我正在使用迁移指南从 Spring Security 3.2.5 升级到 4.0.4 。
我的UserDetailsService看起来像这样:
package com.me.security;
import org.springframework.security.core.userdetails.UserDetailsService;
public class Users implements UserDetailsService {
public Users() {
System.err.println("USERS CONSTRUCTOR");
}
@Override
public UserDetail loadUserByUsername(String name) {
System.err.println("LOAD BY USER NAME " + name);
throw new UsernameNotFoundException("User not found.");
}
}
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我的WEB-INF/applicationContext.xml有这个:
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:security="http://www.springframework.org/schema/security"
xsi:schemaLocation="
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-4.0.xsd
http://www.springframework.org/schema/security
http://www.springframework.org/schema/security/spring-security-4.0.xsd">
<security:http disable-url-rewriting="true" use-expressions="false">
<security:intercept-url pattern="/auth/**" access="ROLE_ANONYMOUS"/>
<security:intercept-url pattern="/dashboard/**" access="ROLE_ADMIN,ROLE_USER"/>
<!-- ...more intercept-urls... -->
<security:access-denied-handler error-page="/pages/general/403.xhtml"/>
<security:form-login login-page="/auth/login.html"
username-parameter="j_username"
password-parameter="j_password"
login-processing-url="/j_spring_security_check" …Run Code Online (Sandbox Code Playgroud) 有没有办法在 kusto 中获得类似于 Java 中的 foreach 循环的行为?例如,假设我有一个不同的服务列表 AF,那么对于这个不同的列表,我想为每个不同的列值取 N 行,有没有办法在单个查询中做到这一点?
我尝试了多种加入方式,但无法以动态方式使其发挥作用。我所说的动态是指我不想编写指定| where service = 'A' | take 30.
ServiceLogs
| where isnotempty( service)
| distinct service
| join kind = rightouter ServiceLogs on $left.service== $right.service
| take 30
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实际结果是它只为不同列表中的单个值返回 30,而不是每个值
正在使用 Spring boot 2.1.1 在使用 org.springframework.web.client.RestTemplate 时出现以下错误
java.lang.ClassNotFoundException: com.fasterxml.jackson.dataformat.xml.XmlMapper
at java.net.URLClassLoader.findClass(Unknown Source) ~[na:1.8.0_191]
at java.lang.ClassLoader.loadClass(Unknown Source) ~[na:1.8.0_191]
at sun.misc.Launcher$AppClassLoader.loadClass(Unknown Source) ~[na:1.8.0_191]
at java.lang.ClassLoader.loadClass(Unknown Source) ~[na:1.8.0_191]
at
org.springframework.http.converter.json
.Jackson2ObjectMapperBuilder.build(Jackson2ObjectMapperBuilder.java:617) ~[spring-web-
5.1.3.RELEASE.jar:5.1.3.RELEASE]
at org.springframework.http.converter.xml.MappingJackson2XmlHttpMessageConverter.<init>
(MappingJackson2XmlHttpMessageConverter.java:50) ~[spring-web-5.1.3.RELEASE.jar:5.1.3.RELEASE]
at
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从我的pom:
<parent>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-parent</artifactId>
<version>2.1.1.RELEASE</version>
<relativePath/> <!-- lookup parent from repository -->
</parent>
<dependency>
<groupId>com.fasterxml.jackson.core</groupId>
<artifactId>jackson-databind</artifactId>
</dependency>
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在创建 org.springframework.web.client.RestTemplate 对象时,我收到此异常
我该如何解决?
我正在制作一个程序来改变 DNA 的序列。找到字母“a”时,将其替换为“t”,并与之相反。但是我在程序运行时遇到了一个问题,它将“a”替换为“t”,然后再次将“t”替换为“a”。
我该如何解决?
编码:
def opposite_nucleotide(dna):
dna = dna.replace("a", "t")
dna = dna.replace("t", "a")
dna = dna.replace("g", "c")
dna = dna.replace("c", "g")
return dna
dna3 = input("Enter the DNA to represent it's oppsite one: ")
dna4 = opposite_nucleotide(dna3)
print(dna4)
Run Code Online (Sandbox Code Playgroud) 我想迭代下面的数组:
var a=[1,2,3,4,5,6,0,11]
for(let value of a)
{
console.log(a[value-1])
}
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我得到的输出如下。我找不到为什么0和11被打印为未定义的原因。
1
2
3
4
5
6
undefined
undefined
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