Laravel 从关系中检查值
Laravel Check for Value from Relation
我有一个看起来像这样的查询,我在其中获取特定位置的各种企业的数据,我需要能够判断每个企业有(或没有)女性员工。
$business = Business::where('location', $location)
->with(['staff'])
->get();
return MiniResource::collection($business);
我的迷你资源是这样的:
return [
'name' => $this->name,
'location' => $this->location,
'staff' => PersonResource::collection($this->whenLoaded('staff')),
];
这是示例响应的样子:
{
"id": 1,
"name": "XYZ Business"
"location": "London",
"staff": [
{
"name": "Abigail",
"gender": "f",
"image": "https://s3.amazonaws.com/xxxx/people/xxxx.png",
"role": "Project Manager",
},
{
"name": "Ben",
"gender": "m",
"image": "https://s3.amazonaws.com/xxxx/people/xxxx.png",
"role": "Chef",
},
]
}
我真的不需要 staff 数组,我只想检查关系中是否存在女性,然后 return 类似这样的东西:
{
"id": 1,
"name": "XYZ Business"
"country": "USA",
"has_female_employee": true;
}
有没有 eloquent 方法来实现这个?
注意:在我的原始代码中,我查询了更多关系,但我不得不将此 post 限制在我的问题范围内。
像下面这样更改您的代码并查看
$business = Business::where('location', $location)
->with(['staff'])
->where('gender', 'f')
->get();
return [
'name' => $this->name,
'location' => $this->location,
'has_female_employee' => empty($this->whenLoaded('staff')) ? false : true,
];
如果你只是在寻找男性或女性工作人员,你可以这样实现:
$someQuery->whereHas('staff', function ($query) {
$query->where('gender', 'f');
})
如果您想要两种性别,我不会在查询中经历实现这一点的麻烦,但建议减少 MiniResource 中的结果集合:
return [
'name' => $this->name,
'location' => $this->location,
'has_female_employee' => $this->whenLoaded('staff')->reduce(
function ($hasFemale, $employee) {
$hasFemale = $hasFemale || ($employee->gender === 'f');
return $hasFemale;
}, false),
];
为了提高可读性,更好的方法是将其创建为 MiniResource 上的方法。
我有一个看起来像这样的查询,我在其中获取特定位置的各种企业的数据,我需要能够判断每个企业有(或没有)女性员工。
$business = Business::where('location', $location)
->with(['staff'])
->get();
return MiniResource::collection($business);
我的迷你资源是这样的:
return [
'name' => $this->name,
'location' => $this->location,
'staff' => PersonResource::collection($this->whenLoaded('staff')),
];
这是示例响应的样子:
{
"id": 1,
"name": "XYZ Business"
"location": "London",
"staff": [
{
"name": "Abigail",
"gender": "f",
"image": "https://s3.amazonaws.com/xxxx/people/xxxx.png",
"role": "Project Manager",
},
{
"name": "Ben",
"gender": "m",
"image": "https://s3.amazonaws.com/xxxx/people/xxxx.png",
"role": "Chef",
},
]
}
我真的不需要 staff 数组,我只想检查关系中是否存在女性,然后 return 类似这样的东西:
{
"id": 1,
"name": "XYZ Business"
"country": "USA",
"has_female_employee": true;
}
有没有 eloquent 方法来实现这个?
注意:在我的原始代码中,我查询了更多关系,但我不得不将此 post 限制在我的问题范围内。
像下面这样更改您的代码并查看
$business = Business::where('location', $location)
->with(['staff'])
->where('gender', 'f')
->get();
return [
'name' => $this->name,
'location' => $this->location,
'has_female_employee' => empty($this->whenLoaded('staff')) ? false : true,
];
如果你只是在寻找男性或女性工作人员,你可以这样实现:
$someQuery->whereHas('staff', function ($query) {
$query->where('gender', 'f');
})
如果您想要两种性别,我不会在查询中经历实现这一点的麻烦,但建议减少 MiniResource 中的结果集合:
return [
'name' => $this->name,
'location' => $this->location,
'has_female_employee' => $this->whenLoaded('staff')->reduce(
function ($hasFemale, $employee) {
$hasFemale = $hasFemale || ($employee->gender === 'f');
return $hasFemale;
}, false),
];
为了提高可读性,更好的方法是将其创建为 MiniResource 上的方法。