looping through an array in jQuery

Question

I'm trying to create an array of textarea values and then loop through them.

Here is a jsFiddle that I am trying to get this to work in:

When I run it, nothing happens.

Any idea what I'm doing wrong?

var textArray = [];
$('[class=objectives]').each(function (i) {
     textArray.push($(this).val());
});

for (var i = 0; i < textArray.length; i++) {
     console.log(textArray[i].value);
}

Just change to console.log(textArray[i]);

Alex K.
– Alex K.

2013-05-21 15:04:01 +00:00
Commented May 21, 2013 at 15:04 — Alex K.
– Alex K., Commented May 21, 2013 at 15:04

techfoobar · Accepted Answer · 2013-05-21 15:03:43Z

5

There is no value property in each element, you can simply do:

for (var i = 0; i < textArray.length; i++) {
     console.log(textArray[i]);
}

Working Demo - Note: jQuery wasn't originally included

answered May 21, 2013 at 15:03

techfoobar

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4 Comments

kaapa Over a year ago

There is a good practise to store array length in a variable, and then use it in for loop

mplungjan Over a year ago

You mean for (var i=0, n=textArray.length; i<n; i++) ?

SkyeBoniwell Over a year ago

for some reason it doesn't show up in my firebug console when I run it in jsfiddle...

techfoobar Over a year ago

@999cm999 - jsfiddle.net/kxkHZ/7 <- This does print four strings in the console.

Igor · Accepted Answer · 2013-05-21 15:04:36Z

4

You are pushing the value of the element into the array, you do not need to call value on it again. Just access the string itself:

console.log(textArray[i]);

answered May 21, 2013 at 15:04

Igor

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Comments

Adil Shaikh · Accepted Answer · 2013-05-21 15:04:14Z

3

Working --> http://jsfiddle.net/kxkHZ/11/

for (var i = 0; i < textArray.length; i++) {
    // textArray[i] itself is the value so textArray[i].value is incorrect 
    console.log(textArray[i]); 
}

You forgot to include jquery -

You should have got an error $ is not defined.

answered May 21, 2013 at 15:04

Adil Shaikh

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1 Comment

Justin Niessner Over a year ago

Still doesn't work. You wind up writing 'undefined' to the console 4 times.

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