So...I have this array:
val['an_element'][0]['another_element'][2]['text']
I want to get rid of the entire "2" node.
Now...I THOUGHT the way to do this would be:
delete val['an_element'][0]['another_element'][2];
BUT...it doesn't actually drop the element, but simply empties it out.
I also tried:
val['an_element'][0]['another_element'][2] = null;
...but that just resulted in my console log nearly bleeding it was so red with errors.
Basically, i want that [2] node to NO LONGER EXIST. Basically, I want it to NOT BE FOUND AT ALL.
What do I do??? And I know the ".splice" method will NOT actually modify the original array, so please don't suggest it. :)
The splice method will, in fact, modify the array. Just try:
val['an_element'][0]['another_element'].splice(2, 1);
From the docs:
Changes the content of an array, adding new elements while removing old elements.
...
If you specify a different number of elements to insert than the number you're removing, the array will have a different length at the end of the call.
Related
I have a jQuery object array that I need to remove the 1st and last item from. I tried using shift() and pop() but it threw an error because I guess a jQuery object is not the same as an array? I then used delete however that doesn't change the length so a loop I have after this gets all kinds of messed up. Finally I found out I should use:
$item.splice(0,1) and $item.splice($item.length-1, 1)
however the object array that splice(0, 1) returns is no longer in the same order as it was previously. Is there a reason splice would return a different order?
Did you try using slice? It works with a jQuery elements collection:
$('div').slice(1, -1);
Also, you can use the above statement without changing your original array. It returns the collection, removing the first and last elements.
In the javascript, there are two arrays:tags[] and tags_java[]. I use .splice to delete certain items, which of the same index in the two arrays. The tags[] works fine, but tags_java doesn't, it seems always delete the last item.
Here is the code and the jsfiddle link.
var tag = $(this).text();
var index = $.inArray(tag, tags);
tags.splice(index,1);
tags_java.splice(index,1);
Nah, both don't work, because you're not actually finding the correct index of your tag.
Why not? Because $(this).text() includes the delete mark you added, × - e.g. "Morning×". Since that's not in your tags array, index will be -1. tags.splice(-1, 1); will remove 1 item from the end of the array.
In general, it's never a good idea to use presentation text (i.e. the text of your tag element) as data (e.g. using that text as a lookup value in an array). It's very likely that it'll be broken when something changes in the presentation - like here. So a suggestion would be to store the data (what you need to look up the tags) as data - e.g. using the jQuery-provided data() API - even if it seems redundant.
Here's a quick example - just adding/replacing two lines, which I've marked with comments starting with "JT": JSFiddle
Now, instead of looking up by $(this).text(), we're looking up by the data value "tagValue" stored with $(this).data() - that way, the lookup value is still bound to the element, but we're not relying on presentation text.
If the tag is not in the tags array, $.inArray will return -1, which would then cause the last item to be deleted.
You have to make sure that the item is actually in the array.
I'm using Knockout-sortable to drag-and-drop/sort records in my table, but I've run into a problem. I have no clue how to keep track of the position in the sort index of an element. (I.e. element A, B and C appear in that order and have 1,2,3 as index respectively, but if B gets dropped above A the correct index would be 2,1,3)
Nothing in my code is custom: I just include knockout-sortable and it's plug and play. I usually always include a code snippet, but I don't feel that's useful. The only thing I know is that I'm probably gonna need a ko.computed(), but I have no idea what to fill it in with.
If you look at example http://jsfiddle.net/rniemeyer/Jr2rE/, you can see that the plug-in works by updating an observable array of data. Because of this, you don't have to keep track of the index value. The order of the records, technically, gives you all the information you need.
That being said, I ran into the same issue in last year. To solve my problem, I added a consecutively numbered index property to each object in my observable array. Then, when the sortable plug-in re-arranged the contents of the observable array, I just had to read out the new index property to know the sort order.
I'm wondering, why do some elements appear like an array and others like HTMLSpanElement. I've attached a picture as I'm not sure how to describe this otherwise.
The following log is made via
log(returner);
log(returner[0]);
Is returner a jQuery object as a result of $() ? $() will always return an array, even if there is one or zero elements inside of it. Without specifying an index in your first console.log, the entire contents of the array are outputted. In the second console.log, you include an array index, so only the element matching that index is outputted.
Because the element that appears like an array IS an array - it's an array of DOM element objects (HTMLSpanElement, etc).
When you log the first element of the array with returner[0], that element is a DOM object, so it logs it as an object.
Because (it looks like) returner is not an element, but an array of elements.
I have an array called as wcs declared using var wcs= new Array();.
I do add items like this, wcs[indx] = value. where i will keep on changing the indx value, so at times, my array will be looking like this
wcs[2] ='a'; wcs[5]=')';
when i call the splice method on this array, all the created indices are re-indexed, meaning they become reset from 0...
how to avoid this in javascript & jQuery
Write your own splice method that works the way you want. If you specify the input, processing and expected output, you might get some help with that.
If you simply want a copy of the array, you may be after the concat method.