我是Laravel的新手,正在为我的练习开发小应用程序。我正在做求职功能。这个错误给了我很多麻烦,让我很困惑。
public function job_search(Request $request) {
$search_skill_set = $request->job_skills;
$search_results = JobPost::whereRaw('FIND_IN_SET(?, job_skills)', $search_skill_set)
->get()
->toArray();
for ($i = 0; $i < count($search_results); $i++) {
$department_id = (int)$search_results[$i]['department_name'];
$department_name = Department::select('department_name')
->where('id', '=', $department_id)
->get()
->toArray();
// the next statement raises an Undefined:offset 1 error
$search_results[$i]['department_name_info'] = $department_name[$i]['department_name'];
}
var_dump($search_results);
}
我没有得到我做错的地方,所以来自给定代码段的任何建议以及代码中的任何修改
答案 0 :(得分:0)
更改此行:
$search_results[$i]['department_name_info'] = $department_name[$i]['department_name'];
到
$search_results[$i]['department_name_info'] = $department_name[0]['department_name'];
答案 1 :(得分:0)
for ($i=0; $i < count($search_results) ; $i++) {
$department_id = (int)$search_results[$i]['department_name'];
//I am getting department id correct here
$department_name = Department::select('department_name')->where('id','=',$department_id)->get()->toArray();
//$depratment_name is also going okay and working
$search_results[$i]['department_name_info'] = $department_name[0]['department_name'];
// This line should have a static index.
}