让我们有一个示例场景:
客户端打开一个网站,找到他从文本框输入的两个数字的总和.然后点击ADD按钮.两个参数是HTTP GET到服务器,其中写入PHP代码以添加数字,结果是echo'ed .
基于这种情况,任何人都可以解释Web服务和应用程序之间的区别吗?
我在hybris 6平台中使用Spring-security-web-3.2.5 jar和Java 8以及Spring-web-mvc 4.1.7.
我在ApplicationFilterProxy和类 AbstractRememberMeServices中得到Nullpointer异常.
当我调试这个类时,我发现logger对象为null,当代码试图执行以下操作时,抛出NullPointer异常
this.logger.debug("Interactive login attempt was unsuccessful.");
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异常StackTrace:
SEVERE: Servlet.service() for servlet [DispatcherServlet] in context with path [] threw exception [Error while processing internal filterchain. Exception occurred at chain position: 9 of 16. Current filter: 'FilterChainProxy[Filter Chains: [[ Ant [pattern='/_ui/**'], []], [ Ant [pattern='/checkout/**'], [org.springframework.security.web.access.channel.ChannelProcessingFilter@f6a714c, org.springframework.security.web.context.SecurityContextPersistenceFilter@3aa9c43e, org.springframework.security.web.context.request.async.WebAsyncManagerIntegrationFilter@148b51b7, org.springframework.security.web.authentication.logout.LogoutFilter@7dc977d4, org.springframework.security.web.authentication.UsernamePasswordAuthenticationFilter@21759930, org.springframework.security.web.savedrequest.RequestCacheAwareFilter@55893114, org.springframework.security.web.servletapi.SecurityContextHolderAwareRequestFilter@1bc1c4fb, org.springframework.security.web.authentication.AnonymousAuthenticationFilter@2d007780, org.springframework.security.web.access.ExceptionTranslationFilter@20b1d709, org.springframework.security.web.access.intercept.FilterSecurityInterceptor@3c2c439a]], [ com.example.storefront.security.ExcludeUrlRequestMatcher@3a6c99ce, [org.springframework.security.web.access.channel.ChannelProcessingFilter@31cb980a, org.springframework.security.web.context.SecurityContextPersistenceFilter@3ef9fc3c, org.springframework.security.web.context.request.async.WebAsyncManagerIntegrationFilter@28aef10c, org.springframework.security.web.authentication.logout.LogoutFilter@7652af3e, org.springframework.security.web.authentication.UsernamePasswordAuthenticationFilter@1bb47236, org.springframework.security.web.savedrequest.RequestCacheAwareFilter@6add7560, org.springframework.security.web.servletapi.SecurityContextHolderAwareRequestFilter@3524cfa6, org.springframework.security.web.authentication.rememberme.RememberMeAuthenticationFilter@566cad63, org.springframework.security.web.authentication.AnonymousAuthenticationFilter@6ef1505a, org.springframework.security.web.access.ExceptionTranslationFilter@6f065e91, …Run Code Online (Sandbox Code Playgroud) 我正在使用最新的 flutter 稳定版本 flutter doctor -v 如下:
\n [\xe2\x9c\x93] Flutter (Channel stable, 2.2.2, on Mac OS X 10.15.7 19H2 darwin-x64, locale en-IN)\n \xe2\x80\xa2 Flutter version 2.2.2 at /Users/daydreamers/Documents/development/flutter\n \xe2\x80\xa2 Framework revision d79295af24 (2 weeks ago), 2021-06-11 08:56:01 -0700\n \xe2\x80\xa2 Engine revision 91c9fc8fe0\n \xe2\x80\xa2 Dart version 2.13.3\n\n[\xe2\x9c\x93] Android toolchain - develop for Android devices (Android SDK version 30.0.2)\n \xe2\x80\xa2 Android SDK at /Users/daydreamers/Library/Android/sdk\n \xe2\x80\xa2 Platform android-30, build-tools 30.0.2\n \xe2\x80\xa2 ANDROID_HOME = /Users/daydreamers/Library/Android/sdk\n \xe2\x80\xa2 Java binary at: /Applications/Android Studio.app/Contents/jre/jdk/Contents/Home/bin/java\n \xe2\x80\xa2 Java version …Run Code Online (Sandbox Code Playgroud) 我正在尝试使用“一对一”共享主键映射实体,我收到标题中提到的错误,我在这里检查了以前的帖子,但没有解决我的问题。以下是相关代码:
