带有 JSON 数据的 Javascript HTTP POST
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Javascript HTTP POST with JSON data
提问by anaconda_wly
Can I send a request as below? With parameters being assigned with a JSON style object. I only get error. But when I use a REST client and choose RAW data, it's OK. I guess I must have written incorrect code. How to send raw JSON data in JavaScript? Could anyone help me?
我可以发送如下请求吗？使用 JSON 样式对象分配参数。我只得到错误。但是当我使用 REST 客户端并选择 RAW 数据时，就可以了。我想我一定写了不正确的代码。如何在 JavaScript 中发送原始 JSON 数据？有人可以帮助我吗？
xmlhttp = new XMLHttpRequest();
var url = "https://someURL";
xmlhttp.open("POST", url, true);
xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xmlhttp.onreadystatechange = function () { //Call a function when the state changes.
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
alert(xmlhttp.responseText);
}
}
var parameters = {
"username": "myname",
"password": "mypass"
};
// Neither was accepted when I set with parameters="username=myname"+"&password=mypass" as the server may not accept that
xmlhttp.send(parameters);
回答by Quentin
No. The send()method can take a number of different argument types, but a plain object is not one of them (so it will probably end up having toString()being called on it and being turned into "[Object object]").
不。该send()方法可以采用许多不同的参数类型，但普通对象不是其中之一（因此它可能最终会toString()被调用并变成"[Object object]"）。
If you want to send JSON then you must:
如果要发送 JSON，则必须：
- Say you are sending JSON:
xmlhttp.setRequestHeader("Content-type", "application/json"); - Convert your JavaScript object to a string of JSON text:
var parameters = JSON.stringify({"username":"myname","password":"mypass"}); - Be prepared to accept JSON instead of application/x-www-form-urlencoded data on the server side.
- 假设您要发送 JSON：
xmlhttp.setRequestHeader("Content-type", "application/json"); - 将您的 JavaScript 对象转换为一串 JSON 文本：
var parameters = JSON.stringify({"username":"myname","password":"mypass"}); - 准备好在服务器端接受 JSON 而不是 application/x-www-form-urlencoded 数据。
Also note that, since you are using an absoluteURI, you may run into cross domain issues.
另请注意，由于您使用的是绝对URI，您可能会遇到跨域问题。
回答by mhdnp1234
xmlhttp = new XMLHttpRequest();
xmlhttp.open("POST", AjaxURL, true);
xmlhttp.onreadystatechange = function () { //Call a function when the state changes.
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
cb(xmlhttp.responseText);
}
};
xmlhttp.send(JSON.stringify(Idata));
