使用 jQuery $.ajax() 和 $.post() 将字符串数据发送到 MVC 控制器
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Sending String Data to MVC Controller using jQuery $.ajax() and $.post()
提问by MegaMatt
There's got to be something I'm missing. I've tried using $.ajax() and $.post() to send a string to my ASP.NET MVC Controller, and while the Controller is being reached, the string is null when it gets there. So here is the post method I tried:
一定有我遗漏的东西。我尝试使用 $.ajax() 和 $.post() 将字符串发送到我的 ASP.NET MVC 控制器，当到达控制器时，该字符串在到达那里时为空。所以这是我尝试的post方法：
$.post("/Journal/SaveEntry", JSONstring);
And here is the ajax method I tried:
这是我尝试过的ajax方法：
$.ajax({
url: "/Journal/SaveEntry",
type: "POST",
data: JSONstring
});
Here is my Controller:
这是我的控制器：
public void SaveEntry(string data)
{
string somethingElse = data;
}
For background, I serialized a JSON object using JSON.stringify(), and this has been successful. I'm trying to send it to my Controller to Deserialize() it. But as I said, the string is arriving as null each time. Any ideas?
作为背景，我使用 JSON.stringify() 序列化了一个 JSON 对象，这已经成功了。我正在尝试将它发送到我的控制器以反序列化（）它。但正如我所说，字符串每次都以空值到达。有任何想法吗？
Thanks very much.
非常感谢。
UPDATE:It was answered that my problem was that I was not using a key/value pair as a parameter to $.post(). So I tried this, but the string still arrived at the Controller as null:
更新：有人回答说我的问题是我没有使用键/值对作为 $.post() 的参数。所以我尝试了这个，但字符串仍然作为空值到达控制器：
$.post("/Journal/SaveEntry", { "jsonData": JSONstring });
回答by MegaMatt
Answered. I did not have the variable names set correctly after my first Update. I changed the variable name in the Controller to jsonData, so my new Controller header looks like:
回答。第一次更新后，我没有正确设置变量名称。我将 Controller 中的变量名称更改为 jsonData，因此我的新 Controller 标头如下所示：
public void SaveEntry(string jsonData)
and my post action in JS looks like:
我在 JS 中的 post 操作如下所示：
$.post("/Journal/SaveEntry", { jsonData: JSONstring });
JSONstring is a "stringified" (or "serialized") JSON object that I serialized by using the JSON pluginoffered at json.org. So:
JSONstring 是一个“字符串化”（或“序列化”）JSON 对象，我使用json.org 提供的JSON 插件对其进行序列化。所以：
JSONstring = JSON.stringify(journalEntry); // journalEntry is my JSON object
So the variable names in the $.post, and in the Controller method need to be the same name, or nothing will work. Good to know. Thanks for the answers.
所以 $.post 中的变量名和 Controller 方法中的变量名需要是相同的名称，否则什么都不起作用。很高兴知道。感谢您的回答。
回答by prodigitalson
Final Answer:
最终答案：
It seems that the variable names were not lining up in his post as i suggested in a comment after sorting out the data formatting issues (assuming that was also an issue.
在整理了数据格式问题（假设这也是一个问题）后，我在评论中建议的那样，变量名称似乎没有在他的帖子中排列。
Actually, make sure youre using the right key name that your serverside code is looking for as well as per Olek's example - ie. if youre code is looking for the variable data then you need to use data as your key. – prodigitalson 6 hours ago
@prodigitalson, that worked. The variable names weren't lining up. Will you post a second answer so I can accept it? Thanks. – Mega Matt 6 hours ago
实际上，请确保您使用的是服务器端代码正在寻找的正确密钥名称以及根据 Olek 的示例 - 即。如果您的代码正在寻找可变数据，那么您需要使用数据作为键。– prodigitalson 6 小时前
@prodigitalson，有效。变量名称没有对齐。您会发布第二个答案以便我接受吗？谢谢。– Mega Matt 6 小时前
So he needed to use a key/value pair, and make sure he was grabbing the right variable from the request on the server side.
所以他需要使用一个键/值对，并确保他从服务器端的请求中获取了正确的变量。
the data argument has to be key value pair
数据参数必须是键值对
$.post("/Journal/SaveEntry", {"JSONString": JSONstring});
回答by prodigitalson
It seems dataType is missed. You may also set contentType just in case. Would you try this version?
