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the close parenthesis is there, why does it say it is missing the parenthesis?

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0

Why wouldn't this work? I'm not sure why it says there's a syntax error, it says missing a parenthesis for .append() when the ending parenthesis is clearly there (line 32).

I was told the backslash is used for line breaks in .append()

for(var i = 0; i < works.length; ++i) 
 $('#work').append('
 <div class='col-sm'>
 <img class='img-responsive' src='" + works[i] + "'>
 </div>
 '); //THIS IS THE END PARENTHESIS FOR .APPEND()
 ;

I expect the images in an array to be displayed in my webpage. The array is called 'works'.

jquery

share|improve this question

edited Mar 26 at 16:49

Robert Harvey♦

152k3737 gold badges290290 silver badges426426 bronze badges

asked Mar 26 at 16:45

Winnie CaiWinnie Cai

811 bronze badge

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  unescaped single quotes in value
  
  – Taplar
  Mar 26 at 16:50
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  try using ticks instead of quotes if you want multi line strings - ` (the thing left of the 1), also solves your unescaped quotes problem
  
  – Pete
  Mar 26 at 16:50
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Just put it all on one line. You're not really improving readability all that much with those line continuation characters, and it makes it more difficult to maintain anyway.
  
  – Robert Harvey♦
  Mar 26 at 16:50
  
  

  
  
  
  

add a comment | 

0

Why wouldn't this work? I'm not sure why it says there's a syntax error, it says missing a parenthesis for .append() when the ending parenthesis is clearly there (line 32).

I was told the backslash is used for line breaks in .append()

for(var i = 0; i < works.length; ++i) 
 $('#work').append('
 <div class='col-sm'>
 <img class='img-responsive' src='" + works[i] + "'>
 </div>
 '); //THIS IS THE END PARENTHESIS FOR .APPEND()
 ;

I expect the images in an array to be displayed in my webpage. The array is called 'works'.

jquery

share|improve this question

edited Mar 26 at 16:49

Robert Harvey♦

152k3737 gold badges290290 silver badges426426 bronze badges

asked Mar 26 at 16:45

Winnie CaiWinnie Cai

811 bronze badge

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  unescaped single quotes in value
  
  – Taplar
  Mar 26 at 16:50
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  try using ticks instead of quotes if you want multi line strings - ` (the thing left of the 1), also solves your unescaped quotes problem
  
  – Pete
  Mar 26 at 16:50
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Just put it all on one line. You're not really improving readability all that much with those line continuation characters, and it makes it more difficult to maintain anyway.
  
  – Robert Harvey♦
  Mar 26 at 16:50
  
  

  
  
  
  

add a comment | 

Why wouldn't this work? I'm not sure why it says there's a syntax error, it says missing a parenthesis for .append() when the ending parenthesis is clearly there (line 32).

I was told the backslash is used for line breaks in .append()

for(var i = 0; i < works.length; ++i) 
 $('#work').append('
 <div class='col-sm'>
 <img class='img-responsive' src='" + works[i] + "'>
 </div>
 '); //THIS IS THE END PARENTHESIS FOR .APPEND()
 ;

I expect the images in an array to be displayed in my webpage. The array is called 'works'.

jquery

share|improve this question

edited Mar 26 at 16:49

Robert Harvey♦

152k3737 gold badges290290 silver badges426426 bronze badges

asked Mar 26 at 16:45

Winnie CaiWinnie Cai

811 bronze badge

for(var i = 0; i < works.length; ++i) 
 $('#work').append('
 <div class='col-sm'>
 <img class='img-responsive' src='" + works[i] + "'>
 </div>
 '); //THIS IS THE END PARENTHESIS FOR .APPEND()
 ;

