用方括号替换js

Question

I'm hoping there is a better way to write this? Removing [] square brackets are a problem to me.

alert(CanvasData)//images[]=Base.jpg&images[]=Frame_Clear.png&images[]=Left_Clear.png&images[]=Right_Clear.png&images[]=Lenses_Lenses-Semi-Clear.png&images[]=


var PayName = CanvasData.replace("images[]=", "");
PayName = PayName.replace(/\[.*?\]/g, '');
PayName = PayName.replace(/\&images=/g, ' ');
PayName = PayName.replace(/\.png/g, " &");
PayName = PayName.replace(/\_/g, ' ');
PayName = PayName.substring(8);//remove fist 8 character (Base.jpg)
PayName = PayName.substring(0, PayName.length - 2);//remove last 2 characters // Frame Clear & Left Clear & Right Clear & Lenses Lenses-Semi-Clear &


alert(PayName)// Frame Clear & Left Clear & Right Clear & Lenses Lenses-Semi-Clear 

Thanks

Answer 1

try escaping the brackets, otherwise they take on a special meaning (define a character class) for the regular expression.

CanvasData.replace("images\[\]=", "");

You are already doing the same thing, by the way, in the second line of code in the replace section.

Answer 2

Call replace with three global regular expressions, where the first call lists every alternate that should be replaced with nothing (ie remove all '=', and 'Base.jpg&'), the second lists alternates that should be replaced with a space ('images[]', '_', and '.png'), and the third ties up your loose ends:

 var PayName = CanvasData.replace(/=|Base\.jpg&/g,'')
                         .replace(/images\[\]|_|\.png/g, ' ')
                         .replace(/^\s*|\s*&\s*$/g, '');

 // => "Frame Clear & Left Clear & Right Clear & Lenses Lenses-Semi-Clear"

Answer 3

If the square brackets [] are always empty, you could do:

var Payname = CanvasData.split("[]").join("");

Obviously that doesn't handle the general case.

What this does, is treat [] as a data separator, and turn the string into a parsed array -- just like parsing "1,3,6,4.5,3" except our comma is a [] . Then the array formed by breaking up the string is joined back to a string, with a blank separator. All the [] disappear. But those [occupied] do not.