Question

我有一个文本框，我试图用来搜索数据库，我想要实现的是当我输入出生日期（例如14/06/1996）时textbox（searchdob）它将在数据库中显示具有此出生日期值的用户。我已经尝试过使用我的AppointmentController来执行where条件，以便foreach循环保持整洁。

错误是：非静态方法Symfony \ Component \ HttpFoundation \ Request :: get（）不应该静态调用，假设$ this来自不兼容的上下文

AppointmentController

function addAppointment()
{
    $doctors = Doctor::all();
    $search = Request::get('searchdob');
    $users = User::where('role',  '=',  1)
                   ->where('dateofbirth', 'LIKE', '%'.$search.'%')
                   ->get();

    return view('appointment/addappointmentform',['users'=>$users],['doctors'=>$doctors]);
}

addappointment.blade

<form>
    Insert Patients date of birth
    <input
    type="text"
    name="searchdob"
    id="searchdob"
    placeholder="dd/mm/yyyy"
    onkeyup="
        var v = this.value;
        if (v.match(/^\d{2}$/) !== null) {
            this.value = v + '/';
        } else if (v.match(/^\d{2}\/\d{2}$/) !== null) {
            this.value = v + '/';
        }"
    maxlength="10"
></form>


    <fieldset>
<legend>Select the Patient</legend> 
@foreach($users as $user)
    <div>
        <label for="dirBtn{{$user->id}}">
        <input id="dirBtn{{$user->id}}" type="radio" name="user" value="{{$user->id}}">
        {{$user->firstname}}
        </label>
    </div>
@endforeach
</fieldset>

Answer 1

将其更改为：

$search = request('searchdob');

Laravel-MYSQL搜索数据库中的值，然后显示它们

1 个答案: