mysql命令executescalar在c#中返回null

本文关键字：返回 null 命令 executescalar mysql | 更新日期: 2023-09-27 18:01:58

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name: John的单行实例，其中电话号码为'12%'。

在c#中，我编写了这个语法来获取PersonName变量作为查询结果。

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?

mysql命令executescalar在c#中返回null

我们肯定还遗漏了什么。根据您提供的代码尝试下面的代码示例:

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，错误被其他地方吞噬了。试试上面的方法，然后告诉我你有什么发现。

将(string)command.ExecuteScalar();改为 Convert.ToString(command.ExecuteScalar()); 

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 = Convert.ToString(command.ExecuteScalar());
connection.Close();

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