how to exclude blank values in jquery array?

Question

there are following input fields with datepicker class:

<input type="text" class="datepicker" value="" />
<input type="text" class="datepicker" value="" />
<input type="text" class="datepicker" value="2011-02-15" />
<input type="text" class="datepicker" value="2011-02-16" />

I need to create an array of dates, exclude blank values and return maximum value. Following code doesn't seem to exclude blank values:

var datelist = [];
$(".datepicker").each(function(i) {
    if (this.value!="") {datelist[i] = this.value;}
});
datelist.sort();
datelist.reverse();
alert(datelist); //,,2011-02-16,2011-02-15
alert(datelist[0]); //undefined

What's the catch?

Felix Kling · Accepted Answer · 2011-02-15 14:55:08Z

6

It excludes the empty values. That is why datalist[0] is undefined.

The value with index 0 is empty, hence datelist[0] = this.value; is not executed. The only values set in the array are the ones with keys 2 and 3, making the keys 0, 1 and all >3 undefined.

If the values were included, you would not get undefined but an empty string.

You should not use the index for adding values to the array. push() them:

$(".datepicker").each(function(i) {
    if (this.value!="") {datelist.push(this.value);}
});

That said, a more concise way would be:

var datelist = $('.datepicker[value!=""]').map(function() {
    return this.value;
}).get();

Reference: .map(), .get()

edited Feb 15, 2011 at 14:55

answered Feb 15, 2011 at 14:50

Felix Kling

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Comments

Robin · Accepted Answer · 2011-02-15 14:52:01Z

2

Try

$('.datepicker[value!=""]')

answered Feb 15, 2011 at 14:52

Robin

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Comments

Guffa · Accepted Answer · 2011-02-15 15:02:48Z

1

You can use the .map and $.grep methods to get the array and remove the empty items:

Get all values into an array:

var datelist = $(".datepicker").map(function() { return this.value; }).get();

Remove empty items from the array:

datelist = $.grep(datelist, function(s) { return s.length != 0; });

answered Feb 15, 2011 at 15:02

Guffa

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Comments

Jakub Hampl · Accepted Answer · 2011-02-15 14:51:47Z

0

It's like doing a = []; a[0] = 1; a[2] = 3; - if you skip an index it will be filled with an undefined value. So you could change your code to something like this:

$(".datepicker").each(function() {
    if (this.value != "" || this.value != undefined) {
          datelist[datelist.length] = this.value;
    }
});

answered Feb 15, 2011 at 14:51

Jakub Hampl

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Comments

ThiefMaster · Accepted Answer · 2011-02-15 14:53:20Z

0

First, do not use this in each but the second function argument - see http://fixingthesejquery.com/#slide29

Then, append values to the list instead of using an index (remember, you do not create an array element for some values of i!):

if (this.value) {
    datelist.push(this.value); 
}

So, the code should be like this:

var dateList = [];
$(".datepicker").each(function(i, elem) {
    if (elem.value) {
        dateList[i] = elem.value;
    }
});
dateList.sort();
dateList.reverse();

answered Feb 15, 2011 at 14:53

ThiefMaster

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1 Comment

Šime Vidas Over a year ago

The OP is iterating over a jQuery object, not a regular array - using this is safe.

Šime Vidas · Accepted Answer · 2011-02-15 14:53:35Z

0

Try changing this portion -

var datelist = [];
$(".datepicker").each(function(i) {
    if (this.value!="") {datelist[i] = this.value;}
});

to this -

var datelist = [];
var index = 0;
$(".datepicker").each(function(i) {
    if ( $(this).val() != "") 
    {
        datelist[index++] = $(this).val();
    }
});

edited Feb 15, 2011 at 14:53

Šime Vidas

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answered Feb 15, 2011 at 14:49

MD Sayem Ahmed

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