I have an array in which all the elements are also arrays (of integers), called mainArray. I am trying to use splice() to add and remove its elements (subarrays).

Initially mainArray has a single element (an Array with one integer), I wish to delete it and add 3 new subarrays to mainArray, these are defined in arraysToAdd.

mainArray = new Array(new Array(1));
arraysToAdd = new Array(new Array(1,2), new Array(1,4), new Array(1,7));

alert(arraysToAdd.length); // Returns: 3: as expected

mainArray.splice(0,1,arraysToAdd);

alert(mainArray.length); // Returns: 1: I want this to be 3

I expect the length of mainArray at the end to be 3 (as it should contain 3 subarrays), but it seems splice() is flattening arraysToAdd and so mainArray ends up just being an array of integers.

What am I missing?

I have an array in which all the elements are also arrays (of integers), called mainArray. I am trying to use splice() to add and remove its elements (subarrays).

Initially mainArray has a single element (an Array with one integer), I wish to delete it and add 3 new subarrays to mainArray, these are defined in arraysToAdd.

mainArray = new Array(new Array(1));
arraysToAdd = new Array(new Array(1,2), new Array(1,4), new Array(1,7));

alert(arraysToAdd.length); // Returns: 3: as expected

mainArray.splice(0,1,arraysToAdd);

alert(mainArray.length); // Returns: 1: I want this to be 3

I expect the length of mainArray at the end to be 3 (as it should contain 3 subarrays), but it seems splice() is flattening arraysToAdd and so mainArray ends up just being an array of integers.

What am I missing?

• javascript
• arrays
• splice

• 1 console.log(mainArray) – zerkms Commented Feb 24, 2014 at 0:44
• "but it seems splice() is flattening arraysToAdd" Uhm, how can the length of mainArray be 1 if the other arrays is flattened into mainArray? If mainArray was an array of integers, wouldn't the length be 6? – Felix Kling Commented Feb 24, 2014 at 0:46
• 2 You do know you can just type var mainArray = [[1]];, right? – StackSlave Commented Feb 24, 2014 at 0:51
• @PHPglue brings up a good point. In fact, your mainArray worked because you didn't use the value, but if you had tried to use it, or if you created it with a greater number, you'd get a result that you may not expect. Like if you did new Array(3), you now have an Array with .length === 3, and no actual defined members. – cookie monster Commented Feb 24, 2014 at 1:58

2 Answers 2

What you're missing is that you're adding an Array of Arrays to your Array of Arrays. You want to add each individual Array instead.

You can use .apply() to do this:

mainArray.splice.apply(mainArray, [0,1].concat(arraysToAdd));

So the 0 and 1 arguments you passed are joined with your arraysToAdd to form the arguments that you're going to pass to .splice() via .apply().

Demo: http://jsfiddle/QLwLA/

Without .apply(), you would have needed to add them individually, like this:

mainArray.splice(0, 1, arraysToAdd[0], arraysToAdd[1], arraysToAdd[2]);

Demo: http://jsfiddle/QLwLA/1/

try this:

mainArray.splice(0, 1, ...arraysToAdd)