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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! :)
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
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