jQuery how to get data from json with multiple loop

Question

I have json data like this http://pastebin.com/k3a5UNsL When viewed using the json formater, I want to take the data stored on the object using jquery inside possible_persons object.

I'm trying to loop into the data, but confused how do I get the data names, gender, addresses, etc...

When I use this code

function build_possible_persons(result) {
    var relationship = result['possible_persons'];
    $.each(relationship, function(key, value) {
        $.each(value, function(sec_key, sec_value) {
            alert(sec_value);
        });
    });
}

I get a result of the alert value of the results of the first loop, which as shown in Figure 0:25, 3788825f48716e6c ...., [object], [object], etc ....

How can I do a looping into the objects in it in order to get the value I want? I try

function build_possible_persons(result) {
    var relationship = result['possible_persons'];
    $.each(relationship, function(key, value) {
        $.each(value, function(sec_key, sec_value) {
            $.each(sec_key.names, function(third_key, third_value) {
                alert(third_key);
            });
        });
    });
}

or use

alert(sec_key.names[0].display);

But no luck. The code that I create show no results.

Please help to see the flaws in my code. Thanks

As a side remark, in an otherwise well-formed question I find the images actually distracting. Since your data is plentiful the paste is appropriate but rather than a picture of the tree I'd rather see a minified object that shows the relevant properties — JCall-Engineer
– JCall-Engineer, Commented Aug 14, 2016 at 4:05
I had wanted to make it easier to read the data. But all right, as input for my better future. Thank you for your help @Assimilater — Bertho Joris
– Bertho Joris, Commented Aug 14, 2016 at 4:08

Santiago Hernández · Accepted Answer · 2016-08-14 03:57:25Z

2

It could be simpler with Array.prototype.map()

function build_possible_persons(result) {
    var persons = result.possible_persons;
    alert(persons.map(p => p.addresses ? p.addresses[0].display : 'N/A').join("\n"))
}

EDIT: using arrow functions you could merge the person's data like this:

function build_possible_persons(result) {
  var persons = result.possible_persons;
  var rows = persons.map(p => {
    var address = p.addresses ? p.addresses[0].display : 'N/A';
    var name = p.names ? p.names[0].display : 'N/A'
    return name + ' ' + address; 
  });

  alert(rows.join('\n'));
}

edited Aug 14, 2016 at 3:57

answered Aug 14, 2016 at 3:43

Santiago Hernández

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5 Comments

Bertho Joris Over a year ago

Thanks for your help....I tried and get result. But how do I merge the address data with other data? Eg names. Usually if I use $.each I could create additional mechanisms to use $.append row for each row of data found. If you use the code that you created how do I get the data per one data? Thank you for your help

Santiago Hernández Over a year ago

answer updated ;), keep in mind that p => { in this situation is the same that function(p) { and instead of return name + ' ' + address; you could do $.append()

Bertho Joris Over a year ago

Thanks for your help @Santiago Cheers... ;)

Bertho Joris Over a year ago

Hi @Santiago how to make return value get and looping via $.each ? I tried but always fail

Santiago Hernández Over a year ago

you have to store the values in a variable out of your $.each and return it after $.each is done. Something like this: var result = []; $.each(values, function(i, value){ result.push(value)}); return result;

JCall-Engineer · Accepted Answer · 2016-08-14 04:02:09Z

1

The first problem I see is that in your second loop's sec_key is the text "names" (as well as your other properties. If you want to loop over names replace with:

$.each(value['names'], function(j, name)

Second problem: not all of your relationships have a property names so you need to check for this:

if(value['names']) {
    $.each(value['names'], function(....

finally each "name" is an object that has properties first and last. If you alert an object you're not going to see the actual names. You should make sure of the display property you have set or reconstruct the string

Bringing it all together:

$.each(relationship, function(i, value) {
    if (value['names'] !== undefined) {
        $.each(value['names'], function(j, name) {
            console.log(name.display);
        });
    }
});

answered Aug 14, 2016 at 4:02

JCall-Engineer

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2 Comments

Bertho Joris Over a year ago

Hi @Assimilater , how to query not only name data? I want to get all data, like gender, addresses, etc in one looping. Your sample when I tried only loop for value['names'] object. am I rigth?

JCall-Engineer Over a year ago

@BerthoJoris I had hoped that this would give you enough detail to help you with the rest. The way your data is formatted, gender is an object as opposed to an array like addresses and names. For gender just look at value['gender'].content and for addresses you can copy the inner loop for names. If this isn't enough I'd recommend a new question with more specificity on the desired output

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