Reputation: 17
Please Help me with how to send json object and receive it it views.py in django.
my script:
var obj={ 'search_name':search_name,'search_email':search_email };
jsonobj=JSON.stringify(obj);
//alert(jsonobj);
var xhr=new XMLHttpRequest();
xhr.open('POST',"viewprofile",true);
xhr.setRequestHeader("Content-Type", "application/json;charset=UTF-8");
xhr.send(jsonobj);
Upvotes: 1
Views: 2697
Reputation: 229
Using XMLHttpRequest() in plain JavaScript.
XMLHttpRequest is an API that provides client functionality for transferring data between a client and a server.
xhr = new XMLHttpRequest();
var url = "url";
xhr.open("POST", url, true);
xhr.setRequestHeader("Content-type", "application/json");
xhr.onreadystatechange = function () {
if (xhr.readyState == 4 && xhr.status == 200) {
var json = JSON.parse(xhr.responseText);
console.log(json.email + ", " + json.name)
}
}
var data = JSON.stringify({"email":"[email protected]","name":"LaraCroft"});
xhr.send(data);
Using AJAX Calls (preferred way)
$.ajax({
url: "https://youknowit.com/api/",
type: "POST",
data: { apiKey: "23462", method: "POST", ip: "208.74.35.5" },
dataType: "json",
success: function (result) {
switch (result) {
case true:
processResponse(result);
break;
default:
resultDiv.html(result);
}
},
error: function (xhr, ajaxOptions, thrownError) {
alert(xhr.status);
alert(thrownError);
}
});
Note: Keeping an eye on developer tools or firebug XHR debug process is good way to learn more about the api calls be it in any technology.
Upvotes: 1