使用 POST 请求替代 window.location.href - ASP.NET MVC

Question

我正在用 Javascript 编写下面的代码来检索评论并在重新填充数据后导航同一页面

function MoveItem() {

var empId = document.getElementById('EMP_ID').value;
var commentValue = $("#RESPONSE").val();
if ($.trim(commentToSave).length > 0) {
    showAjaxLoading();
    var empData= "APPROVE";
    var baseControllerUrl = '/Employee/EmpManagement/PushItem';
    window.location.href = baseControllerUrl + "/" + empId + "?comment=" + commentValue + "&empData=" + empData + "&currentItem=" + itemData;

} else {
    aet('Pelase enter comments', 'E');
  }
}

在控制器中方法写成

    [HttpGet]
    public async Task<ActionResult> MoveItem(int id, string comment, string decision, string currentworkflow)
    {
        return RedirectToAction("EditEmpManagemnt", "EmpManagement", new { id = id });
   }

我想将 MoveItem 操作方法转换为 [HttpPost] 类型，在 Javascript 和操作方法中需要进行哪些更改？谁能用示例代码解释一下。

Answer 1

您可以使用fetch 方法发送POST 请求。我们以 JSON 格式发送数据。用这个替换你的javascript：

function MoveItem() {
    var empId = document.getElementById('EMP_ID').value;
    var commentValue = $("#RESPONSE").val();
    if ($.trim(commentToSave).length > 0) {
        showAjaxLoading();
        var empData = "APPROVE";
        var baseControllerUrl = '/Employee/EmpManagement/PushItem';

        var url = baseControllerUrl + "/" + empId;

        //here we put the url as a first parameter
        //then we configure the http method
        //and in the body we pass in the object in JSON format
        fetch(url, {
            method: "POST",
            body: JSON.stringify({
                "comment": commentValue,
                "empData": empData,
                "currentItem": itemData
            })
        }).then(result => {
            //do something with the result you get back
            console.log("Completed with result: " + result);
        }).catch(err => {
            //if any error occured, then show it here
            console.log("There is an error: " + err);
        });
        
    } else {
        aet('Please enter comments', 'E');
    }
}

在您的 C# 代码中，只需将 [HttpGet] 属性替换为 [HttpPost]