如何从数组中删除元素（包括下一个项目） - jquery

Question

I have the following arrays to start with:

var x = '[123_0|345_1|789_3|222_4|358_9]'

var y = '[345~2.0~/pathname.jpg|789~1.25~/pathname2.jpg|222~1.15~pathname3.jpg|090~1.0~/pathname4.jpg]'

What i need to do is:

Check if any of the items from x exist in y, then remove it (including the elements after the ~)

At the end i need an array as this:

var newarray = [222~1.15~/pathname3.jpg|090~1.0~/pathname4.jpg]

I managed to remove the value x from array y by the following code:

var toremove = [];

var z = x.split('|')

$.each(z,function(){
     var newval= this.split('_')[0];
     toremove.push(newval);
 });

The above is returning(as expected):

'[123,345,789,222,358]'

I can easily do an $.inArray to check and remove the above values in y.. But i need to remove the other values (from y) after the "~" as well

Please help on how to achieve this

Answer 1

Its kinda strange what you want to do, but the important tool to use is the Array filter function, try the following and view in fiddler

  var x = '[123_0|345_1|789_3|222_4|358_9]';

  var y = '[345~2.0~/pathname.jpg|789~1.25~/pathname2.jpg|222~1.15~pathname3.jpg|090~1.0~/pathname4.jpg]';


  x = x.substring(1, x.length-1); // remove [ and ]
  var needles = x.split('|')
  y = y.substring(1, y.length-1); // remove [ and 
  var haystack = y.split('|');
  var newArray = haystack.filter(function (value) {
      for (var ii=0; ii<needles.length; ii++) {
          var needle = needles[ii].split('_')[0];
          var needleRegex = new RegExp("^" + needle);
          if(value.match(needleRegex)) {
              return false;
          }
      }
      return true;
  });

  console.log(newArray);
  $('body').append(newArray.join('|'));

This assumes that you do in fact want to filter 222 , because 222_4 is in x ;

http://jsfiddle.net/SLnV2/

Answer 2

Using regular expressions ..

var x = '[123_0|345_1|789_3|222_4|358_9]',
    y = '[345~2.0~/pathname.jpg|789~1.25~/pathname2.jpg|222~1.15~pathname3.jpg|090~1.0~/pathname4.jpg]',
    pat = /(\d+)[^\]|]+\|?/, tmp = '', xx = x ;

while( (tmp = xx.match(pat)) ){
    if(y.indexOf(tmp[1]) != -1){
        y = y.replace(new RegExp(tmp[1]+"[^\\]|]+",'gi'), '');
    }
    xx = xx.replace(pat, '');
};

y = y.replace(/(\[)\|+|(.)\|+(?!\d)/g,'$1');
console.log(y); // prints =>  [090~1.0~/pathname4.jpg]

Answer 3

You can use delete to remove an element from an array but unfortunately it will leave a hole in the array.

You can use splice(index,number) :

1. index is the index of array
2. number` is how many element to delete

EDIT:

example:

numbers is an array as ['zero', 'one', 'sd', 'shi', 'go']

using delete :

delete numbers[2]
// numbers is ['zero', 'one', undefined, 'shi', 'go']

while if I use splice :

numbers.splice(2, 1);
// numbers is ['zero', 'one', 'shi', 'go']

Answer 4

try this , DEMO

$( document ).ready(function() {
    var x = '[123_0|345_1|789_3|222_4|358_9]';

    var y = '[345~2.0~/pathname.jpg|789~1.25~/pathname2.jpg|222~1.15~pathname3.jpg|090~1.0~/pathname4.jpg]';

    x=x.replace('[', '');
    var z = x.split('|');
    var toremove = [];
    $.each(z,function(){
        var newval= this.split('_')[0];
        toremove.push(newval);
    });

    y=y.replace('[', '');
    var zz = y.split('|');
    $.each(zz,function(){
        var newval= this.split('~')[0];
        var appendval= this;
        if(jQuery.inArray( newval, toremove ) != -1)
        {
            zz = jQuery.grep(zz, function(value) {
                return value != appendval;
            });
        }
    });
    alert(zz);
});

more detail about used function : grep , inArray