Question

我有一个样本2表。一个是学者，列出生日期和奖学金表。我在这里创造的是通过获得年龄范围的标准。使用条件 Age_from 和 Age_to 。我想得到/显示在该范围内找到的行。这段代码的错误可能是什么，因为返回总是空的。

非常感谢任何帮助。

数据库学者

scholar_id      Date_of_birth           Ship_level  
1               1991-12-19              Tertiary
2               1990-01-19              Secondary
3               2008-03-19              Primary
4               1992-21-19              Tertiary
5               1991-12-19              Tertiary
6               2000-10-12              Secondary
7               2001-12-23              Secondary
8               2009-12-19              Primary

数据库奖学金

ship_id     Age_from  Age_to    Level
1           10        16        Primary

我的代码

    public function ListOrgaScholar($ship_id)
    {
        $ship = Scholarship::find($ship_id);

        $Agefrom = $ship->Age_from;
        $Ageto   = $ship->Age_to;
        $Level   = $ship->Level;


        $scholars = (new Scholar)->newQuery()->select('*');
        if($Level)
            $scholars->where('Level', '=', $Level);

        $birthDates = $scholars->selectRaw('Date_of_birth');    
        $ag = [];
        foreach($birthDates as $birthDatee){
            list($m,$d,$y)=explode('-',$birthDatee->Date_of_birth);
            $age = Carbon::createFromDate($m,$d,$y)->age;
            array_push($ag,$age);
        }
        $scholars = $scholars->get();
}

Answer 1

您可以尝试使用

//use Carbon\Carbon; - import Carbon in your class/controller

public function ListOrgaScholar($ship_id)
{
    $ship = Scholarship::find($ship_id);


    $fromDate = Carbon::now()->subYears($ship->Age_from)->format('Y-m-d');
    $endDate = Carbon::now()->subYears($ship->Age_to)->format('Y-m-d');

    $scholars = Scholar::where('Ship_level', $ship->Level)
                ->whereBetween('Date_of_birth', [$fromDate, $toDate])
                ->get();        

}

显示与年龄范围条件相对应的行

1 个答案: