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javascript - 如何解决尝试获取非对象错误的属性“出价”

问题描述

我想以数据表的形式显示出价表中的数据。但是如果没有出价，我会收到此错误“尝试获取非对象的财产‘出价’”。出价模型连接到拍卖模型，拍卖模型连接到媒体站点模型。如果没有数据，如何使其显示空白记录。

这是我的控制器：

<?php

namespace App\Http\Controllers;

use App\Auction;
use App\Bid;
use App\User;
use App\Media;
use App\MediaSite;
use Illuminate\Http\Request;
use Illuminate\Support\Facades\DB;

class MediaSiteController extends Controller
{
public function show(MediaSite $mediaSite)
{
$auction = $mediaSite->auction;
$bids = $auction->bids;


return view('admin.media-site.show', ['mediaSite' => $mediaSite,'auction' => $auction], compact('auction'));
    }

我的观点：

<body>
<div id="datatable-bid"></div>
</body>

<script>
$(document).ready(function () {
var datatableBid = $('#datatable-bid').mDatatable({

// datasource definition
data: {
type: 'local',
source: {!! json_encode($auction->bids) !!},
pageSize: 10
},

// layout definition
layout: {
theme: 'default', // datatable theme
class: '', // custom wrapper class
scroll: false, 

footer: false // display/hide footer
},

// column sorting
sortable: true,

pagination: true,

search: {
input: $('#panel-search')
},

// columns definition
columns: [
{

field: "price",
title: "Price",
}, {

field: "month",
title: "Month",
},{

field: "user_id",
title: "User Id",
}
]
});

</script>

这是我的错误： 尝试获取非对象的属性“出价”

标签： javascriptlaraveldatatables

解决方案

在show()函数中进行这些更改

$auction = $mediaSite->auction;
if($auction) {
  $bids = $auction->bids;
} else {
  $bids = [];
}
// now send $bids to view along with $auction
// may be like this
// return view(..., compact($auction, $bids));

然后，在您看来进行此更改

// datasource definition
data: {
type: 'local',
source: {!! json_encode($bids) !!},
pageSize: 10
},

看看这是否有帮助。

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