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JS synchronous not firing in order

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1

I have a JavaScript function which is not firing in order. It fires the CheckForAddRecordsToAddFromDSS method, then the if (check1 || check2) statement before even firing the CheckForMissingRecordsFromSupp method.

Why is this happening?

$.getJSON("/Home/CheckForMissingRecordsFromSupp", function (data) 
 check1 = data;
 ).done($.getJSON("/Home/CheckForAddRecordsToAddFromDSS", function (data) 
 check2 = data;
 ).done(function () )
 );

javascript json

share|improve this question

asked Mar 28 at 18:28

BeengieBeengie

86333 gold badges99 silver badges2727 bronze badges

add a comment
 | 

1

I have a JavaScript function which is not firing in order. It fires the CheckForAddRecordsToAddFromDSS method, then the if (check1 || check2) statement before even firing the CheckForMissingRecordsFromSupp method.

Why is this happening?

$.getJSON("/Home/CheckForMissingRecordsFromSupp", function (data) 
 check1 = data;
 ).done($.getJSON("/Home/CheckForAddRecordsToAddFromDSS", function (data) 
 check2 = data;
 ).done(function () )
 );

javascript json

share|improve this question

asked Mar 28 at 18:28

BeengieBeengie

86333 gold badges99 silver badges2727 bronze badges

add a comment
 | 

I have a JavaScript function which is not firing in order. It fires the CheckForAddRecordsToAddFromDSS method, then the if (check1 || check2) statement before even firing the CheckForMissingRecordsFromSupp method.

Why is this happening?

$.getJSON("/Home/CheckForMissingRecordsFromSupp", function (data) 
 check1 = data;
 ).done($.getJSON("/Home/CheckForAddRecordsToAddFromDSS", function (data) 
 check2 = data;
 ).done(function () )
 );

javascript json

share|improve this question

asked Mar 28 at 18:28

BeengieBeengie

86333 gold badges99 silver badges2727 bronze badges

$.getJSON("/Home/CheckForMissingRecordsFromSupp", function (data) 
 check1 = data;
 ).done($.getJSON("/Home/CheckForAddRecordsToAddFromDSS", function (data) 
 check2 = data;
 ).done(function () )
 );

share|improve this question

asked Mar 28 at 18:28

BeengieBeengie

86333 gold badges99 silver badges2727 bronze badges

share|improve this question

asked Mar 28 at 18:28

BeengieBeengie

86333 gold badges99 silver badges2727 bronze badges

asked Mar 28 at 18:28

BeengieBeengie

86333 gold badges99 silver badges2727 bronze badges

asked Mar 28 at 18:28

BeengieBeengie

86333 gold badges99 silver badges2727 bronze badges

1 Answer
1

active

oldest

votes

5

Why is this happening?

.done expects to be passed a function. You are passing the return value of $.getJSON which is not a function.

Consider the following example:

foo(bar())

Even without knowing anything about foo or bar, we definitely know that bar is executed before foo and bar's return value is passed to foo.

Your code should be:

$.getJSON("/Home/CheckForMissingRecordsFromSupp", function(data) 
 check1 = data;
).done(function() // <- function
 $.getJSON("/Home/CheckForAddRecordsToAddFromDSS", function(data) 
 check2 = data;
 ).done(function() check2) 
 location.reload();
 
 );
);

Having said that, since .done also receives the network response, you don't need to pass a callback to $.getJSON. You can just write:

$.getJSON("/Home/CheckForMissingRecordsFromSupp").done(function(check1) 
 $.getJSON("/Home/CheckForAddRecordsToAddFromDSS").done(function(check2) check2) 
 location.reload();
 
 );
);

share|improve this answer

edited Mar 28 at 18:38

answered Mar 28 at 18:36

Felix KlingFelix Kling

591k136136 gold badges907907 silver badges971971 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Reference: api.jquery.com/deferred.done
  
  – Aravind Voggu
  Mar 28 at 18:38
  

  
  
  
  

add a comment
 | 

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5

Why is this happening?

.done expects to be passed a function. You are passing the return value of $.getJSON which is not a function.

Consider the following example:

foo(bar())

Even without knowing anything about foo or bar, we definitely know that bar is executed before foo and bar's return value is passed to foo.

