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Can't parse JSON returned from MySQL


Can't parse JSON returned from MySQL

By : sebastian.kazmiercza
Date : October 18 2020, 01:08 AM
I hope this helps . You should just be able to access it as an object, so if result has fields name and title you can just access them as:
code :
var name = result.name
var title = result.title


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PHP Script fails to parse JSON returned by Google Maps Geocoder API if multiple results are returned by Geocoder

PHP Script fails to parse JSON returned by Google Maps Geocoder API if multiple results are returned by Geocoder


By : Lift
Date : March 29 2020, 07:55 AM
it should still fix some issue It's quite possible that the response gets over 10k characters, in that case you're trying to decode an invalid JSON string. Make sure you're reading the whole stream before decoding. The simplest way is to use file_get_contents instead of fopen/fread/fclose. On the long run you should switch to curl.
Ext.Error: Unable to parse the JSON returned by the server: You're trying to decode an invalid JSON String

Ext.Error: Unable to parse the JSON returned by the server: You're trying to decode an invalid JSON String


By : LaviJ
Date : March 29 2020, 07:55 AM
wish helps you JSON property names must be quoted to be considered valid. In addition, your data needs to be a property of an object, like {"data":[{"year":2005, "comedy":3400000....
EDIT: What Neil said is correct. Ext.decode does not require property names to be in quotes. However, the JSON standard does require them so you should definitely get in the habit of using them.
Django returned JSON, how to parse JSON Objects in js

Django returned JSON, how to parse JSON Objects in js


By : user2490735
Date : March 29 2020, 07:55 AM
hope this fix your issue Your data is already parsed. You have a list of JavaScript objects, there is no need to parse it again.
You used jQuery to load the data, and jQuery parses JSON responses for you when you use dataType: 'json' on the $.ajax() call; had you omitted that argument, jQuery would have auto-detected JSON responses by their mime type anyway and decoded for you then too.
How to efficiently Parse Json to Object when different variables returned in each json string

How to efficiently Parse Json to Object when different variables returned in each json string


By : MrCamoga
Date : March 29 2020, 07:55 AM
wish helps you I need some advice on JSON parsing in Java. For some real-time update, i get JSON response like following (server returned only the variables that has new values): , Use Google Gson.
code :
public class Product {
    @SerializedName("33") private String code;
    @SerializedName("170") private String name;
    @SerializedName("151") private String pointer;
    @SerializedName("98") private String price;
    @SerializedName("rate") private String rate;
    @SerializedName("132") private String value;
    @SerializedName("45") private String description;
    // setters, getters
}
    String str = "{\"33\":\"7153\", \"170\":\"AA\",\"151\":10}";
    Gson gson = new Gson();
    Product s = gson.fromJson(str, Product.class);
Unable to parse the JSON returned by the server: You're trying to decode an invalid JSON String: <?xml version="

Unable to parse the JSON returned by the server: You're trying to decode an invalid JSON String: <?xml version="


By : Plornt Xhizors
Date : March 29 2020, 07:55 AM
I wish this help you Maybe because you wrote some strange, custom JSON serialization that's probably is not right (removing trailing, ending characters? why?)
.NET JSON serialization is simple as that:
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