似乎缺少 dataType。您也可以设置 contentType 以防万一。你会试试这个版本吗？
$.ajax({
url: '/Journal/SaveEntry',
type: 'POST',
data: JSONstring,
dataType: 'json',
contentType: 'application/json; charset=utf-8'
});
Cheers.
干杯。
回答by ngwanevic
Thanks for answer this solve my nightmare.
感谢您的回答，这解决了我的噩梦。
My grid
我的格子
..
.Selectable()
.ClientEvents(events => events.OnRowSelected("onRowSelected"))
.Render();
<script type="text/javascript">
function onRowSelected(e) {
id = e.row.cells[0].innerHTML;
$.post("/<b>MyController</b>/GridSelectionCommand", { "id": id});
}
</script>
my controller
我的控制器
public ActionResult GridSelectionCommand(string id)
{
//Here i do what ever i need to do
}
回答by Md. Nazrul Islam
The Way is here.
道就在这里。
If you want specify
如果你想指定
dataType: 'json'
数据类型：'json'
Then use,
然后使用，
$('#ddlIssueType').change(function () {
var dataResponse = { itemTypeId: $('#ddlItemType').val(), transactionType: this.value };
$.ajax({
type: 'POST',
url: '@Url.Action("StoreLocationList", "../InventoryDailyTransaction")',
data: { 'itemTypeId': $('#ddlItemType').val(), 'transactionType': this.value },
dataType: 'json',
cache: false,
success: function (data) {
$('#ddlStoreLocation').get(0).options.length = 0;
$('#ddlStoreLocation').get(0).options[0] = new Option('--Select--', '');
$.map(data, function (item) {
$('#ddlStoreLocation').get(0).options[$('#ddlStoreLocation').get(0).options.length] = new Option(item.Display, item.Value);
});
},
error: function () {
alert("Connection Failed. Please Try Again");
}
});
If you do not specify
如果不指定
dataType: 'json'
数据类型：'json'
Then use
然后使用
$('#ddlItemType').change(function () {
$.ajax({
type: 'POST',
url: '@Url.Action("IssueTypeList", "SalesDept")',
data: { itemTypeId: this.value },
cache: false,
success: function (data) {
$('#ddlIssueType').get(0).options.length = 0;
$('#ddlIssueType').get(0).options[0] = new Option('--Select--', '');
$.map(data, function (item) {
$('#ddlIssueType').get(0).options[$('#ddlIssueType').get(0).options.length] = new Option(item.Display, item.Value);
});
},
error: function () {
alert("Connection Failed. Please Try Again");
}
});
If you want specify
如果你想指定
dataType: 'json' and contentType: 'application/json; charset=utf-8'
数据类型：'json' 和内容类型：'应用程序/json；字符集=utf-8'
Then Use
然后使用
$.ajax({
type: 'POST',
url: '@Url.Action("LoadAvailableSerialForItem", "../InventoryDailyTransaction")',
data: "{'itemCode':'" + itemCode + "','storeLocation':'" + storeLocation + "'}",
contentType: "application/json; charset=utf-8",
dataType: 'json',
cache: false,
success: function (data) {
$('#ddlAvailAbleItemSerials').get(0).options.length = 0;
$('#ddlAvailAbleItemSerials').get(0).options[0] = new Option('--Select--', '');
$.map(data, function (item) {
$('#ddlAvailAbleItemSerials').get(0).options[$('#ddlAvailAbleItemSerials').get(0).options.length] = new Option(item.Display, item.Value);
});
},
error: function () {
alert("Connection Failed. Please Try Again.");
}
});
回答by Jakub Lok?a
If you still can't get it to work, try checking the page URLyou are calling the $.postfrom.
如果您仍然无法使其工作，请尝试检查您正在调用的页面 URL$.post。
In my case I was calling this method from localhost:61965/Exampleand my code was:
在我的情况下，我正在调用这个方法localhost:61965/Example，我的代码是：
$.post('Api/Example/New', { jsonData: jsonData });
Firefox sent this request to localhost:61965/Example/Api/Example/New, which is why my request didn't work.
Firefox 将此请求发送到localhost:61965/Example/Api/Example/New，这就是我的请求无效的原因。
