我正在尝试使用以下代码连接到我的远程
MySQL服务器.你能告诉我我做错了什么,因为用变量替换数据库连接信息似乎没有用.
using MysqL.Data.MysqLClient; string db_Server = "10.0.0.0"; string db_Name = "myDatabase"; string db_User = "myUser"; string db_Pass = "myPassword"; // Connection String MysqLConnection myConnection = new MysqLConnection("server = {0}; database = {1}; uid = {2}; pwd = {3}",db_server,db_Name,db_User,db_Pass);
作为PHP开发人员,我更喜欢使用上面的代码而不是下面的故意转换：
MysqLConnection myConnection = new MysqLConnection("server=10.0.0.0; database=myDatabase; uid=myUser; pwd=myPassword");
但正如你在这张图片中看到的那样,我得到了很多红色的波浪形：http://screencast.com/t/xlwoG9by
解决方法
您的参数顺序错误,应该是：
db_server,db_Pass
目前它是：
"server = {0}; database = {1}; uid = {2}; pwd = {3}"
db_Server db_User db_Pass db_Name
所以你的陈述应该是：
MysqLConnection myConnection = new MysqLConnection(string.Format( "server = {0}; database = {1}; uid = {2}; pwd = {3}",db_Server,db_Pass));
编辑：基于评论和讨论,你得到的错误是你正在尝试所有的课程级别的东西.您应该在方法中包含这些行,并在需要的地方调用该方法.就像是：
class MyClass
{
string db_Server = "10.0.0.0";
string db_User = "myUser";
string db_Pass = "myPassword";
string db_Name = "myDatabase";
public MysqLConnection GetConnection()
{
MysqLConnection myConnection = new MysqLConnection(string.Format(
"server = {0}; database = {1}; uid = {2}; pwd = {3}",db_Pass));
return myConnection;
}
}
