Question

我在树查询中有多个值。我希望得到数组的所有3个值并在表格中显示结果..不幸的是我只有第3个记录，我已经甚至在第二个数组中回显表，但什么也没有显示..有什么方法可以做到吗？

 $sql = mysql_query("select userid from dept_user where lead_id=61");

 while($res = mysql_fetch_array($sql)){

  // $user = $res['userid'];
    $a = $res['userid'];

    if ($a !== 0){
      echo "1st: " . $a;
      echo "<br/>";
      $s = mysql_query("select userid from dept_user where lead_id=$a");

      while($res = mysql_fetch_array($sql)) {
        $q = $res['userid'];
        echo "<br/>";

        if($q !==0) {
          echo  "Second: " . $q;
          echo "<br/>";
          $f = mysql_query("select userid from dept_user where lead_id=$q");

          while($res = mysql_fetch_array($f)) {
            $u = $res['userid'];
            $n = mysql_query("Select user.firstname, user.lastname, department.dept_name, job_title.title, role.rolename, user.userid From user Inner Join  dept_user On user.userid = dept_user.userid Inner Join  department On department.dept_id = dept_user.dept_id Inner Join  job_title On job_title.title_id = user.titleid Inner Join  role On role.roleid = user.roleid Where user.userid = $u "); 
            echo "<table border='1'>
<tr>

<th>id</th>
<th>first name</th>
<th>last name</th>
<th>dept name</th>
<th>title</th>
<th>rolename</th>
</tr>";

            while($row = mysql_fetch_array($n)) {

              echo "<tr>";
              echo "<td>" . $row['userid'] . "</td>";
              echo "<td>" . $row['firstname'] . "</td>";
              echo "<td>" . $row['lastname'] . "</td>";
              echo "<td>" . $row['dept_name'] . "</td>";
              echo "<td>" . $row['title'] . "</td>";
              echo "<td>"  . $row['rolename'] . "</td>";
              echo "</tr>";
            }

          echo "</table>";
        }
      }
    }
  }
}

Answer 1

尝试在$ res到$ res1和$ res2

的第二个和第三个循环中重命名变量

在单个表上显示结果

1 个答案: