我有一个 mysql 左连接查询。该查询运行正常。但左连接表没有获取值。请通过以下查询。我正在使用 C# 工作
void getstuinfo()
{
try
{
MySqlCommand com = new MySqlCommand("select stumaster.stuname,"+
"stumaster.lname,"+
"stumaster.fname,"+
"stumaster.mname,"+
"stumaster.fa_calty,"+
"stumaster.sex,"+
"castmaster.castdisp,"+
"stumaster.castcode,"+
"stumaster.nwscs "+
"from stumaster "+
" left join castmaster on stumaster.castcode = castmaster.castcode "+
" where grno = " + Convert.ToInt32(textBox1.Text).ToString(), con_db.con);
MySqlDataReader dr1 = com.ExecuteReader();
if (dr1.HasRows)
{
while (dr1.Read())
{
textBox2.Text = (dr1[("stuname")].ToString());
textBox4.Text = (dr1["lname"]).ToString();
textBox5.Text = (dr1["fname"]).ToString();
textBox6.Text = (dr1["mname"]).ToString();
comboBox5.Text = (dr1["fa_calty"]).ToString();
comboBox1.Text = (dr1["castdisp"]).ToString();
textBox7.Text = (dr1["castcode"]).ToString();
textBox9.Text = (dr1["nwscs"]).ToString();
string wsex = (dr1["sex"]).ToString();
if (wsex == "M")
{
radioButton1.Checked = true;
}
else
{
radioButton2.Checked = true;
}
}
dr1.Close();
}
else
{
MessageBox.Show("Not a Valid G.R.No.", " Student Information ");
dr1.Close();
textBox1.Focus();
return;
}
}
catch (FormatException)
{
MessageBox.Show("Date is Invalid ");
}
}
最佳答案
你必须用 grno 给出表名
即
stumaster.grno = conditiion
关于c# - Left Join Mysql 查询结果错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42969356/