小编Ama*_*ade的帖子

我正在创建一个日志功能,将我的错误记录在一个文件中.

我认为它将包含发生错误的类和方法.

有没有办法记录发生错误的类和方法,所以我不必每次都手动输入？

我遇到了这样的问题:错误代码:1290.MySQL服务器运行时带有--secure-file-priv选项,因此 当我尝试执行MySQL语句时,它无法执行此语句(Windows):

SELECT *
FROM xxxx
WHERE XXX
INTO OUTFILE 'report.csv'
    FIELDS TERMINATED BY '#'
    ENCLOSED BY '"'
    LINES TERMINATED BY '\n'

当我执行它时:

INTO OUTFILE 'report.csv'
    FIELDS TERMINATED BY '#'
    ENCLOSED BY '"'
    LINES TERMINATED BY '\n'

然后它工作.此外,在INTO OUTFILE xxx重新安装MySQL服务器之前,实际工作的相同声明.

有人有想法如何处理这个错误？

通过阅读有关CI中指针和数组的一些细节,有点困惑.一方面,数组可以看作数据类型.另一方面,阵列往往是不可修改的左值.我想编译器会做一些事情,比如用常量地址替换数组的标识符,以及用于计算索引在运行时给出的位置的表达式.

myArray[3] -(compiler)-> AE8349F + 3 * sizeof(<type>)

当说数组是数据类型时,这究竟意味着什么？我希望你能帮助我澄清我对数组实际是什么以及编译器如何处理它的困惑.

我试图将我的Dockerized Wordpress实例的上传限制增加到150M.

[filename] exceeds the maximum upload size for this site.

我创建了一个uploads.ini文件,并按卷将其附加到实例.然而,我仍然无法让Wordpress媒体上传者增加它的限制.

码头工人,compose.yml:

version: '2'
services:
   db:
     image: mysql:5.7
     volumes:
       - db_data:/var/lib/mysql
     restart: always
     environment:
       MYSQL_ROOT_PASSWORD: password1
       MYSQL_DATABASE: wordpress
       MYSQL_USER: wordpress
       MYSQL_PASSWORD: password1

   wordpress:
     depends_on:
       - db
     build: ./wordpress
     ports:
       - "8085:80"
     restart: always
     environment:
       WORDPRESS_DB_HOST: db:3306
       WORDPRESS_DB_PASSWORD: password1
     volumes: 
       - ./uploads.ini:/usr/local/etc/php/conf.d/uploads.ini 
volumes:
    db_data:

uploads.ini:

file_uploads = On
memory_limit = 500M
upload_max_filesize = 500M
post_max_size = 500M
max_execution_time = 600

Docker检查容器:

[
    {
        "Id": "8edc91a51bf8480edc927de2fa5c16624ac43a7beb58c109cd771a143f67746f",
        "Created": "2017-03-15T23:47:54.480206792Z",
        "Path": …

我有一个返回用户定义对象的函数.首先,我想知道该对象是否通过引用返回,如果它是私有的？

另外,我如何将其作为常量(最终)引用返回,因为我不希望有人弄乱它？在返回一个对象和返回object.copy()之间我很困惑; 或object.clone();

我有3个表:posts,tags,post_tag.

每个Post都有很多标签,所以我使用hasMany它们的方法.但是当我在下拉列表中选择3个标签时,我无法添加它们,post_tag因此我无法选择并显示每个帖子的标签.

我的Post模特:

 class Post extends Eloquent{
 public function tag()
         {
           return  $this->hasMany('Tag');
         }
    }

我的Tag模特:

class Tag extends Eloquent{
 public function post()
         {
           return  $this->belongsToMany('Post');
         }

}

我的postController:

class postController extends BaseController{

public function addPost(){

    $post=new Post;

    $post_title=Input::get('post_title');
    $post_content=Input::get('post_content');
    $tag_id=Input::get('tag');

    $post->tag()->sync($tag_id);
    $post->save();

我希望将此post_id保存保存到post_tag带有标签ID的表中,但它不起作用.谢谢你的时间.

