Question

我想使用mysql_fetch_array的结果更新mysql表中列的值

但最后的结果只是插入/更新到列

中的结果

我错在哪里？这是我的代码。提前致谢

    $studentname="some value";
    $course="some value";
    $query=mysql_query("select SABJEK,GRADE,REMARKS from table where STUDENTNAME='$studentname' && STUDENTNUMBER IS NULL") or die(mysql_error());
    while($result=mysql_fetch_array($query))
        {
        $sabjek=$result['SABJEK'];
        $grade=$result['GRADE'];
        $remarks=$result['REMARKS'];
        $msg="$sabjek = $grade - $remarks ";
        $msg1="$studentname $course $msg";
        }
    mysql_query("update table2 set `msg`='$msg1' where studentname='$studentname'") or die(mysql_error());

Answer 1

将mysql_query放入while中。基本上取代

  }
mysql_query("update table2 set `msg`='$msg1' where studentname='$studentname'") or die(mysql_error());

与

    mysql_query("update table2 set `msg`='$msg1' where studentname='$studentname'") or die(mysql_error());
}

它会起作用：）

Answer 2

  If you want to update the column name every-time whenvere you fetch the result just put update query inside the while loop,   
   $studentname="some value";
   $course="some value";
   $query=mysql_query("select SABJEK,GRADE,REMARKS from table where STUDENTNAME='$studentname' && 
   STUDENTNUMBER IS NULL") or die(mysql_error());
   while($result=mysql_fetch_array($query));
     { $sabjek=$result['SABJEK'];
       $grade=$result['GRADE'];
       $remarks=$result['REMARKS'];
       $msg="$sabjek = $grade - $remarks ";
       $msg1="$studentname $course $msg";        
       mysql_query("update table2 set `msg`='$msg1' where studentname='$studentname'") or         
       die(mysql_error());
     }

如果你不期待这个结果，那么请详细解释我，我会更新答案。

感谢。

从mysql_fetch_array结果PHP更新列值

2 个答案: