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Optimizing JS code

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}

-2

$begingroup$

I have two variables that could be the same. In such a case I am assigning one of them to be undefined. Is this code the best way I can write it?

    let folderOwner = $("#change-folder-owner").val();

    let folderUser = $("#user_name_alias").data('value');



    folderOwner = (folderOwner === folderUser) ? undefined : folderOwner;

Can someone please help me with a better way?

javascript

share|improve this question

asked Mar 29 at 14:59

BhargaviBhargavi

1282

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Why should the value be undefined if they are equal?
  $endgroup$
  – dustytrash
  Mar 29 at 17:22
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Using if can be clearer and easier to understand than the ternary operator
  $endgroup$
  – David White
  Mar 29 at 19:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @dustytrash I need to use that undefined elsewhere
  $endgroup$
  – Bhargavi
  Apr 1 at 4:01
  

  
  
  
  

add a comment | 

-2

$begingroup$

I have two variables that could be the same. In such a case I am assigning one of them to be undefined. Is this code the best way I can write it?

    let folderOwner = $("#change-folder-owner").val();

    let folderUser = $("#user_name_alias").data('value');



    folderOwner = (folderOwner === folderUser) ? undefined : folderOwner;

Can someone please help me with a better way?

javascript

share|improve this question

asked Mar 29 at 14:59

BhargaviBhargavi

1282

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Why should the value be undefined if they are equal?
  $endgroup$
  – dustytrash
  Mar 29 at 17:22
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Using if can be clearer and easier to understand than the ternary operator
  $endgroup$
  – David White
  Mar 29 at 19:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @dustytrash I need to use that undefined elsewhere
  $endgroup$
  – Bhargavi
  Apr 1 at 4:01
  

  
  
  
  

add a comment | 

$begingroup$

I have two variables that could be the same. In such a case I am assigning one of them to be undefined. Is this code the best way I can write it?

    let folderOwner = $("#change-folder-owner").val();

    let folderUser = $("#user_name_alias").data('value');



    folderOwner = (folderOwner === folderUser) ? undefined : folderOwner;

Can someone please help me with a better way?

javascript

share|improve this question

asked Mar 29 at 14:59

BhargaviBhargavi

1282

$endgroup$

    let folderOwner = $("#change-folder-owner").val();

    let folderUser = $("#user_name_alias").data('value');



    folderOwner = (folderOwner === folderUser) ? undefined : folderOwner;

share|improve this question

asked Mar 29 at 14:59

BhargaviBhargavi

1282

share|improve this question

asked Mar 29 at 14:59

BhargaviBhargavi

1282

asked Mar 29 at 14:59

BhargaviBhargavi

1282

asked Mar 29 at 14:59

BhargaviBhargavi

1282

2 Answers
2

active

oldest

votes

4

$begingroup$

It would be clearer to use if:

if (folderOwner === folderUser) folderOwner = undefined

share|improve this answer

answered Mar 29 at 15:18

Oh My GoodnessOh My Goodness

2,177315

$endgroup$

add a comment | 

1

$begingroup$

You can avoid re-assigning folderOwner's value. Use two separate variables for the raw values, then evaluate for folderOwner's value.

const ownerValue = ...



const userValue = ...



const folderOwner = (ownerValue === userValue) ? undefined : ownerValue

share|improve this answer

answered Mar 29 at 16:19

JosephJoseph

22.9k21935

$endgroup$

add a comment | 

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4

$begingroup$

It would be clearer to use if:

if (folderOwner === folderUser) folderOwner = undefined

share|improve this answer

answered Mar 29 at 15:18

Oh My GoodnessOh My Goodness

2,177315

$endgroup$

add a comment | 

4

$begingroup$

It would be clearer to use if:

if (folderOwner === folderUser) folderOwner = undefined

share|improve this answer

answered Mar 29 at 15:18

Oh My GoodnessOh My Goodness

2,177315

$endgroup$

add a comment | 

$begingroup$

It would be clearer to use if:

if (folderOwner === folderUser) folderOwner = undefined

share|improve this answer

answered Mar 29 at 15:18

Oh My GoodnessOh My Goodness

2,177315

$endgroup$

if (folderOwner === folderUser) folderOwner = undefined

share|improve this answer

answered Mar 29 at 15:18

Oh My GoodnessOh My Goodness

2,177315

answered Mar 29 at 15:18

Oh My GoodnessOh My Goodness

2,177315

answered Mar 29 at 15:18

Oh My GoodnessOh My Goodness

2,177315

1

$begingroup$

You can avoid re-assigning folderOwner's value. Use two separate variables for the raw values, then evaluate for folderOwner's value.

const ownerValue = ...



const userValue = ...



const folderOwner = (ownerValue === userValue) ? undefined : ownerValue

share|improve this answer

answered Mar 29 at 16:19

JosephJoseph

22.9k21935

$endgroup$

add a comment | 

1

$begingroup$

You can avoid re-assigning folderOwner's value. Use two separate variables for the raw values, then evaluate for folderOwner's value.

const ownerValue = ...



const userValue = ...



const folderOwner = (ownerValue === userValue) ? undefined : ownerValue

share|improve this answer

answered Mar 29 at 16:19

JosephJoseph

22.9k21935

$endgroup$

add a comment | 

$begingroup$

You can avoid re-assigning folderOwner's value. Use two separate variables for the raw values, then evaluate for folderOwner's value.

const ownerValue = ...



const userValue = ...



const folderOwner = (ownerValue === userValue) ? undefined : ownerValue

share|improve this answer

answered Mar 29 at 16:19

JosephJoseph

22.9k21935

$endgroup$

const ownerValue = ...



const userValue = ...



const folderOwner = (ownerValue === userValue) ? undefined : ownerValue

share|improve this answer

answered Mar 29 at 16:19

JosephJoseph

22.9k21935

answered Mar 29 at 16:19

JosephJoseph

22.9k21935

answered Mar 29 at 16:19

JosephJoseph

22.9k21935