Javascript SyntaxError: missing

Question

I'm trying to put 2 scripts in 1 js file and i'm getting:

SyntaxError: missing } after property list
note: { opened at line 9, column 19

But as far as I check all the curly brackets are closed, not sure what the real issue is.

Code

// country
jQuery( document ).ready( function( $ ) {
    $('select[name="country"]').on('change', function() {
      $.ajaxSetup({
          headers: { 'X-CSRF-TOKEN': $('meta[name="_token"]').attr('content') }
      });
       var CountryId = $(this).val();
        if(CountryId) {
            $.ajax({
              url: '{{ url('getprovinces') }}/'+encodeURI(CountryId),
              type: "GET",
              dataType: "json",
              success:function(data) {
                $('select[name="province"]').empty();
                var options = data.map(function(state) {
                    return $("<option class='form-control'>").val(state.id)
                                        .text(state.name);
                });
                $('select[name="province"]').empty().append(options);
              }
            });
        }else{
          $('select[name="province"]').empty().append("<option class='form-control' value='' selected>Select</option>");
        }
    });
});

error comes from this line:

url: '{{ url('getprovinces') }}/'+encodeURI(CountryId),

Any idea?

I think, your code is not in laravel blade, it is in a separate js file. Don't use blade syntax in javascript — Tamil Selvan C
– Tamil Selvan C, Commented Dec 24, 2018 at 4:35
"{{ url('getprovinces') }}/"+encodeURI(CountryId) This will work — Rajender Joshi
– Rajender Joshi, Commented Dec 24, 2018 at 4:38

kingdaro · Accepted Answer · 2018-12-24 04:34:38Z

1

You need to escape the quotes or use a double-quoted string. Otherwise JS thinks the string ends on the quote after url(' then gets confused when a variable name pops up.

answered Dec 24, 2018 at 4:34

kingdaro

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Comments

Pyae Phyoe Shein · Accepted Answer · 2018-12-24 04:42:12Z

0

Try like that:

// country
jQuery( document ).ready( function( $ ) {
    $('select[name="country"]').on('change', function() {
      $.ajaxSetup({
          headers: { 'X-CSRF-TOKEN': $('meta[name="_token"]').attr('content') }
      });
       var CountryId = $(this).val();
        if(CountryId) {
            $.ajax({
              url: '{{' + url('getprovinces') + '}}/' + encodeURI(CountryId),
              type: "GET",
              dataType: "json",
              success:function(data) {
                $('select[name="province"]').empty();
                var options = data.map(function(state) {
                    return $("<option class='form-control'>").val(state.id)
                                        .text(state.name);
                });
                $('select[name="province"]').empty().append(options);
              }
            });
        }else{
          $('select[name="province"]').empty().append("<option class='form-control' value='' selected>Select</option>");
        }
    });
});

edited Dec 24, 2018 at 4:42

answered Dec 24, 2018 at 4:36

Pyae Phyoe Shein

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7 Comments

Pyae Phyoe Shein Over a year ago

what is value of url('getprovinces')?

mafortis Over a year ago

this is the route Route::get('/getprovinces/{countryid}','Front\AddressLists@getprovinces'); it returns provinces as json

Pyae Phyoe Shein Over a year ago

so far, above my code is not problem. there should be problem url('getprovinces')

Pyae Phyoe Shein Over a year ago

wth..!! why did you pass those kinda json data to url? better post method instead for those getprovinces data.

mafortis Over a year ago

so you suggest i change my route to post?

|

holydragon · Accepted Answer · 2018-12-24 04:56:13Z

0

Try

url: url('getprovinces') + '/' + encodeURI(CountryId),

answered Dec 24, 2018 at 4:56

holydragon

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