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I'm looking for some JavaScript code improvement and optimization. Is it possible to optimize this block of code (for example, remove double array loop)?

replace: function(ex, str) {
 var arr = str.match(/{add_([A-Z]+)}/gi);
 if (arr && arr.length) {
 arr.each(function(e){ 
 var id = e.replace('{add_','').replace('}',''),
 o = Core.get_data_by_id(ex.extensions, id); 
 if (o && o.status =="1") {
 str = str.replace('{add_'+id+'}', o.html);
 } 
 });
 } 
 var arr2 = str.match(/{banner_([0-9]+)}/gi);
 if (arr2 && arr2.length) {
 arr2.each(function(e){
 var id = e.replace('{banner_','').replace('}',''),
 b = Core.get_data_by_id(ex.banners, id);
 if (b && b.status =="1" && b.code && b.code.length) {
 var content = '<div class="units-row banner_'+b.id+'"><ul>' +
 b.code.each(function(s){
 content += '<li>'+s.code+'</li>';
 }); 
 content += '</ul></div>';
 str = str.replace('{banner_'+id+'}', content);
 } 
 });
 }
 return str;
}
Jamal
35.2k13 gold badges134 silver badges238 bronze badges
asked Jun 2, 2014 at 10:28
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1 Answer 1

1
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You should be using String.prototype.replace() with a function as the second argument. Example:

str = str.replace(/{add_([A-Z]+)}/gi, function(match, id) {
 var o = Core.get_data_by_id(ex.extensions, id); 
 return (o && o.status == "1") ? o.html : match;
}); 
str = str.replace(/{banner_([0-9]+)}/gi, function(match, id){
 var b = Core.get_data_by_id(ex.banners, id);
 if (b && b.status =="1" && b.code && b.code.length) {
 var content = '<div class="units-row banner_'+b.id+'"><ul>' +
 b.code.each(function(s){
 content += '<li>'+s.code+'</li>';
 }); 
 content += '</ul></div>';
 return content;
 } 
 return match; 
});

EDIT: I've added the second part.

answered Jun 2, 2014 at 11:06
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4
  • \$\begingroup\$ But I will have two regular expressions: /{add_([A-Z]+)}/gi and /{banner_([0-9]+)}/gi \$\endgroup\$ Commented Jun 2, 2014 at 11:07
  • \$\begingroup\$ That was just an example. Your second regular expression is independent from the first one and can be replaced in a similar manner. \$\endgroup\$ Commented Jun 2, 2014 at 11:09
  • \$\begingroup\$ Thanks! Do you know how to combine 2 regular expressions and pass it to replace function? \$\endgroup\$ Commented Jun 2, 2014 at 11:19
  • \$\begingroup\$ I wouldn't do that. The functions a different enough to leave them separate, so leave the regular expressions separate, too. \$\endgroup\$ Commented Jun 2, 2014 at 11:24

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