如何从两个不同的表中选择记录

Question

我有两张桌子

+----------------------------+
| task                       |
+----------------------------+
| id | t_title | t_assign_to |
+----------------------------+

+-----------------------------+
| task_employee               |
+-----------------------------+
| id | emp_name | emp_underon |
+-----------------------------+

我的问题是，我想从带有外键( t_assign_to )的task_employee表中有emp_underon = '2‘的任务表中提取task_employee

<?php                
  $sql = mysql_query("SELECT * FROM task");/// i got all value from here

  $sql=mysql_query("SELECT * FROM task_employee WHERE emp_underon ='2'");// i got all value from here

  //$sql=mysql_query("SELECT * FROM task UNION SELECT * FROM task_employee WHERE emp_underon='2'");
  $count = mysql_num_rows($sql);
  while($row=mysql_fetch_assoc($sql))
  {
    echo $row['t_assign_to'];
  }
?>

Answer 1

我认为t_assign_to是task_employee的外键，所以我把它们连接在一起

方案1

$sql = "SELECT * FROM task t INNER JOIN task_employee te ON t.t_assign_to = te.id";

方案2

$sql = "SELECT * FROM task t,task_employee te WHERE t.t_assign_to = te.id";

如果表中的字段名与其他表中的其他字段相同，只需更改字段名如下：

$sql = "SELECT table.field as new_field_name FROM task t,task_employee te WHERE t.t_assign_to = te.id";

对你来说很方便

$sql = "SELECT * FROM task t INNER JOIN task_employee te ON t.t_assign_to = te.id WHERE te.emp_underon = '2' "
$q = mysql_query($sql);
$c = mysql_num_rows($q);
if($c > 0)
{
  while($r = mysql_fetch_assoc($q))
  {
    echo $r['t_assign_to'];
  }
}

Answer 2

我希望您能够通过这个查询获得您的答案，尽管我没有测试这个SQL。

select task.t_title, task.t_assign_to, task_employee.id as emp_id, task_employee.emp_name from task inner join task_employee on task.t_assign_to = task_employee.emp_underon where  task_employee.emp_underon = 2