通过从两个表中以Laravel中的where条件获取数据，vuejs

Question

[我想通过键入公司名称而不是公司ID通过搜索从员工和公司的两个表中获取数据，这些表与一对多的关系链接。它们的其余搜索功能运行完美。非常感谢您的帮助。

EmployeeController中的代码为：

 public function search(){

       if($search = \Request::get('q')){

             $employees = Employee::where(function($query) use ($search){
                 $companies = Company::Where('id','company_id');
                $searcompany ?????
                foreach($companies AS $company){ 
                     $searcompany = $company->Company;
                }

                $query->Where('BadgeCode','LIKE',"%$search%")              
                 ->orWhere($searcompany,'LIKE',"%$search%")

            })->orderBy('BadgeCode','desc')->paginate(20);
                            // return $users;
        }        else{
           $employees = Employee::latest()->paginate(5);
                     }
         return $employees;

    }

Employee.vue中的HTML代码是：

<div style="margin-left:450px; margin-top:0px;margin-bottom:-10px;">
              <h3 class="card-title">
                <div class="card-tools">
                  <div class="input-group input-group-sm" style="width: 250px;">
                    <input
                      name="table_search"
                      class="form-control float-right"
                      placeholder="Search"
                      @keyup="searchit"
                      v-model="search"
                      type="search"
                      aria-label="Search"
                    />
                    <div class="input-group-append">
                      <button type="submit" class="btn btn-default" @click="searchit">
                        <i class="fas fa-search"></i>
                      </button>
                    </div>
                  </div>
                </div>
              </h3>
            </div>

Employee.vue中的脚本代码是：

    export default {
    data() {
        return {
employees :{},
     search: "",
    },
        methods: {
            searchit: _.debounce(() => {
              Fire.$emit("searching");
            }, 300),
    },

    created() {
        Fire.$on("searching", () => {
          let query = this.search;
          axios
            .get("api/findemployee?q=" + query)
            .then(data => {
              this.employees = data.data;
            })

            .catch(() => {});
        });
    }
    }

API Rout为：

Route::get('findemployee','API\EmployeeController@search');

Answer 1

Controller方法看起来像是过度设计。您应该这样做。

$employees = Employee::with('Company')
    ->when($request->filled('q'), function ($query) use ($request) {
        return $query
           ->where('name', 'like', "%$request->q%")
           ->whereHas('Company', function ($query)  use ($request) {
                return $query->where('name', 'like', "%$request->q%");
            });
    })->paginate(5);

输入您的字段名称。