Your code should be:

$.getJSON("/Home/CheckForMissingRecordsFromSupp", function(data) 
 check1 = data;
).done(function() // <- function
 $.getJSON("/Home/CheckForAddRecordsToAddFromDSS", function(data) 
 check2 = data;
 ).done(function() check2) 
 location.reload();
 
 );
);

Having said that, since .done also receives the network response, you don't need to pass a callback to $.getJSON. You can just write:

$.getJSON("/Home/CheckForMissingRecordsFromSupp").done(function(check1) 
 $.getJSON("/Home/CheckForAddRecordsToAddFromDSS").done(function(check2) check2) 
 location.reload();
 
 );
);

share|improve this answer

edited Mar 28 at 18:38

answered Mar 28 at 18:36

Felix KlingFelix Kling

591k136136 gold badges907907 silver badges971971 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Reference: api.jquery.com/deferred.done
  
  – Aravind Voggu
  Mar 28 at 18:38
  

  
  
  
  

add a comment
 | 

5

Why is this happening?

.done expects to be passed a function. You are passing the return value of $.getJSON which is not a function.

Consider the following example:

foo(bar())

Even without knowing anything about foo or bar, we definitely know that bar is executed before foo and bar's return value is passed to foo.

Your code should be:

$.getJSON("/Home/CheckForMissingRecordsFromSupp", function(data) 
 check1 = data;
).done(function() // <- function
 $.getJSON("/Home/CheckForAddRecordsToAddFromDSS", function(data) 
 check2 = data;
 ).done(function() check2) 
 location.reload();
 
 );
);

Having said that, since .done also receives the network response, you don't need to pass a callback to $.getJSON. You can just write:

$.getJSON("/Home/CheckForMissingRecordsFromSupp").done(function(check1) 
 $.getJSON("/Home/CheckForAddRecordsToAddFromDSS").done(function(check2) check2) 
 location.reload();
 
 );
);

share|improve this answer

edited Mar 28 at 18:38

answered Mar 28 at 18:36

Felix KlingFelix Kling

591k136136 gold badges907907 silver badges971971 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Reference: api.jquery.com/deferred.done
  
  – Aravind Voggu
  Mar 28 at 18:38
  

  
  
  
  

add a comment
 | 

Why is this happening?

.done expects to be passed a function. You are passing the return value of $.getJSON which is not a function.

Consider the following example:

foo(bar())

Even without knowing anything about foo or bar, we definitely know that bar is executed before foo and bar's return value is passed to foo.

Your code should be:

$.getJSON("/Home/CheckForMissingRecordsFromSupp", function(data) 
 check1 = data;
).done(function() // <- function
 $.getJSON("/Home/CheckForAddRecordsToAddFromDSS", function(data) 
 check2 = data;
 ).done(function() check2) 
 location.reload();
 
 );
);

Having said that, since .done also receives the network response, you don't need to pass a callback to $.getJSON. You can just write:

$.getJSON("/Home/CheckForMissingRecordsFromSupp").done(function(check1) 
 $.getJSON("/Home/CheckForAddRecordsToAddFromDSS").done(function(check2) check2) 
 location.reload();
 
 );
);

share|improve this answer

edited Mar 28 at 18:38

answered Mar 28 at 18:36

Felix KlingFelix Kling

591k136136 gold badges907907 silver badges971971 bronze badges

foo(bar())

$.getJSON("/Home/CheckForMissingRecordsFromSupp", function(data) 
 check1 = data;
).done(function() // <- function
 $.getJSON("/Home/CheckForAddRecordsToAddFromDSS", function(data) 
 check2 = data;
 ).done(function() check2) 
 location.reload();
 
 );
);

$.getJSON("/Home/CheckForMissingRecordsFromSupp").done(function(check1) 
 $.getJSON("/Home/CheckForAddRecordsToAddFromDSS").done(function(check2) check2) 
 location.reload();
 
 );
);

share|improve this answer

edited Mar 28 at 18:38

answered Mar 28 at 18:36

Felix KlingFelix Kling

591k136136 gold badges907907 silver badges971971 bronze badges

answered Mar 28 at 18:36

Felix KlingFelix Kling

591k136136 gold badges907907 silver badges971971 bronze badges

answered Mar 28 at 18:36

Felix KlingFelix Kling

591k136136 gold badges907907 silver badges971971 bronze badges