Question

现在它没有显示任何数据；

控制台： finished loading: GET "http://todolist.local/teachers/search?text=a".

当用户在搜索中输入内容时，我试图在tbody中显示结果。

Ajax代码：

<script>
$(document).ready(function(){
        $('#searchname').on('keyup', function(){
                var text = $('#searchname').val();
                $.ajax({
                        type:"GET",
                        url: 'teachers/search',
                        data: {text: $('#searchname').val()},
                        success:function(data){
$('tbody').html(data);
}
                });
        });
});
</script>

web.php：

Route::get('/search', 'TeachersController@ajaxsearch');

搜索控制器：

public function ajaxsearch(){
$searchname = Input::get ( 'searchname' );
  if($searchname != ""){
  $teacher = Teacher::where ( 'efirst', 'LIKE', '%' . $searchname . '%' )->paginate(10);
  return response()->json($teacher);
}
}

视图：

<div class="input-group stylish-input-group">
                        <input type="text" id="searchname" name="searchname" class="form-control"  placeholder="Search..." >
                   <span class="input-group-addon">
                            <button type="submit" class="btn btn-primary">
                                <span class="glyphicon glyphicon-search">Search</span>
                            </button>
                        </span>
                    </div>

Answer 1

这应该可以解决问题。

HTML：

  <div id="datasearch"></div>

JS：

$(function(){
    $('#searchname').on('keyup', function(){
        $.get('/teachers/search/'+$(this).val(), function(response){
            $('#datasearch').html(response);
        });
    });
});

控制器：

public function ajaxsearch(string $value = null){
    return $value ? Teacher::whereRaw("UPPER(efirst) LIKE '%".strtoupper($value)."%'")->paginate(10) : [];
}

路线：

Route::get('/teachers/search/{value?}', 'TeachersController@ajaxsearch');

聊天后版本

HTML：

  <ul id="datasearch"></ul>

JS：

$(function(){
    var $datasearch=$('#datasearch');
    $('#searchname').on('keyup', function(){
        $.get('/teachers/search/'+$(this).val(), function(teachers){
            $datasearch.empty();
            for (var i=0; i<teachers.length; i++){
                $datasearch.append('<li>'+teachers[i].efirst+' <a href="/teachers/edit/'+teachers[i].id+'">edit</a></li>');
            }
        });
    });
});

控制器：

public function ajaxsearch(string $value = null){
    return $value ? Teacher::select('id','efirst')->whereRaw("UPPER(efirst) LIKE '".strtoupper($value)."%'")->offset(0)->limit(10)->get() : [];
}

路线：

Route::get('/teachers/search/{value?}', 'TeachersController@ajaxsearch');

Answer 2

您知道jquery ajax中的错误功能吗？

$(document).ready(function(){
        $('#searchname').on('keyup', function(){
                var text = $('#searchname').val();
                $.ajax({
                        type:"GET",
                        url: 'teachers/search',
                        data: {text: $('#searchname').val()},
                        success:function(data){$('tbody').html(data);},
                        error:function(jqXHR){alert(jqXHR.status);} 
                });
        });
});

您可以尝试执行此操作，如果有一些键，并且文本应该以逗号分隔，则它将显示错误，因为它是键值对。

Answer 3

在搜索控制器中，您正在访问错误的输入名称。应该是这样

public function ajaxsearch(){
$searchname = Input::get ( 'text' );
  if($searchname != ""){
  $teacher = Teacher::where ( 'efirst', 'LIKE', '%' . $searchname . '%' )->paginate(10);
  return response()->json($teacher);
}
}

Answer 4

尝试通过此密码令牌和正确的路线

  <script>
    $(document).ready(function(){
            $('#searchname').on('keyup', function(){
                    var text = $('#searchname').val();
                    $.ajax({
                            type:"GET",
                            headers: {
                              'X-CSRF-TOKEN': '{{ csrf_token() }}'
                            },
                            url: "{{ route('teachers/search') }}",
                            data: {text: $('#searchname').val()},
                            success:function(data){
    $('tbody').html(data);
    }
                    });
            });
    });
    </script>

以及您的控制器部分

public function ajaxsearch(){
    $searchname = Input::get ( 'searchname' );
      if($searchname != ""){
      $teacher = Teacher::where ( 'efirst', 'LIKE', '%' . $searchname . '%' )->paginate(10);
     $teacher->appends(['searchname' => $searchname ]);
      return response()->json($teacher);
    }
    }

Answer 5

首先，只需手动转到URL并输入参数。
http://todolist.local/teachers/search?text=a
如果您有任何回应。这意味着您的PHP运行良好。

如果遇到错误，则说明您使用的是GET方法。请在Route中传递变量参数

Route::get('/search/{searchName}', 'TeachersController@ajaxsearch');

，请更正您的控制器

public function ajaxsearch($searchname){
   if($searchname != ""){
       $teacher = Teacher::where ( 'efirst', 'LIKE', '%' . $searchname.'%' )->paginate(10);
       return response()->json($teacher);
   }
}

第二，请在jQuery中使用$ .get方法获取GET方法AJAX

<script>
    $(document).ready(function(){
        $('#searchname').on('keyup', function(){
           var text = $('#searchname').val();
            $.get(urlHere, function(response){
                console.log(response);
            });
        });
     });
</script>

查看控制台标签，如果得到响应

Laravel搜索功能中的Ajax无法正常工作

5 个答案: