Comparing Textbox values in JQUERY

Question

Am comparing 2 values in text box

say $('#Low'.val()) > $('#High'.val()) //Where low and high are textbox id

But it is fails in few scenarios as it is comparing string.

i have 2 questions here

1.how to convert to integer or float and check it in correct way

2.When i compare the string as per my above code "99" > "1000" is returning true. Why it succeeds ?

Thanks

PSR · Accepted Answer · 2013-05-30 06:47:11Z

3

parseInt($('#Low'.val(),10)) > parseInt($('#High'.val(),10))

or

 parseFloat($('#Low'.val())) > parseFloat($('#High'.val())

answered May 30, 2013 at 6:47

PSR

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1 Comment

Atber Over a year ago

And "parseFloat" similarly

Siva Charan · Accepted Answer · 2013-05-30 06:48:31Z

1

You have used function as $('#Low'.val()) which is wrong. It should be $('#Low').val()

Do this way:-

parseInt($('#Low').val(), 10) > parseInt($('#High').val(), 10)

answered May 30, 2013 at 6:48

Siva Charan

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Comments

Milind Anantwar · Accepted Answer · 2013-05-30 06:50:11Z

1

Try This:

(parseInt($("#Low").val()) > parseInt($('#High').val()))

answered May 30, 2013 at 6:50

Milind Anantwar

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Comments

Supra · Accepted Answer · 2018-10-03 22:36:43Z

1

Even better is to use parseFloat.

parseFloat($('#Low').val())) > parseFloat($('#High').val())

edited Oct 3, 2018 at 22:36

answered May 30, 2013 at 6:48

Supra

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Comments

zkanoca · Accepted Answer · 2013-05-30 06:52:38Z

0

Alphabetically 9 comes after 1. So "99" > "1000" may return true.

if(parseInt($('#Low'.val())) > parseInt($('#High'.val()))
{
    //Yes, #Low is bigger than #High
}
else
{
    //No, #Low is not bigger than #High yet.
}

must work.

answered May 30, 2013 at 6:52

zkanoca

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