Reputation: 235
I'm banging my noob head against the wall on this one...
I have the following code:
var guns2 =
[
["Model 17", "Glock"],
["Model 19", "Glock"],
["PPQ", "Walther"],
["P2000", "HK"],
["Model 92", "Beretta"],
["Model 34", "Glock"]
]
var gunsMake = function () {
for (i=0; i<guns2.length; i++){
var make = guns2[i][1];
if (make ==="Glock"){
}
else {
guns2.splice(i,1);
}
};
};
gunsMake();
console.log(guns2);
The result I get in the console is as follows:
[["Model 17", "Glock"], ["Model 19", "Glock"], ["P2000", "HK"], ["Model 34", "Glock"]]
What I'd like to see is:
[["Model 17", "Glock"], ["Model 19", "Glock"], ["Model 34", "Glock"]]
"["P2000", "HK"]" shouldn't be there...I have a feeling it has something to do with the "guns2.length" argument in the for loop..it seems like it's skipping the subsequent array every time it splices, but I can't quite wrap my brain around a fix.
Please, someone steer me right :)
Upvotes: 2
Views: 6323
Reputation:
It usually isn't a good idea to modify an array while you're iterating over it, since it becomes hard to keep track of indices and exit conditions. Either insert desired results into a separate array, or use the native filter
method to return a filtered array.
var gunsMake = function (guns, desiredMake) {
return guns.filter(function(v,i,a){
return v[1] == desiredMake;
});
};
guns2 = gunsMake(guns2, "Glock");
console.log(guns2);
More on the Array filter
method on MDN: Array filter method
Upvotes: 6
Reputation: 24352
It's not always awful to modify the array in-place. If that's what you decide to do, decrement i
when you remove an element.
var guns2 =
[
["Model 17", "Glock"],
["Model 19", "Glock"],
["PPQ", "Walther"],
["P2000", "HK"],
["Model 92", "Beretta"],
["Model 34", "Glock"]
]
var gunsMake = function () {
for (i=0; i<guns2.length; i++){
var make = guns2[i][1];
if (make ==="Glock"){
}
else {
guns2.splice(i--,1); // Decrement i here
}
};
};
gunsMake();
console.log(guns2);
Upvotes: 3
Reputation: 2419
You're deleting (splicing) nodes out of your array while you're looping through it with a fixed index. So, you're not checking all of the elements.
If you check element #1 and decide to delete it, then when you check #2 you're actually checking #3.
Upvotes: 1