jquery ajax获取示例
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jquery ajax get example
提问by user48408
At the moment I'm using the post method like this
目前我正在使用这样的 post 方法
$.ajax({
type: "POST",
url: "Servicename.asmx/DoSomeCalculation",
data: "{param1ID:"+ param1Val+"}",
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function(msg) {
UseReturnedData(msg.d);
},
error: function(err) {
alert(err.toString());
if (err.status == 200) {
ParseResult(err);
}
else { alert('Error:' + err.responseText + ' Status: ' + err.status); }
}
});
Am I correct in believing that if I use a GET request instead of POST the behavior will change to being a synchronous request i.e. the execution will wait until the response has been received from the server??
我是否认为如果我使用 GET 请求而不是 POST 行为将更改为同步请求，即执行将等到从服务器收到响应？
Can somebody show me a jquery GET example calling a webmethod of a web service directly?
有人可以向我展示一个直接调用 Web 服务的 webmethod 的 jquery GET 示例吗？
UPDATE: Using the async flag as suggested below is really all i needed to do so this works for me. I'm still curious as to what work needs to be done to the code above to make it a GET request. Changing type: "GET" doesn't have the desired effect!
更新：使用下面建议的 async 标志真的是我需要做的一切，这对我有用。我仍然很好奇需要对上面的代码做哪些工作才能使其成为 GET 请求。更改类型：“GET”没有预期的效果！
采纳答案by peirix
You can decide if you want the ajax call to be async or not using this:
您可以决定是否希望 ajax 调用异步或不使用：
$.ajax({
async: false/true,
//more options
});
回答by Greg
To answer your first point, no: GET and POST are independent of synchronous / asynchronous.
要回答您的第一点，否：GET 和 POST 独立于同步/异步。
You can use the boolean asyncmethod to control this.
您可以使用布尔async方法来控制它。
回答by Moulesh Kumar
There is a "async" flag for making the ajax call synchronous or asynchronous. You can define it as:
有一个“异步”标志用于使 ajax 调用同步或异步。您可以将其定义为：
$.ajax({
async: false/true,
//rest of code
});
$.ajax({
async: false/true,
//rest of code
});
回答by Ashouri
look this sample maybe help you
看看这个样本可能对你有帮助
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("myDiv").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","ajax_info.txt",true);
xmlhttp.send();
