1
你好,這是我在laravel中的控制器,我沒有在這裏找到問題。它會檢查文件是否手動輸入值爲1的數據集,但是如果我使用系統執行該操作,則不會更新該值,並且在未選中框的情況下該框會保持爲0。laravel中的複選框控制器
我想要做的是勾選框,如果用戶選中它並保留框勾選,直到他勾選它,再加上更新數據庫值。謝謝你提前
public function tickoffUpload($id, $type, User $user) {
$uploads = $this->upload->get($id);
// if($this->request->isMethod('get')) {
// if($user->can('untick', $uploads)) {
// $uploads = $this->upload = 0;
// } else {
// return $uploads = $this->upload = 1;
// }
// }
if($this->request->isMethod('get')) {
if($type == 0) {
$uploads = $this->upload = 1;
} else if($type == 1){
$uploads = $this->upload = 0;
}
}
if(isset($_POST["uploaded"]) && !empty($_POST["uploaded"])) {
$uploads->uploaded = $this->request->uploaded;
$uploads->save();
}
return redirect('/home');
}
我覺得你的問題是從這一行來:'$上傳= $這個 - >上傳= 1;',您正在分配值到$ uploads這是一個對象而不是變量,所以你應該做這樣的事情:'$ uploads-> variable_to_update = $ this-> upload = 1;'。應該用'$ uploads-> uploaded = $ this-> request-> uploaded;' – d3r1ck
'中的相應變量替換,它仍然不能正常工作......它不更新數據庫中的值,它仍然爲0當我點擊複選框,並且它仍然不會與數據庫值爲0.這就是我做的 '$ uploads-> uploaded = $ this-> upload = 1;'因爲上傳的是分配0和1的值我的View.php '' –