share|improve this question

edited Mar 26 at 16:49

Robert Harvey♦

152k3737 gold badges290290 silver badges426426 bronze badges

asked Mar 26 at 16:45

Winnie CaiWinnie Cai

811 bronze badge

share|improve this question

edited Mar 26 at 16:49

Robert Harvey♦

152k3737 gold badges290290 silver badges426426 bronze badges

asked Mar 26 at 16:45

Winnie CaiWinnie Cai

811 bronze badge

edited Mar 26 at 16:49

Robert Harvey♦

152k3737 gold badges290290 silver badges426426 bronze badges

edited Mar 26 at 16:49

Robert Harvey♦

152k3737 gold badges290290 silver badges426426 bronze badges

asked Mar 26 at 16:45

Winnie CaiWinnie Cai

811 bronze badge

asked Mar 26 at 16:45

Winnie CaiWinnie Cai

811 bronze badge

3 Answers
3

active

oldest

votes

0

If you are using single quotes to enclose your appended string, you should use double quotes inside of those or you need to escape the other single quotes contained in the opening and closing single quotes, I would go with the latter way of doing it, and change to this:

$('#work').append('<div class="col-sm"><img class="img-responsive" src="' + works[i] + '"></div>'); //THIS IS THE END PARENTHESIS FOR .APPEND()

share|improve this answer

edited Mar 26 at 16:57

answered Mar 26 at 16:51

Ryan WilsonRyan Wilson

4,50911 gold badge1010 silver badges2424 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Not the down vote, but the use of `` is for escaping the new line, which can be a valid approach.
  
  – Taplar
  Mar 26 at 16:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Taplar Yeah, I realized that after I read a comment above and updated my answer, thanks for the information, I've never used that before, typically if I append to a string on multiple lines I just use the + sign at the end of a line
  
  – Ryan Wilson
  Mar 26 at 16:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  stackoverflow.com/questions/805107/…
  
  – j08691
  Mar 26 at 16:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Pete thanks, I missed that last one when editing.
  
  – Ryan Wilson
  Mar 26 at 16:57
  

  
  
  
  

add a comment | 

3

If you are in a position to use ES6, the template literal is here to remove the complexity/annoyance of multi-line string escaping.

const works = ['http://placekitten.com/150/150', 'http://placekitten.com/200/200']

for (let i = 0; i < works.length; ++i) 
 $('#work').append(`
 <div class='col - sm '>
 <img class='img - responsive' src='$works[i]' alt=''>
 </div>
 `);
;

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="work"></div>

share|improve this answer

answered Mar 26 at 17:02

Andy HoffmanAndy Hoffman

11.2k33 gold badges2020 silver badges4141 bronze badges

add a comment | 

1

Take a look at the syntax highlight, you're closing the quotes and the interpreter is reading variables next to your string literals therefore throwing syntax errors, I'd encourage you to use ES6 Template literals or pay close attention to how you are using single and double quotes

Try this

for(var i = 0; i < works.length; ++i) 
 $('#work').append(`
 <div class='col-sm'>
 <img class='img-responsive' src='$works[i]>
 </div>
 `); //THIS IS THE END PARENTHESIS FOR .APPEND()
;

share|improve this answer

answered Mar 26 at 16:58

Gerardo SabettaGerardo Sabetta

2644 bronze badges

add a comment | 

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0

If you are using single quotes to enclose your appended string, you should use double quotes inside of those or you need to escape the other single quotes contained in the opening and closing single quotes, I would go with the latter way of doing it, and change to this:

$('#work').append('<div class="col-sm"><img class="img-responsive" src="' + works[i] + '"></div>'); //THIS IS THE END PARENTHESIS FOR .APPEND()

share|improve this answer

edited Mar 26 at 16:57

answered Mar 26 at 16:51

Ryan WilsonRyan Wilson

4,50911 gold badge1010 silver badges2424 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Not the down vote, but the use of `` is for escaping the new line, which can be a valid approach.
  
  – Taplar
  Mar 26 at 16:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Taplar Yeah, I realized that after I read a comment above and updated my answer, thanks for the information, I've never used that before, typically if I append to a string on multiple lines I just use the + sign at the end of a line
  
  – Ryan Wilson
  Mar 26 at 16:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  stackoverflow.com/questions/805107/…
  
  – j08691
  Mar 26 at 16:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Pete thanks, I missed that last one when editing.
  
  – Ryan Wilson
  Mar 26 at 16:57
  

  
  
  
  

add a comment | 

0

If you are using single quotes to enclose your appended string, you should use double quotes inside of those or you need to escape the other single quotes contained in the opening and closing single quotes, I would go with the latter way of doing it, and change to this:

$('#work').append('<div class="col-sm"><img class="img-responsive" src="' + works[i] + '"></div>'); //THIS IS THE END PARENTHESIS FOR .APPEND()

share|improve this answer

edited Mar 26 at 16:57

answered Mar 26 at 16:51

Ryan WilsonRyan Wilson

4,50911 gold badge1010 silver badges2424 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Not the down vote, but the use of `` is for escaping the new line, which can be a valid approach.
  
  – Taplar
  Mar 26 at 16:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Taplar Yeah, I realized that after I read a comment above and updated my answer, thanks for the information, I've never used that before, typically if I append to a string on multiple lines I just use the + sign at the end of a line
  
  – Ryan Wilson
  Mar 26 at 16:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  stackoverflow.com/questions/805107/…
  
  – j08691
  Mar 26 at 16:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Pete thanks, I missed that last one when editing.
  
  – Ryan Wilson
  Mar 26 at 16:57
  

  
  
  
  

add a comment | 

If you are using single quotes to enclose your appended string, you should use double quotes inside of those or you need to escape the other single quotes contained in the opening and closing single quotes, I would go with the latter way of doing it, and change to this:

$('#work').append('<div class="col-sm"><img class="img-responsive" src="' + works[i] + '"></div>'); //THIS IS THE END PARENTHESIS FOR .APPEND()

share|improve this answer

edited Mar 26 at 16:57

answered Mar 26 at 16:51

Ryan WilsonRyan Wilson

4,50911 gold badge1010 silver badges2424 bronze badges

$('#work').append('<div class="col-sm"><img class="img-responsive" src="' + works[i] + '"></div>'); //THIS IS THE END PARENTHESIS FOR .APPEND()

share|improve this answer

edited Mar 26 at 16:57

answered Mar 26 at 16:51

Ryan WilsonRyan Wilson

4,50911 gold badge1010 silver badges2424 bronze badges

answered Mar 26 at 16:51

Ryan WilsonRyan Wilson

4,50911 gold badge1010 silver badges2424 bronze badges

answered Mar 26 at 16:51

Ryan WilsonRyan Wilson

4,50911 gold badge1010 silver badges2424 bronze badges

3

If you are in a position to use ES6, the template literal is here to remove the complexity/annoyance of multi-line string escaping.

const works = ['http://placekitten.com/150/150', 'http://placekitten.com/200/200']

for (let i = 0; i < works.length; ++i) 
 $('#work').append(`
 <div class='col - sm '>
 <img class='img - responsive' src='$works[i]' alt=''>
 </div>
 `);
;

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="work"></div>

share|improve this answer

answered Mar 26 at 17:02

Andy HoffmanAndy Hoffman

11.2k33 gold badges2020 silver badges4141 bronze badges

add a comment | 

3

If you are in a position to use ES6, the template literal is here to remove the complexity/annoyance of multi-line string escaping.

const works = ['http://placekitten.com/150/150', 'http://placekitten.com/200/200']

for (let i = 0; i < works.length; ++i) 
 $('#work').append(`
 <div class='col - sm '>
 <img class='img - responsive' src='$works[i]' alt=''>
 </div>
 `);
;

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="work"></div>

share|improve this answer

answered Mar 26 at 17:02

Andy HoffmanAndy Hoffman

11.2k33 gold badges2020 silver badges4141 bronze badges

add a comment | 

If you are in a position to use ES6, the template literal is here to remove the complexity/annoyance of multi-line string escaping.

const works = ['http://placekitten.com/150/150', 'http://placekitten.com/200/200']

for (let i = 0; i < works.length; ++i) 
 $('#work').append(`
 <div class='col - sm '>
 <img class='img - responsive' src='$works[i]' alt=''>
 </div>
 `);
;

