how to remove object from JS array based on values from second array

Question

Hey guys I'm trying to remove all objects from an array that matches the values from a second array. Allow me to explain with some examples!

var originalObj = [
{"id":"88","name":"Lets go testing"},
{"id":"88","name":"Have fun boys and girls"},
{"id":"108","name":"You are awesome!"},
{"id":"99","name":"Have fun boys and girls"},
{"id":"108","name":"You are awesome!"},
{"id":"99","name":"Have fun boys and girls2"},
{"id":"108","name":"You are awesome!"}
]

Currently with my javascript code I can only remove ONE specified value from object and any occurrences with said value

var updatedObj= $.grep(originalObj , function(e){ 
             return e.id!= '88'; 
        });

console.log(updatedObj)
[
    {"id":"108","name":"You are awesome!"},
    {"id":"99","name":"Have fun boys and girls"},
    {"id":"108","name":"You are awesome!"},
    {"id":"99","name":"Have fun boys and girls2"},
    {"id":"108","name":"You are awesome!"}
]

What I want to accomplish is have the .grep function look up a second array that looks something like this.

 var filterID = ["88","99"];

so the result should look something like this

//Needs updating
//var updatedObj= $.grep(originalObj , function(e){ 
//           return e.id!= '88'; <-- needs to match all filterID values instead
//      });

console.log(updatedObj)
[
    {"id":"108","name":"You are awesome!"},
    {"id":"108","name":"You are awesome!"},
    {"id":"108","name":"You are awesome!"}
]

Any help would be greatly appreciated!

Thank you.

Jnatalzia · Accepted Answer · 2014-07-02 18:46:28Z

2

You could iterate over all the values in your array and filter that way.

var updatedObj = originalObj;
for (var i in filterID) {
  updatedObj = $.grep(updatedObj , function(e){ 
    return e.id!= filterID[i];
  });
}

answered Jul 2, 2014 at 18:46

Jnatalzia

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2 Comments

Jnatalzia Over a year ago

If it's overwriting it every time, the value in that current iteration would be stripped out each time, resulting in all values stripped out by the end of the iteration...

Patrick Evans Over a year ago

Yea i see that now, sorry

Alnitak · Accepted Answer · 2014-07-02 18:53:36Z

If you use an object to represent the list of keys to filter this becomes simple:

var toRemove = {
    88: true, 99: true;  // keys significant, values thereof irrelevant
};

var updatedObj = $.grep(originalObj, function(e) { 
    return !(e.id in toRemove);
});

For short arrays this is about the same efficiency as a doubly-nested loop, that being approximately O(n * m), where n is the size of the array being filtered and m is the number of elements to remove).

However if the array of values to filter is substantial then this will be O(n) (assuming that the in operator performs in constant time)

Patrick Evans · Accepted Answer · 2014-07-02 18:49:07Z

0

use the indexOf array function

var filterID = ["88","99"];
var updatedObj= $.grep(originalObj , function(e){ 
    return filterID.indexOf(e.id) === -1; 
});

answered Jul 2, 2014 at 18:49

Patrick Evans

42.8k6 gold badges81 silver badges94 bronze badges

Comments

Indranil Mondal · Accepted Answer · 2014-07-02 19:04:01Z

You could iterate over all the values in your array and filter that way.

           var updatedObj = [
                {"id":"108","name":"You are awesome!"},
                {"id":"99","name":"Have fun boys and girls"},
                {"id":"108","name":"You are awesome!"},
                {"id":"99","name":"Have fun boys and girls2"},
                {"id":"108","name":"You are awesome!"}
            ],
            filterId = ["89","99"],result=[];
            $.each(updatedObj,function(index,item){
                if(filterId.indexOf(item.id) === -1){
                   result.push(updatedObj[index])
                }
            })
            console.log(result);

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