我已经安装了项目,vue cli 3但随着项目的增长,组件的导入变得越来越难看,我最终导入了一个组件
import Component from '../../../../components/folder/Component.vue'
我只想给 src 文件夹取别名然后做
import Component from '@components/folder/Component.vue'
我确实读到我必须修改 vue.config.js,我已经完成了但错误是一样的
Module not found: Error: Can't resolve '@components/Permissions/PermissionsTable'
这是我的 vue.config.js
const path = require("path");
const vueSrc = "./src";
module.exports = {
runtimeCompiler: true,
css: {
modules: true
},
configureWebpack: {
resolve: {
alias: {
"@": path.join(__dirname, vueSrc)
}
}
}
};
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我错过了什么吗?我还应该做什么?
我正在使用 vue 2.6.x 创建一个类似聊天的文本框,用户可以在其中写入,然后如果他按 Enter 键,则会发送消息,知道我希望能够在shift+enter按下按键时进行换行
这是我的text field
<v-text-field
@click="answer"
v-model="text"
append-icon="send"
:append-icon-cb="sendMessage"
label="Write your message..."
@keyup.native.enter="sendMessage()"
:multi-line="true"
:auto-grow="true"
:rows="1"
:hide-details="true"
flat
solo>
<v-icon slot="append-icon">send</v-icon>
</v-text-field>
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shift + enter仅同时按下按键时如何监听事件?
我正在使用 laravel 5.6 resources controllers,form request问题是我有一些已required创建的输入,但在编辑时是可选的,例如文件输入。所以我有这个表格请求
<?php
namespace App\Http\Requests;
use Illuminate\Foundation\Http\FormRequest;
class ProgramRequest extends FormRequest
{
/**
* Determine if the user is authorized to make this request.
*
* @return bool
*/
public function authorize()
{
return true;
}
/**
* Get the validation rules that apply to the request.
*
* @return array
*/
public function rules()
{
return [
//
'name.*' => 'required',
'description.*' => 'required',
'logo' => 'required|image|max:3000',
'logo_alt' => 'required|image|max:3000' …Run Code Online (Sandbox Code Playgroud) laravel laravel-5 laravel-validation laravel-request laravel-5.6
我有 2 个模型,business在documents一段1:n关系中,我想用business两种方式过滤,
business其中documentseverydocument.due_balance大于 0business其中documentseverydocument.due_balance等于 0我想做这样的东西
select
A.name, B.due_balance, sum(B.due_balance) as total_due_balance
from
business A
inner join documents B ON A.id = B.business_id
group by A.id
having total_due_balance > 0;
select
A.name, B.due_balance, sum(B.due_balance) as total_due_balance
from
business A
inner join documents B ON A.id = B.business_id
group by A.id
having total_due_balance = 0;
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这些会让我得到我想要的,问题是,以前的代码是用sequelize ORM制作的,我无法改变它,像这样
constbusinesss=awaitdb.business.paginate({attributes:[...],where:{... //一堆where},page:parseInt(params.page,10)||1,paginate:parseInt( params.limit, 10) …
I get all the records from a model, but i need to take chunks from it, i only need the values of the email so i map it
$forwadings = \App\Models\ForwardingEmail::where('company_id', $companyId)
->where('status', 1)
->get(['email']);
->map(function($forwarding) {
return $forwarding['email'];
})
return $forwadings;
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the return display this
[
"email12@example.com",
"email13@example.com",
"email1@example.com",
"email2@example.com",
"email3@example.com",
"email6@example.com",
"email7@example.com",
"email8@example.com",
"email4@example.com",
"email5@example.com",
"email9@example.com",
"email10@example.com"
]
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the problem here is when i do return $forwardings->chunk(10), it show this:
[
[
"email12@example.com",
"email13@example.com",
"email1@example.com",
"email2@example.com",
"email3@example.com", …Run Code Online (Sandbox Code Playgroud) 我在 etl 工作(第一次),我需要从客户端的 SFTP 中提取一些文件。我的问题是,文件数量是可变的,所以我需要检查,如果文件中存在,并得到它,该文件格式,如“_文件YYYY-MM-DD -number- ñ ”,其中YYYY-MM-DD是当前的日期和n是文件的编号,因此如果有 7 个文件,我必须查找:
直到现在我发现我可以做这样的事情
cnopts = pysftp.CnOpts()
with pysftp.Connection(host=host, port=port, username=username, password=password, cnopts=cnopts) as sftp:
files = sftp.listdir(directory)
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我如何在那里的文件中找到?
我想提出一个Ajax请求有axios一个laravel controller,我想用一个middleware检查,如果request与制作ajax,但问题是,当我做出ajax request,的middleware罚球false始终。
我这样打电话
axios.post('/api/contact/send', {
...
data : data
}).then((response) => {
Do somethings
}).catch(err => {
Do somethings
})
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我的 api 路由
Route::namespace('Home')->middleware('IsAjaxRequest')->domain(env('HOST'))->group(function(){
....
Route::post('contact/send','ContactController@postContact');
});
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该IsAjaxRequest中间件
if(!$request->ajax()) {
abort(403, 'Unauthorized action.');
}
return $next($request);
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和控制器
<?php
namespace App\Http\Controllers\Home;
use Illuminate\Http\Request;
use App\Events\Home\ContactMail;
use App\Http\Controllers\Controller;
use App\Http\Requests\ContactRequest;
class ContactController extends Controller
{
//
public function postContact(ContactRequest $request)
{
$data = …Run Code Online (Sandbox Code Playgroud) 我有一个vue项目,laravel作为a back-end,我必须检查用户是否具有访问资源或视图的权限,在后端,我使用laravel权限来完成此操作,并且工作正常,我现在遇到的问题在前面结束。
我在登录时获取用户权限并将其保存在上localStorage,现在如果该用户没有权限,该如何阻止用户输入特定路由?
例如我有这些路线
{
path: 'users',
name: 'Users',
component: Users,
meta : {
permissions: 'read_users'
}
},
{
path: 'roles-permissions',
name: 'RolesPermissions',
component: RolesPermissions,
meta : {
permissions: 'read_roles'
}
},
{
path: 'roles-permissions/create',
name: 'CreateRolesPermissions',
component: CreateRolesPermissions,
meta : {
permissions: 'create_roles'
}
},
{
path: 'roles-permissions/:id/edit',
name: 'EditRolesPermissions',
component: EditRolesPermissions,
meta : {
permissions: 'edit_roles'
}
},
{
path: 'customers',
name: 'Clientes',
component: CustomersList …Run Code Online (Sandbox Code Playgroud) laravel ×4
javascript ×2
vue.js ×2
ajax ×1
axios ×1
eloquent ×1
laravel-5 ×1
laravel-5.6 ×1
middleware ×1
mysql ×1
node.js ×1
orm ×1
permissions ×1
php ×1
pysftp ×1
python ×1
sequelize.js ×1
sftp ×1
sql ×1
vue-cli-3 ×1
vue-router ×1
vuejs2 ×1
webpack ×1