parsererror SyntaxError: JSON.parse: unexpected non-whitespace character after JSON data at line 1 column 190 of the JSON data

Question

When I click button I have above JSON error. It doesn't return any values. can you please help me to solve this,

Here my Jquery code,

$(document).ready(function(){
  $('.package_details').click(function(){

    $('.txt-form').slideDown("slow");
    var package = $(this).attr("id");

    $.ajax({
      type:"post",
      url:"aaa.php", 
      dataType: 'json',
      data:"pack="+pack,
      success:function(html){

        $('.package_name').val(html.package_name);
        $('.package_price').val(html.package_price);

        $('.discount_percentage').val(html.discount);
        $('.amount_after_discount').val(html.amnt_discount);
        $('.tax_percentage').val(html.tax);
        $('.amount_after_tax').val(html.amnt_tax);

        if(html.package_type == 'Contact Visible'){
          $('.contact').attr('checked', 'checked');
        }else if(html.package_type == 'Job post'){
          $('.job').attr('checked', 'checked');
        }else if(html.package_type == 'Both'){
          $('.both').attr('checked', 'checked');
        }

        $('.contact_visible').val(html.count);
        $('.job_post').val(html.job_post);
        $('.valitity_day').val(html.valitity);
      },
    error: function (jqXHR, textStatus, errorThrown) {
        alert('Not done - ' + textStatus + ' ' + errorThrown);
    }
    })
  })
})

And Here my PHP code,

echo json_encode(array("package_name"=>$package_name,"package_type"=>$package_type,"discount"=>$discount,"amnt_discount"=>$amnt_discount,"tax"=>$tax,"amnt_tax"=>$amnt_tax,"count"=>$count,"job_post"=>$job_post,"package_price"=>$package_price,"valitity"=>$valitity));

see what is there in html by using console.log(html) in success function — Shahrukh
– Shahrukh, Commented Apr 21, 2017 at 6:32
these are all textbox values when ajax function go to the success handler that values may replaces json values — dhinesh balu
– dhinesh balu, Commented Apr 21, 2017 at 6:36

ManiMuthuPandi · Accepted Answer · 2017-04-21 06:36:40Z

0

Create json variable like below then post this in data field

post_data =  {  "pack": pack};

$.ajax({
      type:"post",
      url:"aaa.php", 
      dataType: 'json',
      data:post_data,
.
.
.

answered Apr 21, 2017 at 6:36

ManiMuthuPandi

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1 Comment

dhinesh balu Over a year ago

I tried as per your code. But still now i got same error.

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