这是在 Localsession 工厂中设置的休眠属性:
<prop key="hibernate.dialect">org.hibernate.dialect.MySQLDialect</prop>
<prop key="hibernate.show_sql">true</prop>
<prop key="hibernate.hbm2ddl.auto">update</prop>
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Bean类如下:
@Entity
@Table(name="userdetails")
public class UserSignUpBean {
public UserSignUpBean(){}
//fields
@Id @GeneratedValue
@Column(name="id",unique=true,nullable=false)
private long id;
@OneToOne(cascade=javax.persistence.CascadeType.ALL)
@PrimaryKeyJoinColumn
private UserAuthorityBean userauthority;
public UserAuthorityBean getUserauthority() {
return userauthority;
}
public void setUserauthority(UserAuthorityBean userauthority) {
this.userauthority = userauthority;
}
//getters and setters for other fields
}
@Entity
@Table(name="userauthority")
public class UserAuthorityBean {
public UserAuthorityBean(){}
@Id
@GeneratedValue(generator="gen")
@GenericGenerator(name="gen",strategy="foreign",parameters=@Parameter(name="property",value="userdetails"))
@Column(name="id",nullable=false,unique=true)
private long id;
@OneToOne(mappedBy="userauthority",cascade=CascadeType.ALL)
private UserSignUpBean userdetails;
//other fields
public UserSignUpBean getUserdetails() {
return …Run Code Online (Sandbox Code Playgroud) 我在我创建的网站上有一个登录表单.我使用PHP验证密码,密码存储在MYSQL数据库中.我遇到的问题是密码不区分大小写.它接受字符而不考虑案例我正在使用Kubuntu linux OS
有谁可以帮忙解决这个问题?.我应该在mysql/my.cnf中为COLLATION使用什么值来避免这个?或者还有其他解决办法吗?
任何帮助是极大的赞赏.
谢谢.
我有以下代码:
fieldsToFilter.stream()
.map(e -> Arrays.asList(
Filters.ne(e, ""),
Filters.exists(e, true)
))
.toArray(Bson[]::new))
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在执行时我得到了 java.lang.ArrayStoreException: java.util.Arrays$ArrayList
Filters.exists并Filters.ne返回新的Bson实例.查看文档.
任何帮助表示赞赏
我正在传递PHP变量值,如下所示:
<input type="button" name="delete" value="Remove" onclick="del(<?echo $arr1[0]?>);"/>
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del()是一个Javascript函数,例如:$arr1[0]=234-675-90$ arr1 [0]的值是从MYSQL数据库获得的,它在mysql中的数据类型是varchar.
我正在使用mysql_query获取值.
我面临的问题是,当值传递给JavaScript函数时,它会错误地接受它(例如:代替234-675-99作为-876).是否有任何转换要将值从PHP传递给JavaScript?
任何帮助是极大的赞赏.
我真的很想了解这些是什么实时使用:
1.什么是Java概念的使用 - "超类变量可以引用子类对象"? 尽管通过这样做,SuperClass变量只能用于访问由SuperClass定义的对象的那些部分,但是不能访问子类成员.甚至可以通过子类对象来实现.
2.动态方法调度需要什么?我有一个例子如下:
class A
{
void disp()
{
System.out.println("in A");
}
}
class B extends A
{
void disp()
{
System.out.println("in B");
}
}
class C extends A
{
void disp()
{
System.out.println("in C");
}
}
public class mainclass
{
public static void main(String[] args)
{
A a= new A();
B b=new B();
C c=new C();
A ref;
ref=b;
ref.disp();//calls class B method
ref=c;
ref.disp();//calls class C method
ref=a;
ref.disp();//calls class A method
}
}
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上面的代码通过将不同的子类对象分配给超类引用变量来使用动态方法分派.我的问题是为什么使用"ref"并为其分配对象然后调用函数?我们甚至可以在不使用"ref"和对象赋值(动态调度)的情况下调用子类函数. …
我有一个网站,我在服务器端使用PHP,mysql作为数据库.我使用以下脚本从数据库中检索数据.有人能让我知道这段代码是否容易受到注入攻击?如果是这样,请你解决一下吗?
<?php
// PHP script
$usrname=$_POST['usrname'];
$_SESSION['usremail']=$usrname;
$usrpassword=$_POST['passwd'];
$db=mysql_select_db('mydb',$connection);
$result=mysql_query("select usrfname,usrlname from userinformation where usremail='$usrname' and usrpassword='$usrpassword'") or die('failed to login');
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任何帮助是极大的赞赏.
谢谢
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