我有一个web服务，并在其中有两种方法(select,insertdata)。我想用jquery在sql中插入一条记录。我怎样才能做到这一点？我已经制作了该代码，但它不起作用。请帮助我。我如何添加项目在sql中使用jQuery(ajax)通过web服务

我的web服务

Sqlconeection dd = new Sqlconeection();

int rowsInserted = 0;

[WebMethod]

[ScriptMethod(ResponseFormat = ResponseFormat.Json)]

public string insertRecord(Int64 mbn,string name,double amt,bool Notify)

{

SqlConnection connection = dd.getconnection();

SqlCommand cmd = new SqlCommand("InsertData");

cmd.Connection = connection;

if (connection.State == ConnectionState.Closed)

connection.Open();

// string inser = "insert into expensive(mobileNumber,Name,Amount,Notify)values('" + mbn + "','" + name + "','" + amt + "','" + Notify + "')";

//cmd.CommandText = inser;

cmd.Parameters.AddWithValue("@mobile", mbn);

cmd.Parameters.AddWithValue("@Name", name);

cmd.Parameters.AddWithValue("@Amount", amt);

cmd.Parameters.AddWithValue("@Notify", Notify);

cmd.CommandType = CommandType.StoredProcedure;

rowsInserted= cmd.ExecuteNonQuery();

return string.Format("Thank you, {0} number of rows inserted!", rowsInserted);

}

[WebMethod(Description = "Returns all Products")]

[ScriptMethod(ResponseFormat = ResponseFormat.Json)]

public DataTable selectdata(Int64 mnb)

{

DataTable dt = new DataTable();

SqlConnection conn = dd.getconnection();

SqlCommand cmd = new SqlCommand();

cmd.Connection = conn;

if (conn.State == ConnectionState.Closed)

conn.Open();

string select = "select mobileNumber,Name,Amount,Notify from expensive where mobileNumber='" + mnb + "'";

cmd.CommandText = select;

SqlDataReader dr = cmd.ExecuteReader();

dt.TableName = "expensive";

dt.Load(dr);

return dt;

}

我API代码

$(function() {

$('#btnSubmit').click(function() {

var mob = $('#txtmo').val();

var Nm = $('#txtName').val();

var amout = $('#txtAmt').val();

var notify = $('#txtnoty').val();

$.ajax({

type: "POST",

url: "InsertData.asmx/insertRecord",

data: "{ mob: '" + mob + "', Nm: '" + Nm + "',amount:'" + amout + "',notify:'" + notify + "'}",

contentType: "application/json; charset=utf-8",

dataType: "jsonp",

success: function (data) {

var obj = data.d;

if (obj == 'true') {

$('#txtmo').val('');

$('#txtName').val('');

$('#txtAmt').val('');

$('#txtnoty').val('');

$('#lblData').html(JSON.stringify(data.d));

}

},

error: function (r) {

console.log(r);

},

});

});

});

2015-10-15

Raj Gola

+0

请提供一些更多的细节，并显示为代码剪断，而不是整个程序。 –

+0

您的数据对象属性名称与您webmethod的启动项不匹配。 –

+0

同样在您的成功中，您检查webmethod是否返回'true'，但它实际上会返回成功消息。 –