nsub = nsub.split(search[x]).join("[cod"+fincods[x]+"]");

and that sort of thing occurs in four places, so there's a significant drop in size, and you probably can even drop that __regexp_escape function [sorry Simon :(]. You probably should keep him in the credits though, even if you take the function out.

#11. Kevin on 13 April 2008

@ Jonas Raoni: Hi there, how have you been? Thats a brutal approach :) However the long code we have now supports some features that your approach does not. For more information see the comments here. Thanks for this insightful way though!

#10. Jonas Raoni on 12 April 2008

For the str_replace the best and fastest method I know it's:

"Jonas Raoni".split("a").join("X");

#9. Kevin on 31 March 2008

@ Philip: Cool, I've included Simon as well. Thanks again.

#8. Philip on 31 March 2008

No problem!

Also: Credits to Simon Willison (http://simonwillison.net/2006/Jan/20/escape/) for the regexp_escape function.

#7. Kevin on 31 March 2008

@ Philip: A seemingly straightforward function like str_replace turns out to be more complex than e.g. mktime :) I've indented your code and included the extra function (had to 'var' it for namespacing compatibility) in str_replace.

Thanks A LOT for seriously improving this function!

#6. Philip on 31 March 2008

Not necessarily; what if a result produced by the replacement of one of the shorter strings produced the values of one of the longer searches? Here's a code that works, but I haven't indented the part that I made yet, or commented it, and it requires a second external function... I dunno if you can fix that or not.

// http://kevin.vanzonneveld.net
    // +   original by: Kevin van Zonneveld (http://kevin.vanzonneveld.net)
    // +   improved by: Gabriel Paderni
    // +   improved by: Philip Peterson
    // *     example 1: str_replace(' ', '.', 'Kevin van Zonneveld');
    // *     returns 1: 'Kevin.van.Zonneveld'
 
function regexp_escape(text) {
  if (!arguments.callee.sRE) {
    var specials = [
      '/', '.', '*', '+', '?', '|',
      '(', ')', '[', ']', '{', '}', '\\'
    ];
    arguments.callee.sRE = new RegExp(
      '(\\' + specials.join('|\\') + ')', 'g'
    );
  }
  return text.replace(arguments.callee.sRE, '\\$1');
}
 
function str_replace(search, replace, subject)
{
 
    if(!(replace instanceof Array)){
        replace=new Array(replace);
        if(search instanceof Array){//If search    is an array and replace    is a string, then 
 
this replacement string is used for every value of search
            while(search.length>replace.length){
                replace[replace.length]=replace[0];
            }
        }
    }
 
    if(!(search instanceof Array))search=new Array(search);
    while(search.length>replace.length){//If replace    has fewer values than search , then an 
 
empty string is used for the rest of replacement values
        replace[replace.length]='';
    }
 
    if(subject instanceof Array){//If subject is an array, then the search and replace is 
 
performed with every entry of subject , and the return value is an array as well.
        for(k in subject){
            subject[k]=str_replace(search,replace,subject[k]);
        }
        return subject;
    }
 
 
 
 
numreplx=search.length;
numon=0;
fincods=new Array();
while(fincods.length<numreplx)
{
nsub=subject;
for(x=0;x<fincods.length;x++)
{
nsub=nsub.replace(new RegExp(regexp_escape(search[x]), "g"), "[cod"+fincods[x]+"]");
}
for(x=0;x<fincods.length;x++)
{
nsub=nsub.replace(new RegExp(regexp_escape("[cod"+fincods[x]+"]"), "g"), replace[x]);
}
if(nsub.indexOf("[cod"+numon+"]") == -1)
{
fincods[fincods.length]=numon;
}
numon++;
}
for(x=0;x<fincods.length;x++)
{
subject=subject.replace(new RegExp(regexp_escape(search[x]), "g"), "[cod"+fincods[x]+"]");
}
for(x=0;x<fincods.length;x++)
{
subject=subject.replace(new RegExp(regexp_escape("[cod"+fincods[x]+"]"), "g"), replace[x]);
}
return subject;
}

#5. Kevin on 31 March 2008

@ Philip: Thanks for noticing. My first thought is to order the search array by length, ascending. This way 'l' will be searched & replaced before hello will be.
Do you agree that that would solve the issue?

#4. Philip on 30 March 2008

There's only a slight problem with this code... in PHP, each entry is replaced all at once, rather than one after another. This creates a problem in the following example:

function parseInf(kyl){
 
kyl=str_replace([
"{name}",
"l"
],
[
"hello",
"m"
],
kyl);
 
 
nev=kyl.split(/\r{0,1}\n-\r{0,1}\n/);
 
 
return nev;
 
 
}
 
alert(parseInf("{name}, lars"));

Theoretically, the code should return "hello, mars", but instead it returns "hemmo, mars". It completes the first iteration before the second, rather than all at once.

#3. Kevin on 02 March 2008

@ G.Paderni: Good point! Thank you for contributing!

#2. G.Paderni on 02 March 2008

oh, sorry there's an error please remove my last comment this one is ok:

function str_replace(search,replace,subject) {
  // by: Gabriel Paderni
 
  if(!(replace instanceof Array)){
    replace=new Array(replace);
    if(search instanceof Array){//If search  is an array and replace  is a string, then this replacement string is used for every value of search
      while(search.length>replace.length){
        replace[replace.length]=replace[0];
      }
    }
  }
 
  if(!(search instanceof Array))search=new Array(search);
  while(search.length>replace.length){//If replace  has fewer values than search , then an empty string is used for the rest of replacement values
    replace[replace.length]='';
  }
 
  if(subject instanceof Array){//If subject is an array, then the search and replace is performed with every entry of subject , and the return value is an array as well.
    for(k in subject){
      subject[k]=str_replace(search,replace,subject[k]);
    }
    return subject;
  }
  
  for(var k=0; k<search.length; k++){
    var i = subject.indexOf(search[k]);
    while(i>-1){
      subject = subject.replace(search[k], replace[k]);
      i = subject.indexOf(search[k],i);
    }
  }
 
  return subject;
}

#1. sbsweb on 19 February 2008

Increible, ese .js es de lo más útil que he visto.