小编Fri*_*dez的帖子

即使输入仅包含空格,JQuery 验证也会接受

我无法验证输入字段是否仅包含空格。我已为必填字段指定了必填规则,但如果我仅键入空格,它仍然有效。另外,用户名字段的 noSpace 规则似乎不起作用。它禁用其他字段的验证。这是我的标记。

<form method=post class="form-auth-small" action="../php/controller/registration_controller.php">
  <div class="form-group row m-t-10px">
    <div class="col-sm-6 text-left m-b-5px">
      <label><h5>RANK: </h5></label>
      <select class="form-control" name="rank" required="required">
        <option value="rank1">NUP</option>
        <option value="rank2">PO1</option>
        <option value="rank3">PO2</option>
        <option value="rank4">PO3</option>
        <option value="rank5">SPO1</option>
        <option value="rank6">SPO2</option>
        <option value="rank7">SPO3</option>
        <option value="rank8">SPO4</option>
        <option value="rank9">PINSP</option>
        <option value="rank10">PSINSP</option>
        <option value="rank11">PCINSP</option>
        <option value="rank12">PSUPT</option>
        <option value="rank13">PSSUPT</option>
        <option value="rank14">PCSUPT</option>
        <option value="rank15">PDIR</option>
        <option value="rank16">PDDG</option>
        <option value="rank17">PDG</option>
      </select>
    </div>
    <div class="col-sm-6 text-left m-b-5px">
      <label><h5>USERNAME: </h5></label>
      <input class="form-control" name="username" placeholder="Username must only contain characters [A-Z] and numbers [0-9]" type="text" minlength="1" required="required">
    </div>
    <div …
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html javascript validation jquery

5
推荐指数
1
解决办法
3104
查看次数

Laravel - 导致404错误的事件

我正在制作实时通知,并偶然发现这个奇怪的错误.我在我的模型中有一个启动方法,它触发一个名为SendNotificationData(no listener)的事件.它会在有新通知时处理.

试用控制器

<?php

namespace App\Http\Controllers\Notification;

use App\Http\Controllers\Controller;
use Illuminate\Http\Request;

use App\Http\Requests;
use App\Models\Notification;

class NotificationController extends Controller
{  
    /**
     * Trigger event to display notifications. This displays 404 error page
     *
     * @return none
     */
    public function displayNotification()
    {
        $notification = new Notification();
        $notification->EmployeeID = "EMP-00001";
        $notification->NotificationText =  "There is a new notification";
        $notification->NotificationStatus = "unread";
        $notification->NotificationType = "trial";
        $notification->save();
    }
}
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通知模型启动方法:

/**
 * Handle booting of model.
 *
 * @var string
 */
 public static function …
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php laravel-5.2

5
推荐指数
1
解决办法
586
查看次数

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