0

假设我有一个如下所示的查询结果：

ID    NAME    Phone
----  ----    -----
1     John    123456
2     John    125678
3     John    345678
4     Abby    456789
5     Abby    567890

我只想返回名称的单行实例：John，其中电话号码类似于“12%”。

在 c# 中，我编写了这个语法来获取 PersonName 变量作为查询的结果。

MySqlConnection connection = new MySqlConnection("SERVER=" + "localhost" + ";" + "DATABASE=" + "testdb" + ";" + "UID=" + "root" + ";" + "PASSWORD=" + "" + ";");
MySqlCommand command = new MySqlCommand();    

    connection.Open();
    string selectQuery = "SELECT NAME FROM testtable WHERE Phone LIKE '12%' ORDER BY ID LIMIT 1";
    command.Connection = connection;
    command.CommandText = selectQuery;
    string PersonName = (string)command.ExecuteScalar();
    connection.Close();

我不知道我的代码有什么问题，但 PersonName 返回 null。我做错了什么？

4

2 回答 2

1

我们必须在这里遗漏其他东西。根据您提供的内容尝试以下代码示例：

try {
MySqlConnection connection = new MySqlConnection("SERVER=localhost;DATABASE=testdb;UID=root;PASSWORD=;");
MySqlCommand command = new MySqlCommand();    

connection.Open();
string selectQuery = "SELECT NAME FROM testtable WHERE Phone LIKE '12%' ORDER BY ID LIMIT 1";
command.Connection = connection;
command.CommandText = selectQuery;
string PersonName = (string)command.ExecuteScalar();
}
catch (Exception ex) {
MessageBox.Show(ex.Message);
}
finally {
    connection.Close();
}

我有一种感觉，由于某种原因，对 .Open() 的调用失败了，并且错误正在其他地方被吞没。试试上面的方法，让我知道你发现了什么。

于 2015-07-13T16:15:28.830 回答
0

(string)command.ExecuteScalar();这样做：改变这个 Convert.ToString(command.ExecuteScalar());

MySqlConnection connection = new MySqlConnection("SERVER=" + "localhost" + ";" + "DATABASE=" + "testdb" + ";" + "UID=" + "root" + ";" + "PASSWORD=" + "" + ";");

MySqlCommand 命令 = new MySqlCommand();

connection.Open();
string selectQuery = "SELECT NAME FROM testtable WHERE Phone LIKE '12%' ORDER BY ID LIMIT 1";
command.Connection = connection;
command.CommandText = selectQuery;
string PersonName = Convert.ToString(command.ExecuteScalar());
connection.Close();
于 2021-04-09T21:00:36.387 回答