小编Pau*_*lar的帖子

如何在使用Cucumber JVM时测试是否抛出了正确的异常？使用JUnit时,我会做这样的事情:

@Test(expected = NullPointerException.class)
public void testExceptionThrown(){
    taskCreater.createTask(null);
}

如您所见,这非常优雅.但是,当使用黄瓜JVM时,我怎样才能达到同样的优雅？我的测试现在看起来像这样:

@Then("the user gets a Null pointer exception$")
public void null_exception_thrown() {
    boolean result = false;
    try {
        taskCreater.createTask(null);
    } catch (NullPointerException e) {
        result = true;
    }
    assertTrue(result);
}

注意需要一个try.. catch后跟assertTrue一个标志.

我必须在webview中加载url并发送一些cookie.你想实现这个目标吗？

我正在做以下代码..

CookieManager cookieManager;

CookieSyncManager.createInstance(PrivacyActivity.this);
cookieManager = CookieManager.getInstance();

cookieManager.setCookie("param", "value");
CookieSyncManager.getInstance().sync();

WebSettings webSettings = wv.getSettings();
webSettings.setJavaScriptEnabled(true);
webSettings.setBuiltInZoomControls(true);
wv.setWebViewClient(new WebViewClient() {
    @Override
    public boolean shouldOverrideUrlLoading(WebView view, String url) {
        return super.shouldOverrideUrlLoading(view, url);
    }
});
wv.loadUrl("https://example.com");

但没有得到正确的结果.只是得到" https://example.com ".Cookie不起作用..

<?php
          error_reporting(0);
          $link = mysqli_connect("localhost", "root", "", "checksql"); 

          if($link === false){
            die("ERROR: Could not connect. " . mysqli_connect_error());
          }

          $myemailaddress=$_POST['useremail']; 
          $mypassword=$_POST['userpassword'];

          $sql = mysqli_query("SELECT * FROM register WHERE Email = '$myemailaddress' ");
          $count = mysqli_num_rows($sql);
          echo $count;

          if($count > 0){
            echo "success";
          } else{
            echo "failed";
          }
       ?>

我正在尝试检查数据库中是否存在电子邮件.我在stackoverflow上搜索了不同的线程并试图纠正它但失败了.即使是$ count的回声也没有表现出它的价值.有没有其他方法来检查它？