小编LJT*_*LJT的帖子

我有以下代码,其中我有2个电子邮件输入字段,我需要验证它们是相同的,使用jQuery验证equalTo成功.

<div class="control-group">
 <label class="control-label" for="inputEmail"><strong>Email Address</strong></label>
   <div class="controls">
    <input type="email" name="inputEmail" placeholder="jane.smith@email.com" id="inputEmail" required>
   </div>
    <label class="control-label" for="inputEmailConfirm"><strong>Confirm Email Address</strong></label>
     <div class="controls">
    <input type="email" name="inputEmailConfirm" placeholder="jane.smith@email.com" id="inputEmailConfirm" required>
    </div>
    </div>
  <button type="submit" id="emailSubmit" disabled="disabled" class="btn btn-danger" data-toggle="tooltip" data-placement="bottom" title="Click me to buy">Credit Card Checkout &raquo; </button>

验证成功后,我希望启用该按钮,但似乎无法弄清楚如何执行此操作.

 $('#ccSelectForm').validate({
    rules: {
      inputEmail: {
        required: true,
        email: true
      },
      inputEmailConfirm: {
        required: true,
        email: true,
        equalTo: '#inputEmail'
      },
    }
  });

理想的情况是作为奖金,我想设置窗体控件以利用Bootstrap3验证状态,详细在这里- http://getbootstrap.com/css/#forms-control-validation

小提琴在这里http://jsfiddle.net/4wU5m/

我正在尝试使用一个简单的LAMP站点,但似乎无法让我的本地Apache实例接收Zend框架.我在Windows上开始使用XAMPP,然后尝试使用手动Apache/PHP安装的Centos 6 VM,但仍然在下面的两个版本上都得到了相同的错误.phpinfo()工作正常,网站的其他部分也是如此.

Fatal error: Class 'Zend\Log\Logger' not found in /var/www/html/site/public/test.php on line 20

我的网站代码是一个调用Zend框架记录器的相当简单的测试,如下所示

use Zend\Log\Logger;
use Zend\Log\Writer;
echo "<p>Hello world</p>";
echo $_POST["VIN"]; 
phpinfo();
$logger = new Zend\Log\Logger;
$writter = new Zend\Log\Writer\Stream('php://output');
$logger->addWriter($writer);
?>

我的Apache linux httpd.conf是

<VirtualHost *:80>

DocumentRoot /var/www/html/site/public
<Directory /var/www/html/site/public>              
    DirectoryIndex test.php 
    AllowOverride All
    Order allow,deny
    Allow from all
</Directory>

我的Zend框架位于/ var/www/html/site/library/Zend,我还将/ var/www/html/site/library添加到php.ini include中.

我正在尝试在Zend框架2中创建一个SOAP客户端,我创建了下面的正确返回数据

try {
  $client = new Zend\Soap\Client("http://www.webservicex.net/country.asmx?WSDL");
  $result = $client->GetCountries();      
  print_r($result);
} catch (SoapFault $s) {
  die('ERROR: [' . $s->faultcode . '] ' . $s->faultstring);
} catch (Exception $e) {
  die('ERROR: ' . $e->getMessage());
}

但是,当我尝试将数据发送到Web服务时,例如使用

try {
  $client = new Zend\Soap\Client("http://www.webservicex.net/country.asmx?WSDL");
  $result = $client->GetCurrencyByCountry('Australia');
  print_r($result);
} catch (SoapFault $s) {
  die('ERROR: [' . $s->faultcode . '] ' . $s->faultstring);
} catch (Exception $e) {
  die('ERROR: ' . $e->getMessage());
}

我刚收到以下消息

ERROR: [soap:Receiver] System.Web.Services.Protocols.SoapException: Server was unable to process request. …