如何在AJAX成功连接中调用函数背后的代码
[英]How to call code behind function in success junction of AJAX
问题描述
I'm consuming a web service using ajax in web form 3.5, then save it to database. 
我正在使用Web表单3.5中的Ajax使用Web服务，然后将其保存到数据库中。 I'm getting a success response from the web service. 我从Web服务获得成功响应。 However, how will I call my code behind function to save the data after success in ajax? 但是，在ajax成功后，我将如何调用函数背后的代码来保存数据？ I'm new to ajax. 我是ajax的新手。
<body>
<form id="form1" runat="server">
<div id ="divName" runat="server" class="form-group">
<label class="control-label">Name</label>
<asp:TextBox ID="txtname" CssClass="form-control" runat="server" ></asp:TextBox>
</div>
<input id="btnRegister" type="button" value="Register" /><br />
</form>
<script type ="text/javascript">
$(document).ready(function () {
$("#btnRegister").click(function () {
debugger;
var namee = document.getElementById("<%=txtname.ClientID %>").value;
Register(namee);
});
function Register(name) {
debugger;
var id =
$.ajax({
type: "POST",
url: "https://webservice.com/register",
data: JSON.stringify({
"name": name
}),
contentType: "application/json",
datatype: "json",
success: function (response) {
//call code behind function in saving data.
},
failure: function (response) {
alert("ERROR");
}
});
}
});
</script>
</body>
4 个解决方案
解决方案1
0 已采纳 2019-09-11 06:35:01
you just should to call another ajax again in success section. 您只应该在成功部分再次调用另一个ajax。 like this : 像这样 ：
function Register(name) {
debugger;
var id =
$.ajax({
type: "POST",
url: "https://webservice.com/register",
data: JSON.stringify({
"name": name
}),
contentType: "application/json",
datatype: "json",
success: function (response) { MySaveDataFunc(response) },
failure: function (response) {
alert("ERROR");
}
});
}
var MySaveDataFunc=function(mydata)
{
$.ajax({
type: "POST",
url: "Your SaveData Url",
data: mydata
contentType: "application/json",
datatype: "json",
success: function (response) {
},
failure: function (response) {
}
});
}
I hope it has been helpful . 希望对您有所帮助。
解决方案2
0 2019-09-11 06:41:00
Ajax code: Ajax代码：
$(document).ready(function () { $.ajax({ url: "Registration.aspx/GetData", type:"POST", contentType: "application/json; charset=utf-8", dataType: "json", success: function (data) { $('#responseContent').text(data.d); } }); });
解决方案3
0 2019-09-11 06:53:20
For these type of problems. 对于这些类型的问题。 It's good to create a separate Js file as it keeps the code clean and understandable. 最好创建一个单独的Js文件，因为它可以使代码清晰易懂。 And please be more precise with the questions. 并且请更精确地回答问题。
From my understanding => if this ie " https://webservice.com/register " one sends a success response, you need to save it to say " https://localhost/register " in your own system. 以我的理解=>如果此“ https://webservice.com/register ”发送成功响应，则需要将其保存为在您自己的系统中说“ https：// localhost / register ”。 create a varible url in the global scope of your field and set that url to your ajax req. 在您的字段的全局范围内创建一个可变网址，并将该网址设置为您的ajax请求。
var yourUrl = "something"; //on ajax send method url: yourUrl //on success method, call the same method but change the url to your new url, this way you save both the time and redundant code Register(name,yourUrl); //add a second parameter to take url in your register function
解决方案4
0 2019-09-11 07:05:40
You can directly use shor-hand method of ajax for post request. 您可以直接使用ajax的shor-hand方法进行发布请求。 $(selector).post(URL,data,function(data,status,xhr),dataType); $（selector）.post（URL，data，function（data，status，xhr），dataType）; For Example in your case: 例如您的情况：
$.post(url:"https://webservice.com/register"
,data:JSON.stringify({"name": name})
,function(data,status,xhr){
//Your callback function
//Example: alert(data.Message);
});
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