creating a javascript recursive filter function

Question

Is there any way of making this function recursive so that I do not need to create a switch for each length of filter criteria ?

var data = [
  {a:'aaa',b:'bbb',c:'ccc',d:'ddd',e:'eee'},
  {a:'aaa',b:'bbb',c:'ccc',d:'eee',e:'fff'},
  {a:'xxx',b:'bbb',c:'ccc',d:'ddd',e:'fff'}
]


function select(data,where){
  return data.filter(function(e){ 
    var k = Object.keys(where); 
    switch(k.length){     
      case 1: return (e[k[0]] == where[k[0]]);
      case 2: return (e[k[0]] == where[k[0]] && e[k[1]] == where[k[1]]);
      case 3: return (e[k[0]] == where[k[0]] && e[k[1]] == where[k[1]] && e[k[2]] == where[k[2]]);
      case 4: return (e[k[0]] == where[k[0]] && e[k[1]] == where[k[1]] && e[k[2]] == where[k[2]] && e[k[3]] == where[k[3]]);
      case 5: return (e[k[0]] == where[k[0]] && e[k[1]] == where[k[1]] && e[k[2]] == where[k[2]] && e[k[3]] == where[k[3]] && e[k[4]] == where[k[4]]);
    }
  })  
}

var where = {a:'aaa',b:'bbb'}

console.log(select(data,where));

This is so ugly, it looks almost like VHDL. Sorry, nothin' personal.. — MightyPork
– MightyPork, Commented Jan 10, 2015 at 0:07
Exactly, that's why I am asking the question, this works but it is very ugly - I don't think it's fair to give me negatices on this, my code is complete and my question is clear ! — crankshaft
– crankshaft, Commented Jan 10, 2015 at 0:10
Well it's very messy, I have trouble understanding what is the purpose / goal. Could you describe the desired behavior? — MightyPork
– MightyPork, Commented Jan 10, 2015 at 0:11
simple, select data from an object where the elements match the filter criteria, similar to SQL SELECT * FROM TABLE WHERE aaa ='aaa' and bbb = 'bbb' and.... — crankshaft
– crankshaft, Commented Jan 10, 2015 at 0:13

Barmar · Accepted Answer · 2016-07-25 00:23:32Z

1

It doesn't need to be recursive (I'm not sure you understand what that means), you just need to loop on the elements in where:

function select(data, where) {
  return data.filter(function(e) {
    var k = Object.keys(where);
    return k.every(function(key) {
      return e[key] == where[key];
    });
  })
}

var data = [
  {a:'aaa',b:'bbb',c:'ccc',d:'ddd',e:'eee'},
  {a:'aaa',b:'bbb',c:'ccc',d:'eee',e:'fff'},
  {a:'xxx',b:'bbb',c:'ccc',d:'ddd',e:'fff'}
]

var where = {a:'aaa',b:'bbb'}

console.log(select(data,where));

edited Jul 25, 2016 at 0:23

user6307369

answered Jan 10, 2015 at 0:11

Barmar

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1 Comment

crankshaft Over a year ago

Thanks, it seems obvious now, also I have never used every() before, that's new to me.

Olim Saidov · Accepted Answer · 2015-01-10 00:13:20Z

0

Try this code:

function select(data, where) {
    return data.filter(function (e) {
        for (var key in where) {
            if (where.hasOwnProperty(key)) {
                if (e.hasOwnProperty(key)) {
                    if (e[key] != where[key]) {
                        return false;
                    }
                }
                else {
                    return false
                }
            }
        }

        return true;
    })
}

answered Jan 10, 2015 at 0:13

Olim Saidov

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