如何通过 jquery ajax 以 json 格式保存数据

Question

空数据是通过jquery从html页面传输的，请帮助我 这是我的 api 代码 asp.net 核心，它工作正常

[HttpPost]
public async Task<ActionResult<LeaveSubmit>> Submitdata()
{
    try
    {
        LeaveSubmit lobj = new LeaveSubmit();
        //  string myobj = Convert.ToString(lobj);
        JObject jsonObject = JObject.FromObject(lobj);
        if (jsonObject == null)
        {
            return BadRequest();
        }
        else
        {
            var json = System.IO.File.ReadAllText(jsonFile);
            var jsonObj = JObject.Parse(json);
            var experienceArrary = jsonObj.GetValue("Notes") as JArray;
            var newnotes = jsonObject;
            experienceArrary.Add(newnotes);
            jsonObj["Notes"] = experienceArrary;
            string newJsonResult = Newtonsoft.Json.JsonConvert.SerializeObject(jsonObj, Newtonsoft.Json.Formatting.Indented);
            System.IO.File.WriteAllText(jsonFile, newJsonResult);

            //  var response = await newJsonResult;
            return Ok("Success");
        }
    }
    catch (Exception ex)
    {
        this.telemetryClient.TrackException(ex);
        return this.StatusCode(StatusCodes.Status500InternalServerError, ex.Message);
    }
}

当我点击提交按钮发送到 api 时的 jquery 代码为空，因此空数据保存在 json 页面中 //数据保存。//`$("#btnSubmit").click(function () { 数据传递（）； });

    function datapass() {
        var values = {};

        var values = $("#notesval").val();

        var myvalue = JSON.stringify(values);

        $.ajax({
            url: window.location.origin + "/api/ServiceApi/Submitdata",
            type: 'POST',
            dataType: 'json',
            data: myvalue,
            contentType: 'application/json; charset=utf-8',
            success: function () {

                //console.log("Savesuccessful");
                //console.log(data);
                alert("Record Save Succesfully");

                //$(".loader").hide();
            },
            error: function (error) {
                alert("Error is occured");
                //console.log("My errror values:", error);

            }
        });
    }
   // Date save close.//`
    

Answer 1

您需要执行 3 个步骤：

1.您的 C# 代码应如下所示：

[HttpPost]
public async Task<ActionResult<LeaveSubmit>> Submitdata([FromBody] MyClass myModel)
{
    try
    {
        LeaveSubmit lobj = new LeaveSubmit();
        //  string myobj = Convert.ToString(lobj);
        JObject jsonObject = JObject.FromObject(lobj);
        if (jsonObject == null)
        {
            return BadRequest();
        }
        else
        {
            var json = System.IO.File.ReadAllText(jsonFile);
            var jsonObj = JObject.Parse(json);
            var experienceArrary = jsonObj.GetValue("Notes") as JArray;
            var newnotes = jsonObject;
            experienceArrary.Add(newnotes);
            jsonObj["Notes"] = experienceArrary;
            string newJsonResult = Newtonsoft.Json.JsonConvert.SerializeObject(jsonObj, Newtonsoft.Json.Formatting.Indented);
            System.IO.File.WriteAllText(jsonFile, newJsonResult);

            //  var response = await newJsonResult;
            return Ok("Success");
        }
    }
    catch (Exception ex)
    {
        this.telemetryClient.TrackException(ex);
        return this.StatusCode(StatusCodes.Status500InternalServerError, ex.Message);
    }
}

基本上添加一个参数来保存输入。

2.您需要添加一个表示输入的类，如下所示：

public class MyClass
{
      public string MyProperty {get; set;}
}

3.您的 ajax 调用应如下所示：

function datapass() {
        var values = { MyProperty:$("#notesval").val() };               

        var myvalue = JSON.stringify(values);

        $.ajax({
            url: window.location.origin + "/api/ServiceApi/Submitdata",
            type: 'POST',
            dataType: 'json',
            data: myvalue,
            contentType: 'application/json; charset=utf-8',
            success: function () {

                //console.log("Savesuccessful");
                //console.log(data);
                alert("Record Save Succesfully");

                //$(".loader").hide();
            },
            error: function (error) {
                alert("Error is occured");
                //console.log("My errror values:", error);

            }
        });
    }