在 laravel 中动态无限制地添加字段
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【中文标题】在 laravel 中动态无限制地添加字段【英文标题】:Add fields in laravel dynamically and unlimited 【发布时间】:2018-02-16 19:56:35 【问题描述】:我的网站有一部分用户或管理员可以在其中添加餐厅列表(真的很像帖子,只是命名不同)
有一些固定输入,例如(标题、描述和地图) .
所以我需要的是 + 按钮,人们可以在其中添加字段并用另一个字段命名菜单项,以显示每个项目的价格。
所以我的问题是如何实现这个选项?
我现在有什么?
餐厅迁移:
<?php
use Illuminate\Support\Facades\Schema;
use Illuminate\Database\Schema\Blueprint;
use Illuminate\Database\Migrations\Migration;
class CreateRestaurantsTable extends Migration
/**
* Run the migrations.
*
* @return void
*/
public function up()
Schema::create('restaurants', function (Blueprint $table)
$table->increments('id');
$table->string('title')->unique();
$table->string('slug')->unique();
$table->string('description')->nullable();
$table->string('image')->nullable();
$table->string('menu')->nullable();
$table->string('address')->nullable();
$table->integer('worktimes_id')->unsigned();
$table->integer('workday_id')->unsigned();
$table->integer('user_id')->unsigned();
$table->string('verified')->default(0);
$table->string('status')->default(0);
$table->timestamps();
);
Schema::table('restaurants', function($table)
$table->foreign('worktimes_id')->references('id')->on('worktimes');
$table->foreign('workday_id')->references('id')->on('workdays');
$table->foreign('user_id')->references('id')->on('users');
);
/**
* Reverse the migrations.
*
* @return void
*/
public function down()
Schema::dropIfExists('restaurants');
就是这样,我仍然没有为餐厅创建 CRUD 控制器,因为我坚持这个选项和你的意见。
谢谢。
更新
存储方法:
public function store(Request $request)
//Validating title and body field
$this->validate($request, array(
'title'=>'required|max:225',
'slug' =>'required|max:255',
'image' =>'sometimes|image',
'description' => 'required|max:100000',
'address' => 'sometimes|max:500',
'user_id' => 'required|numeric',
'verified' => 'sometimes',
'status' => 'required|numeric',
));
$restaurant = new Restaurant;
$restaurant->title = $request->input('title');
$restaurant->slug = $request->input('slug');
$restaurant->description = $request->input('description');
$restaurant->address = $request->input('address');
$restaurant->user_id = $request->input('user_id');
$restaurant->verified = $request->input('verified');
$restaurant->status = $request->input('status');
if ($request->hasFile('image'))
$image = $request->file('image');
$filename = 'restaurant' . '-' . time() . '.' . $image->getClientOriginalExtension();
$location = public_path('images/');
$request->file('image')->move($location, $filename);
$restaurant->image = $filename;
// menu
$newArray = array();
$menuArray = $request->custom_menu; //Contains an array of Menu Values
$priceArray = $request->custom_price; //Contains an array of Price Values
//Creating new array with ARRAY KEY : MENU VALUES and ARRAY VALUE: PRICE VALUES
foreach ($menuArray as $key => $singleMenu)
$newArray[$singleMenu] = $priceArray[$key];
//Output : array("Menu01" => "Price01", "Menu02" => "Price 02", "Menu03" => "Price 04", "Menu04" => "Price 05")
//Converting array to json format to store in your table row 'custom_menu_price'
$jsonFormatData = json_encode($newArray);
//Output like: "Menu01":"Price01","Menu02":"Price 02","Menu03":"Price 04","Menu04":"Price 05"
// Save in DB
//
//
//
// To retrieve back from DB to MENU and PRICE values as ARRAY
$CustomArray = json_decode($jsonFormatData, TRUE);
foreach ($CustomArray as $menu => $price)
echo "Menu:".$menu."<br>";
echo "Price:".$price."<br>";
// menu
$restaurant->save();
$restaurant->workdays()->sync($request->workdays, false);
$restaurant->worktimes()->sync($request->worktimes, false);
//Display a successful message upon save
Session::flash('flash_message', 'Restaurant, '. $restaurant->title.' created');
return redirect()->route('restaurants.index');
【问题讨论】:
您可以将菜单作为 json 格式存储在数据库中,这样每个餐厅都可以拥有自己的 json 对象和不同的食物选项,即使您可以在其中存储 html 属性 @AnarBayramov 我不知道该怎么做!你能帮我给我一个样品吗? w3schools.com/js/js_json_php.asp 您好,这是您想要实现的目标吗?如果没有,请更准确地告诉我细节.. embed.plnkr.co/HQOuNvDfDwkR2uLD632u @demonyowh 嗨,兄弟,这正是我的表单所需要的 + 现在我正在使用 sreejith bs 代码来存储有效的数据,但也可以保存额外的空输入。所以现在我需要这些:停止额外的空输入以保存并在我的表单中使用您的方法。请在 sreejith 上查看我的 cmets 答案,了解我对额外空输入的含义。 【参考方案1】:你能做的是
1) 在您的迁移文件中为custom_menu_price
添加另一个表格行
$table->string('custom_menu_price')->nullable();
2) 修改你的form
<form method="POST" action=" ...... ">
csrf_field()
//I'm Looping the input fields 5 times here
@for($i=0; $i<5; $i++)
Enter Menu $i : <input type="text" name="custom_menu[]"> //**Assign name as ARRAY
Enter Price $i : <input type="text" name="custom_price[]"> //**Assign name as ARRAY
<br><br>
@endfor
<input type="submit" name="submit">
</form>
3) 在你的controller
public function store(Request $request)
//Validating title and body field
$this->validate($request, array(
'title'=>'required|max:225',
'slug' =>'required|max:255',
'image' =>'sometimes|image',
'description' => 'required|max:100000',
'address' => 'sometimes|max:500',
'user_id' => 'required|numeric',
'verified' => 'sometimes',
'status' => 'required|numeric',
));
$restaurant = new Restaurant;
$restaurant->title = $request->input('title');
$restaurant->slug = $request->input('slug');
$restaurant->description = $request->input('description');
$restaurant->address = $request->input('address');
$restaurant->user_id = $request->input('user_id');
$restaurant->verified = $request->input('verified');
$restaurant->status = $request->input('status');
if ($request->hasFile('image'))
$image = $request->file('image');
$filename = 'restaurant' . '-' . time() . '.' . $image->getClientOriginalExtension();
$location = public_path('images/');
$request->file('image')->move($location, $filename);
$restaurant->image = $filename;
// menu
$newArray = array();
$menuArray = $request->custom_menu; //Contains an array of Menu Values
$priceArray = $request->custom_price; //Contains an array of Price Values
//Creating new array with ARRAY KEY : MENU VALUES and ARRAY VALUE: PRICE VALUES
foreach ($menuArray as $key => $singleMenu)
$newArray[$singleMenu] = $priceArray[$key];
//Output : array("Menu01" => "Price01", "Menu02" => "Price 02", "Menu03" => "Price 04", "Menu04" => "Price 05")
//Converting array to json format to store in your table row 'custom_menu_price'
$jsonFormatData = json_encode($newArray);
//Output like: "Menu01":"Price01","Menu02":"Price 02","Menu03":"Price 04","Menu04":"Price 05"
// Save in DB
$restaurant->custom_menu_price = $jsonFormatData;
// menu
$restaurant->save();
$restaurant->workdays()->sync($request->workdays, false);
$restaurant->worktimes()->sync($request->worktimes, false);
//Display a successful message upon save
Session::flash('flash_message', 'Restaurant, '. $restaurant->title.' created');
return redirect()->route('restaurants.index');
在您的front.restaurantshow
视图中:
@php
// To retrieve back from DB to MENU and PRICE values as ARRAY
$CustomArray = json_decode($restaurant->custom_menu_price, TRUE);
@endphp
@foreach ($CustomArray as $menu => $price)
Menu Name: $menu <br>
Menu Price: $price <br><br>
@endforeach
希望它有意义。
【讨论】:
这部分// Save in DB
我假设我必须用menu
和price
的行创建新表对吗?
不需要。只需将该 json 保存在一个表格列中,例如 custom_menu_price
这样您以后可以轻松地从 DB 中检索为 KEY PAIR 值。
我做了所有这些,我在存储方法上没有错误,现在请告诉我如何查看该 json 文件保存的位置,以及如何在前端显示它?我总共是 0 :)
您是否在迁移中添加了额外的行?以上是关于在 laravel 中动态无限制地添加字段的主要内容,如果未能解决你的问题,请参考以下文章