Question

之前我曾问过这个问题，但答案并不重要，无法帮助我。

    $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
            }
        }
    }

我想让老师的ID通过，所以我可以把它放在代码底部附近的链接中。

这绝对是必要的，但是如果没有我的头疼，我似乎无法全部解决这个问题。

任何帮助将不胜感激！：）

Answer 1

我会将教师ID添加到他们的名字中：

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

然后当您分解教师姓名时，您将在索引2（姓氏，名字，身份证）上获得他们的ID。

通过一系列数组传递值

1 个答案: