Javascript array manipulation

Question

I want to create string (delimited by pipe '||') from the contents of an array. I want to remover $ from the item and add || between two items. But I don't want to add || after the last array item. Easiest way to do this?

This is what I have done so far:

var array = ["$db1", "$db2", "$db3", "$db4"];
var dbs = "";
for(var i = 0; i < array.length, i++){
 if(array[i].charAt(0) == "$"){
 dbs += array[i].replace("$","") + "||";
 alert(dbs);
 }
}

Answer 1

Here you go:

array.join('||').replace(/\$/g,'')

Live demo: http://jsfiddle.net/ZrgFV/

• JSRookie: Thanks for the quick reply! Can you explain the pattern for replace?
• Šime Vidas: @JSRookie Read here. In short, /\$/g is a regular expression which matches all $ characters in the string. The second argument is an empty string ('') which means that all $ characters are replaced by an empty string (= they are deleted from the string).

Answer 2

var array = ["$db1", "$db2", "$db3", "$db4"];
var dbs = array.map(function(x) { return x.substring(1); }).join('||');

It requires the relatively new Array.map, so also include:

if(![].map) Array.prototype.map = function(f) {
 var r = [], i = 0;
 for(; i < this.length; i++) r.push(f(this[i]));
 return r;
};

I took it that you meant "remove a leading $" because they're all at the beginning. If not, use:

var array = ["$db1", "$db2", "$db3", "$db4"];
var dbs = array.join('||').replace(/\$/g, '');

• Robert: Your Array.prototype.map won't do anything but copy the array, I think you meant to r.push(f(this[i]))

Answer 3

array.join('||').replace(/(^|(\|\|))\$/g, '1ドル');

Join with ||, then annihilate any $ following either the beginning of the string or the separator. Works as long as your strings do not contain || (in which case, I think you have bigger problems).

Answer 4

var array = ["$db1", "$db2", "$db3", "$db4"];
var arrStr = array.join("||");
var dbs = arrStr.replace(/\$/g, "");

Grr sorry forgot to add the \g switch to replace all.

• Ry-: This fails miserably with more than one item. Did you actually test it? .replace with a string only replaces the first occurrence of the string.
• tomfumb: without a global regex your last line will only replace the first $
• scrappedcola: Yea i saw that and not it's fixed
• Ry-: Now it does. Notifying @tomfumb?

Answer 5

var dbs = array.join('||').replace(/$/gi, '');

This.

• tomfumb: case-insensitive seems a little redundant here!
• Felix Kling: $ is a special regex character and i is unnecessary.