Question

我一直在研究一个简单的Laravel CRUD应用程序。我想为它实现搜索功能，所以我使用了ajax调用。但我的代码无效。

web.php

  Route::post('/search',[
     'as' => 'searchbox',
     'uses' => 'HomeController@postSearch'
     ]);

HomeController.php

public function postSearch(Request $request)
    {
# code...
   $items = Item::where('title', 'like', '%' . 'title1' . '%')->paginate(10);
  // return response()->json(['msg-body'=>$request['body']],200);
    return view('welcome')->with(['items'=>$items]);

 }

welcome.blade.php

@extends('layouts.master')
@section('title')
Welcome
@endsection
@section('content')
<div class="row" >
    <div class="col-lg-12">
        <div class="col-lg-3">
            <h4>Crud</h4>
        </div>
        <div class="col-lg-4 col-lg-offset-3">
          <form method="POST" action=>
            <input class="search" type ="text" name ="search" id="search" placeholder="search"  >

          </form>
        </div>
        <div class="pull-right">

            <a href="{{route('create')}}"class="btn btn-primary"  >Create new item</a>
        </div>
    </div>
     <div class="col-lg-12">
        <table class="table">
            <thead>
              <tr>
                <th>Id</th>
                <th>Title</th>
                <th>Description</th>
                <th style="width: width="280px";">Action</th>
              </tr>
            </thead>
            <tbody>
                @foreach($items as $item)
              <tr ng-model="search_query">
                <td>{{$item->id}}</td>
                <td>{{$item->title}}</td>
                <td>{{$item->description}}</td>
                <td><a href="{{route('showdetails',['id'=>$item->id])}}" class="btn btn-info">show</a><a href="{{route('editdetails',['id'=>$item->id])}}" class="btn btn-primary">Edit</a><a href="{{route('delete',['id'=>$item->id])}}" class="btn btn-danger">Delete</a></td>
              </tr>
              @endforeach
            </tbody>
         </table>
     </div>
</div>
<script type="text/javascript">
var token = '{{Session::token()}}';
var url = '{{route('searchbox')}}';
</script>
{{ $items->render()}}
@endsection

style.js

 $('.search').on('change',function(){
 $.ajax({
        method: 'post',
        url:url,
        data:{body: $("#search").val(),_token:token}
    }).done(function(msg){
        console.log($("#search").val());
    })
  });

有人可以建议我......

Answer 1

在{!! csrf_field() !!}元素中添加<form>..</form>。所有post/put/patch请求都应该发送csrf_token或请求将失败。

Answer 2

检查行

$items = Item::where('title', 'like', '%' . 'title1' . '%')->paginate(10);`,

我认为它应该是

$items = Item::where('title', 'like', '%' . $request->input('body') . '%')->paginate(10);

现在您只需搜索字符串title1

在laravel中使用ajax进行搜索实现的问题

2 个答案: