Cannot read property 'length' of null (javascript)

Question

While trying to debug I am get the 'length' null error with this line. It is written just like the book instructed, so I don't understand why it is giving me the error? Thanks, =)

if (capital.length < 1) {

( here is the full code as requested.. SORRY)

<script type="text/javascript">
var capital = window.prompt("What is the capital of Missouri?","")

if (capital.length < 1) {
    document.getElementById("firstdiv").innerHTML="Sorry you don't feel like playing.<br /> The Capital of Missouri is Jefferson City.";
}
else {
    if (!window.confirm("Is that your final answer?")){ return true;

        document.getElementById("firstdiv").innerHTML = "The capital of Missouri is: <bold>" + capital + "</bold>, so says you.";
    }
    else{
        return false;
    }
}
</script>

post full code please

frickskit
– frickskit

2013-03-31 16:20:09 +00:00
Commented Mar 31, 2013 at 16:20 — frickskit
– frickskit, Commented Mar 31, 2013 at 16:20
What language is this? JavaScript?

user1634845
– user1634845

2013-03-31 16:20:46 +00:00
Commented Mar 31, 2013 at 16:20 — user1634845
– user1634845, Commented Mar 31, 2013 at 16:20
Because you never set capital.

SLaks
– SLaks

2013-03-31 16:21:01 +00:00
Commented Mar 31, 2013 at 16:21 — SLaks
– SLaks, Commented Mar 31, 2013 at 16:21
The variable capital is null.

Tamil Selvan C
– Tamil Selvan C

2013-03-31 16:24:20 +00:00
Commented Mar 31, 2013 at 16:24 — Tamil Selvan C
– Tamil Selvan C, Commented Mar 31, 2013 at 16:24
Done =) sorry about that

JBSAMOM
– JBSAMOM

2013-03-31 17:22:16 +00:00
Commented Mar 31, 2013 at 17:22 — JBSAMOM
– JBSAMOM, Commented Mar 31, 2013 at 17:22

tirdadc · Accepted Answer · 2015-10-22 18:15:21Z

88

The proper test is:

if (capital != null && capital.length < 1) {

This ensures that capital is always non null, when you perform the length check.

Also, as the comments suggest, capital is null because you never initialize it.

edited Oct 22, 2015 at 18:15

tirdadc

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answered Mar 31, 2013 at 16:22

Hunter McMillen

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2 Comments

JBSAMOM Over a year ago

I am so sorry.. Yes it is in JavaScript

Hunter McMillen Over a year ago

I would probably write this as if (capital && capital.length < 1) now and not explicitly reference null.

Kamil Naja · Accepted Answer · 2019-06-13 19:55:49Z

4

This also works - evaluate, if capital is defined. If not, this means, that capital is undefined or null (or other value, that evaluates to false in js)

if (capital && capital.length < 1) {do your stuff}

answered Jun 13, 2019 at 19:55

Kamil Naja

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2 Comments

joanis Over a year ago

If capital is non-null but points to the empty string, will the logic be correct with this solution?

Kamil Naja Over a year ago

probably not, in 2021 is better use nullish coalesce operator ?? developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

secretformula · Accepted Answer · 2013-03-31 16:21:59Z

1

From the code that you have provided, not knowing the language that you are programming in. The variable capital is null. When you are trying to read the property length, the system cant as it is trying to deference a null variable. You need to define capital.

answered Mar 31, 2013 at 16:21

secretformula

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Comments

Pankaj Faliyachi · Accepted Answer · 2019-02-08 07:55:02Z

1

if (capital.touched && capital != undefined && capital.length < 1 ) { 
//capital does exists
}

edited Feb 8, 2019 at 7:55

Pankaj Faliyachi

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answered Apr 8, 2015 at 13:53

Naramsim

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Comments

Aldo Geovani · Accepted Answer · 2016-07-06 17:58:42Z

0

I tried this:

if(capital !== null){ 
//Capital has something 
}

answered Jul 6, 2016 at 17:58

Aldo Geovani

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Comments

Gopal Gautam · Accepted Answer · 2022-11-12 05:50:11Z

0

Use Try cache

try{

//Your Code

} catch(err){

console.log(err)

}

answered Nov 12, 2022 at 5:50

Gopal Gautam

394 bronze badges

1 Comment

Harrison Over a year ago

Could you expand your answer to help OP a little as to why this answer may work for them?

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