在Config/app.phplaravel源中,实际使用的是url什么?
它说要使用的应用程序URL artisan command line tool,那么实际应该是什么?
我的意思是应该是http://mydomainname.com或应该是/var/www/laravel/或/var/www/laravel/public
当前配置
/*
|--------------------------------------------------------------------------
| Application URL
|--------------------------------------------------------------------------
|
| This URL is used by the console to properly generate URLs when using
| the Artisan command line tool. You should set this to the root of
| your application so that it is used when running Artisan tasks.
|
*/
'url' => 'http://localhost/',
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如果我的应用程序源位于/var/www/目录并且laravel公用文件夹是/var/www/laravel/public
,http://mydomainname.com则指向在/var/www/laravel/public目录中解析
使用案例: …
我对opps和laravel都很陌生.所以,将值插入我的users和profiles表中的hav OneToOne关系,这是我的store()方法看起来像
public function store(Requests\StoreNewUser $request)
{
// crate an objct of user model
$user = new \App\User;
// now request and assign validated input to array of column names in user table
$user->first_name = $request->input('first_name');
$user->last_name = $request->input('last_name');
$user->email = $request->input('email');
$user->password = $request->input('password');
/* want to assign request input to profile table's columns in one go
*/
$user->profile()->user_id = $user->id; // foreign key in profiles table
$user->profile()->mobile_no = $request->input('mobile');
dd($user); // …Run Code Online (Sandbox Code Playgroud) 基于compact刀片内部方法所提供的Model变量上的某些选择过滤器,检索列值的正确方法是什么。(Larevl 5)
我读到从视图查询数据库的正确做法是不正确的,因此我遵循惯例使用所需的数据compact来查看视图
但是,在需要根据第一张表的刀片内的foreach循环中返回的某些列值查询另一张表的情况下,我无法找出正确的方法
示例:我有两个模型User和Group
架构 User表
id,name,email,group_id
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计划团体表
id,groupname
这是UserController-> compact方法
$users = \App\User::all(array('id','name','email','group_id'));
$groups = \App\Group::all(array('id','group_name'));
return view('user.index', compact('users','groups'));
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刀片如何需要它们
@foreach ($users as $user)
<tr>
<th>{{$user->id}}</th>
<th>{{$user->name}}</th>
<th>{{$user->email}}</th>
<th>
<!-- Here i need to run something like
select group_name from group where id = $user->id -->
{{$groups->where('id','=',$user->group_id) }}
</th>
<th>Actions</th>
</tr>
@endforeach
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我知道这会返回一个数组,我在这里有两个问题
group_name从基于组的“组模型”中获取列的值。id= $user->id在foreach循环中编辑1:
我将最后一个组查询修改为
<th>@if($groups->where('id','=',$user->group_id))
@foreach($groups as $group)
{{$group->group_name}} …Run Code Online (Sandbox Code Playgroud) 我想过滤 WP_Query 以检索(post_type = shop_order)具有超过 15 分钟的特定状态的所有 woocommerce 订单。
即所有上次修改日期或之前的帖子(现在 - 15 分钟)// 希望我清楚。
下面是我尝试过的
function filter_where($where = '') {
//posts in the last 15 minutes
$where .= " AND post_modified > '" . date('Y-m-d', strtotime('INTERVAL 15 MINUTE')) . "'";
return $where;
}
add_filter('posts_where', 'filter_where');
$args = array(
'post_type' => 'shop_order',
'post_status' => 'publish',
'posts_per_page' => 10,
'tax_query' => array(
array(
'taxonomy' => 'shop_order_status',
'field' => 'slug',
'terms' => array('completed')
)
)
);
$loop = new WP_Query( $args …Run Code Online (Sandbox Code Playgroud) 我扩展了默认迁移,为我的用户表包含了一些额外的表字段.
我希望拥有created_at和作为价值观的updated_at领域timestamps.
这是我的代码
<?php
use Illuminate\Database\Schema\Blueprint;
use Illuminate\Database\Migrations\Migration;
class CreateUsersTable extends Migration {
/**
* Run the migrations.
*
* @return void
*/
public function up()
{
Schema::create('users', function(Blueprint $table)
{
$table->bigIncrements('id');
$table->bigInteger('group_id');
$table->string('fname',255);
$table->string('lname',255);
$table->string('email',255)->unique();
$table->string('password', 60);
$table->boolean('active');
$table->string('gravtar',255)->nullable();
$table->rememberToken();
$table->timestamps('created_at');
$table->timestamps('updated_at');
});
}
/**
* Reverse the migrations.
*
* @return void
*/
public function down()
{
Schema::drop('users');
}
}
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障碍是使用两个时间戳列不迁移表并抛出此异常
[PDOException]
SQLSTATE[42S21]: Column already exists: 1060 Duplicate column name 'created_at'
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看,我只有一个名称列 …
行,
这是我的index()方法UserController
public function index()
{
$name = 'echoashu';
return view('home', compact('name'));
}
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就这么简单,这里是我的home.blade.php代码
<span class="info-box-number">{{$name}} </span>
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这必须按照文档理想工作,但它返回未定义的变量错误
Undefined variable: name (View: C:\xampp\htdocs\laravel1\resources\views\home.blade.php)
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任何猜测?