3 javascript django ajax json fetch-api
In my Project when conditions are insufficient my Django app send JSON response with message.
I use for this JsonResponse() directive,
Code:
data = {
'is_taken_email': email
}
return JsonResponse(data)
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Now I want using Javascript fetch API receive this JSON response and for example show alert.
I don't know how to use fetch API to do this. I want to write a listener who will be waiting for my JSON response from Django App.
I try:
function reqListener() {
var stack = JSON.parse(data);
console.log(stack);
}
var oReq = new XMLHttpRequest();
oReq.onload = reqListener;
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I want to compare JSON from my Django app with hardcoded JSON:
For example:
fetch( 'is_taken_email': email) - > then make something
OR
receive JSON from my Django app and as AJAX make it:
success: function(data) { if (data.is_taken_email) { make something; }
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Thanks in advance!
javascript 在全局窗口范围内提供了 fetch API,第一个参数是 API 的 URL,它是基于 Promise 的机制。
简单的例子
// url (required)
fetch('URL_OF_YOUR_API', {//options => (optional)
method: 'get' //Get / POST / ...
}).then(function(response) {
//response
}).catch(function(err) {
// Called if the server returns any errors
console.log("Error:"+err);
});
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在你的情况下,如果你想接收JSON 响应
fetch('YOUR_URL')
.then(function(response){
// response is a json string
return response.json();// convert it to a pure JavaScript object
})
.then(function(data){
//Process Your data
if (data.is_taken_email)
alert(data);
})
.catch(function(err) {
console.log(err);
});
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使用基于侦听器的示例XMLHttpRequest
// url (required)
fetch('URL_OF_YOUR_API', {//options => (optional)
method: 'get' //Get / POST / ...
}).then(function(response) {
//response
}).catch(function(err) {
// Called if the server returns any errors
console.log("Error:"+err);
});
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使用 Listener 的示例setInterval(我不确定你想要做这样的事情,只是与你分享)
fetch('YOUR_URL')
.then(function(response){
// response is a json string
return response.json();// convert it to a pure JavaScript object
})
.then(function(data){
//Process Your data
if (data.is_taken_email)
alert(data);
})
.catch(function(err) {
console.log(err);
});
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我对 Django 不太熟悉,但希望这可以帮助你。
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