2011-02-13 51 views
0

我以前問過這個問題,但答案對幫助我並不重要。通過一系列數組傳遞價值

$sql = 'SELECT `name`, `course`, `id` FROM `teacher` ORDER BY `id` ASC'; 
    $rows = $mysql_conn->fetch_array($sql); 
    // Teacher's Table (id/name/link /course) 
    // Course default = 1-1-1-1-1-1-1-1 

    foreach($rows as $record) { 
     $result[$record['name']] = $record['course']; 
     //$result["Moore,Tyler"] = "1-1-1-1-1-1-1-1"; 
     //$result["Craig,Joey"] = "1-2-2-2-1-1-1-1"; 
     //$result["Degra,Tina"] = "2-1-1-1-2-1-1-1"; 
    } 

    foreach($result as $teacher=>$courses){ 
     $result[$teacher] = explode('-',$courses); // Remove -'s from courses and separate the array into sections 
     //$result["Craig,Joey"][0] = 1; 
     //$result["Craig,Joey"][1] = 2; 
     //$result["Craig,Joey"][2] = 2; 
     //$result["Craig,Joey"][3] = 2; 
     //$result["Craig,Joey"][4] = 1; 
     //$result["Craig,Joey"][5] = 1; 
     //$result["Craig,Joey"][6] = 1; 
     //$result["Craig,Joey"][7] = 1; 
    } 

    foreach($result as $teacher=>$courses){ 
     foreach($courses as $period => $course){ 
      if($course == $id) { // If course is equal to course page (selected course) record the period 1-8 
       $name = explode(',', $teacher); // $name[0] = 'Craig'/$name[1] = 'Joey'; 
       $result[$period][] = '<a href="?page=teacher&id=">'.$name[0].'<br />'.$name[1].'</a>'; 
       // I want id= to get an id passed to it from the query 
      } 
     } 
    } 

我想讓老師的ID通過,所以我可以將它放入代碼底部附近的鏈接中。

這是絕對必要的,但我似乎無法想象它全部沒有我的頭受傷。

任何幫助將不勝感激! :)

+2

您應該通過編輯和評論繼續跟蹤您的原始問題,無論是哪一個。 – BoltClock 2011-02-13 21:10:23

+0

原來是相當混亂的,所以我想我會清理它並重新發布。 – blanknamefornow 2011-02-13 21:11:54

回答

1

我想老師ID添加到他們的名字:

$sql = 'SELECT CONCAT(`name`, ",", `id`) AS `name`, `course` FROM `teacher` ORDER BY `id` ASC'; 

,然後當你爆老師的名字你就會有他們的索引2(姓氏,姓名,身份證)的標識。