Php 如何在laravel中转换特定fromat
进入 在数据库中看起来是这样的。ofc可以有两个以上Php 如何在laravel中转换特定fromat,php,database,laravel,date,eloquent,Php,Database,Laravel,Date,Eloquent,进入 在数据库中看起来是这样的。ofc可以有两个以上 1 "2021/05/10" 2 "2021/05/14" 3 "2021/05/17" 4 "2021/05/20" 假设它是一个集合,您可以使用collapse()方法 $holiday = new Utility; $h = $request->get('holiday'); -> [ { "from": "20
1 "2021/05/10"
2 "2021/05/14"
3 "2021/05/17"
4 "2021/05/20"
假设它是一个集合,您可以使用
collapse()
方法
$holiday = new Utility;
$h = $request->get('holiday'); -> [ { "from": "2021/05/10"..
$h->map(函数($utility){return[$utility->from$utility->to];});
/*
说明\支持\收集{
全部:[
[
"2021/05/10",
"2021/05/14",
],
[
"2021/05/17",
"2021/05/20",
],
],
}
*/
$h->map(函数($utility){return[$utility->from$utility->to];})
->塌陷();
/*
说明\支持\收集{
全部:[
"2021/05/10",
"2021/05/14",
"2021/05/17",
"2021/05/20",
],
}
*/
$h->map(函数($utility){return[$utility->from$utility->to];})
->崩溃()
->全部();
/*
[
"2021/05/10",
"2021/05/14",
"2021/05/17",
"2021/05/20",
],
*/
如果
$h=$request->get('holiday')
不是一个集合,而是一个json字符串,如下所示
$h = $request->get('holiday');
/*
Illuminate\Support\Collection {
all: [
Utility {
+"from": "2021/05/10",
+"to": "2021/05/14",
},
Utility {
+"from": "2021/05/17",
+"to": "2021/05/20",
},
],
}
*/
您可以使用collect(json\u decode($h))
将其转换为一个集合,并执行相同的操作
->map(…)->collapse()->all()
序列
如果您想在刀片文件中回显这些值(带有编号),可以这样做:
[{"from":"2021/05/10","to":"2021/05/14"},{"from":"2021/05/17","to":"2021/05/20"}]
@foreach($h作为$index=>$value)
{{$index+1}{{$value}}
@endforeach
如何将其保存到数据库中?取决于您拥有的表/模型。它对数组$h=$request->get('holiday')上的成员函数map()抛出错误调用;是一个数组集合($h)->映射(。。。
[{"from":"2021/05/10","to":"2021/05/14"},{"from":"2021/05/17","to":"2021/05/20"}]
@foreach ($h as $index => $value)
{{ $index + 1 }} {{ $value }}<br>
@endforeach