我正在使用vue-i18n进行本地化。当我想翻译输入占位符时，如：

<input type="text" v-model="someValue" :placeholder="$t('translation.string')">

我必须使用$t()在每次重新渲染时执行的函数（也提到了库文档）。这在我的简单预订表单中添加了数千个不必要的函数调用，我想避免这种情况。

有没有办法只绑定一次属性？翻译后的值在 Vue 实例的整个生命周期中都不会改变。v-once不是我要找的，因为我想保持组件/节点的反应性，并且只对属性进行“硬编码”。

我知道我可以通过简单地将翻译后的字符串存储在数据对象中来实现我所需要的，但我想知道是否有替代的、更简单的解决方案（不需要大量的代码重复）。

我曾经用于Paginator::make对表中的记录进行分页.在视图中,我正在获取分页链接,但每个链接都包含其中的所有记录.如何将其限制为perPage项目？

    $datas = Paginator::make($paginator, count($paginator), $perPage);
    return $datas;

代码输出:

{"total":10,"per_page":5,"current_page":1,"last_page":2,"from":1,"to":5,"data":      
[{"id":"10","languages":"ds","created_at":"2014-05-23 
11:59:02.000","created_by":"1","updated_at":"2014-05-23 
11:59:02.000","updated_by":"1","is_active":"1"},
{"id":"9","languages":"urdu","created_at":"2014-05-23 
11:57:24.000","created_by":"1","updated_at":"2014-05-23 
11:57:24.000","updated_by":"1","is_active":"1"},
{"id":"8","languages":"were","created_at":"2014-05-23 
11:55:49.000","created_by":"1","updated_at":"2014-05-23 
11:55:49.000","updated_by":"1","is_active":"1"},  
{"id":"7","languages":"delete","created_at":"2014-05-23  
11:54:57.000","created_by":"1","updated_at":"2014-05-24  
06:02:46.000","updated_by":"1","is_active":"1"},
{"id":"6","languages":"sdf","created_at":"2014-05-23 
11:53:11.000","created_by":"1","updated_at":"2014-05-23 
11:53:11.000","updated_by":"1","is_active":"1"},
{"id":"5","languages":"dada","created_at":"2014-05-23 
11:51:33.000","created_by":"1","updated_at":"2014-05-24 
05:44:34.000","updated_by":"1","is_active":"1"},
{"id":"4","languages":"English","created_at":"2014-05-23 
11:49:49.000","created_by":"1","updated_at":"2014-05-23 
11:49:49.000","updated_by":"1","is_active":"1"},
{"id":"3","languages":"asdfgf","created_at":"2014-05-23 
11:48:20.000","created_by":"1","updated_at":"2014-05-23 
11:48:20.000","updated_by":"1","is_active":"1"},
{"id":"2","languages":"Tamil","created_at":"2014-05-23 
10:55:50.000","created_by":"1","updated_at":"2014-05-23 
10:55:50.000","updated_by":"1","is_active":"1"},
{"id":"1","languages":"Tamil","created_at":"2014-05-23 
10:51:42.000","created_by":"1","updated_at":"2014-05-26 
04:41:27.000","updated_by":"1","is_active":"1"}]}

如何将 Eloquent 查询的结果存储到 tinker 中的变量，而不将结果打印到控制台？

例如，当我运行时，$threads = App\Thread::all()我只想将结果存储在变量中$threads而不查看所有线程。修补匠中有类似“安静模式”的东西吗？

我试图过滤一个分页的雄辩集合,但每当我使用任何收集方法时,我都会失去分页.

$models = User::orderBy('first_name','asc')->paginate(20);

$models = $models->each(function($model) use ($filters) {
    if(!is_null($filters['type'])) {
        if($model->type == $filters['type'])
            return $model;
    }

    if(!is_null($filters['state_id'])) {
        if($model->profile->state_id == $filters['state_id'])
            return $model;
    }

    if(!is_null($filters['city_id'])) {
        if($model->profile->city_id == $filters['city_id'])
            return $model;
    }
});

return $models;

我正在使用Laravel 4.2,有没有办法坚持分页？