Php Laravel+Ajax实现搜索栏
以下是我对ajax部分的看法Php Laravel+Ajax实现搜索栏,php,ajax,laravel,searchbar,Php,Ajax,Laravel,Searchbar,以下是我对ajax部分的看法 function getSearch(){ var sr_s_id= $('#sr_s_id').val(); var sr_name= $('#sr_name').val(); var sr_department= $('#sr_department').val(); var sr_semester= $('#sr_semester').val();
function getSearch(){
var sr_s_id= $('#sr_s_id').val();
var sr_name= $('#sr_name').val();
var sr_department= $('#sr_department').val();
var sr_semester= $('#sr_semester').val();
var request = new XMLHttpRequest();
$.ajaxSetup({
headers: {
'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content')
}
});
$.ajax({
type :"POST",
url :"{{url('/management/sutdent')}}",
dataType:"json",
data :{ s_id:sr_s_id,name:sr_name,department:sr_department,semester:sr_semester },
success :function(response) {
alert(response);
},
error: function(response) {
alert(response);
}
});
}
这是我的控制器
public function getSearch(Request $request){
$s_id =$request->s_id;
$name =$request->name;
$department = $request->department;
$semester =$request->semester;
$student_list=$this->UserRepository->getSearchdata($s_id,$name,$department,$semester);
//echo $student_list;
return response()->json($student_list);
}
这里是存储库
public function getSearchdata($s_id=null,$name=null,$department=null,$semester=null){
return $this->user
->select('s_id','name','email','user_info.updated_at as editDate','semester','USN','phone','address','Department')
->leftjoin('user_info','s_id','user_info.school_id')
->where(['r_id','=','1'],['s_id','like',$s_id],['name','like',$name],['department','like',$department],['semester','like',$semester])
->orderBy('s_id','asc')->get();
}
关于我的代码,我有两个问题。
首先,当我点击搜索栏按钮时,它会显示
消息:“数组到字符串转换”,异常:“ErrorException”,…}
我想这就是问题所在,但我不知道怎么回事
第二,是否有人知道repository.php中的请求值是否为null?然后,我需要使用laravel格式将结果推送到where
在通常的php文件中,我们将使用字符串组合。像
$Where ="something=".$a
if($something!=null)
{
$Where.=",something=".$something;
}
但我不知道如何用拉雷维尔语。
有谁能帮我解决这两个问题吗
新问题:Json输出。 再次显示第一个查询是否正常? 我是说
{s_id: "ym1234", name: "one", email: "test@test.com", editDate: "2018-04-12 05:54:32",…},…]
0
:
{s_id: "ym1234", name: "one", email: "test@test.com", editDate: "2018-04-12 05:54:32",…}
Department
:
"test"
USN
:
"student"
address
:
"CS123"
editDate
:
"2018-04-12 05:54:32"
email
:
"test@test.com"
name
:
"one"
phone
:
null
s_id
:
"ym1234"
semester
:
"105"
1
:
{s_id: "ym321", name: "two", email: "test@test.com", editDate: "2018-04-07 23:53:29",…}
Department
:
"test"
USN
:
"student"
address
:
"CS123"
editDate
:
"2018-04-07 23:53:29"
email
:
"test@test.com"
name
:
"two"
phone
:
null
s_id
:
"ym321"
semester
:
"106"}]
我想您忘记在where条件中添加
[]
将您的函数更改为
public function getSearchdata($s_id=null,$name=null,$department=null,$semester=null){
return $this->user
->select('s_id','name','email','user_info.updated_at as editDate','semester','USN','phone','address','Department')
->leftjoin('user_info','s_id','user_info.school_id')
->where([['r_id','=','1'],['s_id','like',$s_id],['name','like',$name],['department','like',$department],['semester','like',$semester]])
->orderBy('s_id','asc')->get();
}
我认为您需要将所有条件组合在一个数组中,用于where条件,如
where(['r_id','=','1',['s_id','like','s_id','like','s_id','name','like','name','department','like','department','sement','like','semterm].
第二,客户端代码中存在错误,可能会生成相同的错误,因为您将向json格式的对象发出警报。您指定的dataType:json
表示响应转换为json格式。如果您提醒JSON对象,那么错误是显而易见的
第二个问题是-
这就是我在代码片段中使用的内容-假设您应该在getSearchdata
函数中包含以下代码
$query = $this->user
->select('s_id','name','email','user_info.updated_at as editDate','semester','USN','phone','address','Department')
->leftjoin('user_info','s_id','user_info.school_id')
->where('r_id','=','1');
if($s_id)
{
$query->where('s_id','=',$s_id);
}
if($name)
{
$query->where('name','like','%'.$name.'%');
}
if($department)
{
$query->where('department','like','%'.$department.'%');
}
if($semester)
{
$query->where('semester','like','%'.$semester.'%');
}
return $query->orderBy('s_id','asc')->get();
让我知道,它是否适用于您…我认为您的
getSearchdata
功能不正确:
public function getSearchdata($s_id = null, $name = null, $department = null, $semester = null)
{
return $this->user
->select('s_id', 'name', 'email', 'user_info.updated_at as editDate', 'semester', 'USN', 'phone', 'address', 'Department')
->leftJoin('user_info', 's_id', 'user_info.school_id')
->where('r_id', '=', '1')
->where('s_id', 'like', $s_id)
->where('name', 'like', $name)
->where('department', 'like', $department)
->where('semester', 'like', $semester)
->orderBy('s_id', 'asc')->get();
}
错误显示的行是什么?消息:“数组到字符串转换”,异常:“ErrorException”,…}异常:“ErrorException”文件:“D:\XAMPP\htdocs\YangMing567\vendor\laravel\framework\src\light\Database\Query\Grammars\Grammar.php”行:198消息:“数组到字符串转换”只有这些表明你已经解决了所有可能的问题对不起,我两天前不在家。对两个问题都解决了。谢谢你的帮助!!这是工作,谢谢你的想法。@Zong感谢你接受回答并表示欢迎。你把我的答案投给…@VeshrajJoshi HI~我还有一个问题!关于我的乔森输出请。。。请随意