#37. Kevin on 21 February 2009

@ avel: Be sure to checkout http://phpjs.org, I'm making some progress there.

#36. avel on 20 February 2009

Thanks a lot for this blog. Is really usefull.

i'm from Barcelona and i'm a baby php programer and this php.js library look like so good.

note: my website is nearly in net.

#35. jonze on 30 January 2009

hello and congratulations for this article(and script also). I needed something like str_replace from php very fast done with javscript and seems that your script does that very well!

Regards!

#34. Brett Zamir on 22 January 2009

acimeha, yes, using the native JavaScript replace works fine for most cases, but this project aims to support PHP's behavior, which also allows arrays as inputs.

#33. acimeha on 22 January 2009

hi there !
consider this as solution for str_replace

str = str.replace(new RegExp("\n\r", 'gi'), "\n");

... [more]
best regards

#32. Kevin on 01 December 2008

@ Onno Marsman & T. Wild: It's not my preferred coding style either, we may need to refactor it some day so it will be more consistent to other code found in php.js

#31. Onno Marsman on 27 November 2008

No trouble at all! I guess you've misunderstood me. Like I said I also don't like that whole shorthand stuff. It's very hard to read indeed.

#30. T.Wild on 26 November 2008

IN REPLY TO Onno Marsman.
As I said 'I'm not the best JavaScript programmer in the world.' and what I submitted worked for me (well, almost, r+='' should have been r = r+'' etc) in my limited tests, and I was having trouble seeing what the original code was doing anyway - I don't particularly like the shorthand ways of writing ifs etc - that's why I converted s,r and f to strings to be sure.

Your code fix works fine and sorry for the trouble.

#29. Onno Marsman on 26 November 2008

To be clear: I've only changed the s[i] to (s[i]+'') in that line.

#28. Onno Marsman on 26 November 2008

T.Wild: I don't have time for testing right now but doesn't the following do the trick a bit better?
When you do pass an array you'd also want every element to be a string.
I don't see any need to convert r to string(s), but I could be wrong. I find the code a bit hard to read.
Why the whole variable name replacement thing anyway? I thought we have a compiler to do this. This only results in more code after compilation.

while (s[i] = (s[i]+'').split(f[j]).join(ra ? r[j] || "" : r[0]), ++j in f){};

#27. T.Wild on 26 November 2008

OOPS, I leaped before I looked at the code properly, this may be a better idea but I'm not the best JavaScript programmer in the world.

var f = search, r = replace, s = subject;
 
    var ra = r instanceof Array, sa = s instanceof Array, f = [].concat(f), r = [].concat(r), i = (s = [].concat(s)).length;
 
    if(!ra){r+='';}
    if(!sa){s+='';}
    if(!(f instanceof Array)){f+='';}

#26. T.Wild on 26 November 2008

I think the following line should should probably be changed from:

var f = search, r = replace, s = subject;

TO

var f = search+'', r = replace+'', s = subject+'';

Found this problem (if anyone is interested) when trying:

str_replace("foo","baa",location);
//You get the same problem with
str_replace(34,9,123456);

#25. Kevin on 29 September 2008

@ Onno & Anton: Both excellent points. Function updated. Thank you!

#24. Onno Marsman on 25 September 2008

Why be dependant on is_array when you could just use "instanceof Array" ? This probably saves a bit of performance too.

#23. Anton Ongsono on 25 September 2008

i have try to use this, but i five me error in FF 2.0.0.16 :

s[i] has no properties
[Break on this error] while (s[i] = s[i].split(f[j])....in(ra ? r[j] || "" : r[0]), ++j in f){};

so that i tried to fixed like this :

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

#22. Anton Ongsono on 25 September 2008

i have try to use this, but i five me error in FF 2 :

s[i] has no properties
[Break on this error] while (s[i] = s[i].split(f[j])....in(ra ? r[j] || "" : r[0]), ++j in f){};

so that i tried to fixed like this :

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

#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.