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Returning ID with information MySQL/PHP

开发者 https://www.devze.com 2023-02-10 10:24 出处:网络
$sql= \'SELECT `name`, `course`, `id` FROM `teacher`\'; $rows = $mysql_conn->fetch_array($sql); foreach($rows as $record) {
$sql  = 'SELECT `name`, `course`, `id` FROM `teacher`';
$rows = $mysql_conn->fetch_array($sql);
foreach($rows as $record) {
    $search_results[$record['name']] = $record['course'];
}

foreach($search_results as $teacher=>$string_courses){
    $array_courses = explode('-',$string_courses);
    $search_results[$teacher] = $array_courses;
}

foreach($search_results as $teacher=>$courses){
    foreach($courses as $period => $course){
        if($course == $id) {
            $name               = explode(',', $teacher);
            $results开发者_开发技巧[$period][] = '<a href="?page=teacher&id='.ID HERE.'">'.$name[0].'<br />'.$name[1].'</a>';
        }
    }
}

Where it says ID HERE, I want the ID returned for each teacher inserted there but I do not know how to insert it so that it passes through.


You're building your array in a weird way, using $record['name'] as a key, change that to $record['id'] and you have a better start.

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