将JSON响应放入var中并输出

Question

I have a form which sends Zip code + housenumber to an external page and the external page sends me a json response with the address information back.

This all works but how can i grab this json response put it into a variable and output it again on my page.

Here i do a post to a page

function validate() {
  var postc = $('#postc').val();
  var huisnr = $('#huisnr').val();
  $.ajax({
    url: 'postcodecheck.php',
    type: 'POST',
    dataType: 'json',
    data: 'postc=' + postc + '&huisnr=' + huisnr,
    cache: false
  });

and i get this response in the header of postcodecheck.php

{"location":[{"city":"Amsterdam","postcode":"1012NX","straat":"Kalverstraat","nummer":1}]}

How can i grab this reponse and put it into a variable ?

Any help would be highly appriciated.

Answer 1

$.ajax({

    url: 'postcodecheck.php',
    type: 'POST',
    dataType: 'json',
    data: 'postc=' + postc + '&huisnr=' + huisnr,
    cache: false,
    success: function(data) {
        $.each(data.location, function(index, value) {

            console.log("City " + value.city)
            console.log("postcode " + value.postcode)
            console.log("straat " + value.straat)
        })
    }
});

Add success and inside success you can iterate on the data

Demo

Answer 2

you can use request callback to get the retrieved data & put it into a variable

$.ajax({ 
            url: 'postcodecheck.php',
            type: 'POST',
            dataType: 'json',
            data: 'postc=' + postc + '&huisnr=' + huisnr,
                    cache: false
})
.done(function(data) {
console.log(data);});

Answer 3

The ajax call have another attribute called success you need to use it.

var store = '';
$.ajax({
    url: 'postcodecheck.php',
    type: 'POST',
    dataType: 'json',
    data: 'postc=' + postc + '&huisnr=' + huisnr,
    cache: false,
    success: function(result){
            store = result;
        }
});

Result

console.log(store); //{"location":[{"city":"Amsterdam","postcode":"1012NX","straat":"Kalverstraat","nummer":1}]}

Answer 4

You need to use the done method and pass data as an argument to the callback function that will then contain the response from the server.

 $.ajax({
     url: 'postcodecheck.php',
     type: 'POST',
     dataType: 'json',
     data: 'postc=' + postc + '&huisnr=' + huisnr,
     cache: false
   }).done(function(data) {
        var result = data;
        alert(result);
   });