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="work"></div>

share|improve this answer

answered Mar 26 at 17:02

Andy HoffmanAndy Hoffman

11.2k33 gold badges2020 silver badges4141 bronze badges

const works = ['http://placekitten.com/150/150', 'http://placekitten.com/200/200']

for (let i = 0; i < works.length; ++i) 
 $('#work').append(`
 <div class='col - sm '>
 <img class='img - responsive' src='$works[i]' alt=''>
 </div>
 `);
;

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="work"></div>

const works = ['http://placekitten.com/150/150', 'http://placekitten.com/200/200']

for (let i = 0; i < works.length; ++i) 
 $('#work').append(`
 <div class='col - sm '>
 <img class='img - responsive' src='$works[i]' alt=''>
 </div>
 `);
;

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="work"></div>

const works = ['http://placekitten.com/150/150', 'http://placekitten.com/200/200']

for (let i = 0; i < works.length; ++i) 
 $('#work').append(`
 <div class='col - sm '>
 <img class='img - responsive' src='$works[i]' alt=''>
 </div>
 `);
;

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="work"></div>

share|improve this answer

answered Mar 26 at 17:02

Andy HoffmanAndy Hoffman

11.2k33 gold badges2020 silver badges4141 bronze badges

answered Mar 26 at 17:02

Andy HoffmanAndy Hoffman

11.2k33 gold badges2020 silver badges4141 bronze badges

answered Mar 26 at 17:02

Andy HoffmanAndy Hoffman

11.2k33 gold badges2020 silver badges4141 bronze badges

1

Take a look at the syntax highlight, you're closing the quotes and the interpreter is reading variables next to your string literals therefore throwing syntax errors, I'd encourage you to use ES6 Template literals or pay close attention to how you are using single and double quotes

Try this

for(var i = 0; i < works.length; ++i) 
 $('#work').append(`
 <div class='col-sm'>
 <img class='img-responsive' src='$works[i]>
 </div>
 `); //THIS IS THE END PARENTHESIS FOR .APPEND()
;

share|improve this answer

answered Mar 26 at 16:58

Gerardo SabettaGerardo Sabetta

2644 bronze badges

add a comment | 

1

Take a look at the syntax highlight, you're closing the quotes and the interpreter is reading variables next to your string literals therefore throwing syntax errors, I'd encourage you to use ES6 Template literals or pay close attention to how you are using single and double quotes

Try this

for(var i = 0; i < works.length; ++i) 
 $('#work').append(`
 <div class='col-sm'>
 <img class='img-responsive' src='$works[i]>
 </div>
 `); //THIS IS THE END PARENTHESIS FOR .APPEND()
;

share|improve this answer

answered Mar 26 at 16:58

Gerardo SabettaGerardo Sabetta

2644 bronze badges

add a comment | 

Take a look at the syntax highlight, you're closing the quotes and the interpreter is reading variables next to your string literals therefore throwing syntax errors, I'd encourage you to use ES6 Template literals or pay close attention to how you are using single and double quotes

Try this

for(var i = 0; i < works.length; ++i) 
 $('#work').append(`
 <div class='col-sm'>
 <img class='img-responsive' src='$works[i]>
 </div>
 `); //THIS IS THE END PARENTHESIS FOR .APPEND()
;

share|improve this answer

answered Mar 26 at 16:58

Gerardo SabettaGerardo Sabetta

2644 bronze badges

for(var i = 0; i < works.length; ++i) 
 $('#work').append(`
 <div class='col-sm'>
 <img class='img-responsive' src='$works[i]>
 </div>
 `); //THIS IS THE END PARENTHESIS FOR .APPEND()
;

share|improve this answer

answered Mar 26 at 16:58

Gerardo SabettaGerardo Sabetta

2644 bronze badges

answered Mar 26 at 16:58

Gerardo SabettaGerardo Sabetta

2644 bronze badges

answered Mar 26 at 16:58

Gerardo SabettaGerardo Sabetta

2644 bronze badges