如何迭代谓词列表
How to iterate a list of Predicates
我有一个 spring 引导应用程序,我们有一个 application.yml 上面有一组功能标志
featureFlag:
featureOne:true
featureTwo:true
featureThree:true
featureFour:false
然后这个文件被这个class
读取
@Configuration
@ConfigurationProperties(prefix="featureFlag")
public class FeatureFlag{
private Boolean featureOne;
private Boolean featureTwo;
private Boolean featureThree;
private Boolean featureFour;
/*The predicates based on the feature flags*/
private Predicate<FeatureFlag> isFeatureFlagOneEnabled = featureFlag.isFeatureOne();
private Predicate<FeatureFlag> isFeatureFlagTwoEnabled = featureFlag.isFeatureTwo();
private Predicate<FeatureFlag> isFeatureFlagThreeEnabled = featureFlag.isFeatureThree();
private Predicate<FeatureFlag> isFeatureFlagFourEnabled = featureFlag.isFeatureFour();
}
我想传递实际的谓词并迭代它们中的每一个,但我想知道我是否可以做一个通用函数来传递谓词列表及其要测试的值以及它们是否都为真函数 return 为 true 否则为 false
然后在这个 class 添加一些这样的代码,因为我想按需生成这个列表,例如我有一个客户 x 购买了 featureOne 和 featureTwo,在这个例子中我创建了一个这样的列表
Set<Predicate<FeatureFlag>> rulesForClientX = new HashSet<>();
rulesForClientX.add(isFeatureFlagOneEnabled);
rulesForClientX.add(isFeatureFlagTwoEnabled);
然后我想为该客户端创建一个特定的逻辑并将之前创建的谓词列表传递给它,但我想我需要这样的东西
Function<List<Predicate<FeatureFlag>>, Boolean> iteratePredicates = (predicates) -> {
//test each predicate and return true if all of them are true otherwise return false
}
您可以创建一个接受 Set<Predicate<FeatureFlag>>
和值的方法,然后您可以流式传输谓词集并使用 allMatch
public boolean testPredicates(Set<Predicate<FeatureFlag>> predicates, Integer value) {
return predicates.stream().allMatch(pre->pre.test(value));
}
连锁Predicate
s
您应该期待的是链接您拥有的谓词。
I pass the list of Predicates with its value to be tested and if all
of them are true the function return me a true otherwise false
根据您的要求,这应该是这样的;
public Predicate<FeatureFlag> chainPredicates(Set<Predicate<FeatureFlag>> predicates) {
return predicates.stream()
.reduce(Predicate::and) // all true
.orElse(p -> false); // or false
}
消费Predicate
此外,您现在可以像
一样轻松地消费这个 Predicate
boolean testFeatureFlag(Set<Predicate<FeatureFlag>> predicates, FeatureFlag value) {
return chainPredicates(predicates).test(value);
}
或在集合上说 List<FeatureFlag>
以过滤掉特定的 FeatureFlag
为:
List<FeatureFlag> selectiveFeatures(Set<Predicate<FeatureFlag>> predicates, List<FeatureFlag> featureFlags) {
Predicate<FeatureFlag> flagPredicate = chainPredicates(predicates);
return featureFlags.stream()
.filter(flagPredicate)
.collect(Collectors.toList());
}
我有一个 spring 引导应用程序,我们有一个 application.yml 上面有一组功能标志
featureFlag:
featureOne:true
featureTwo:true
featureThree:true
featureFour:false
然后这个文件被这个class
读取@Configuration
@ConfigurationProperties(prefix="featureFlag")
public class FeatureFlag{
private Boolean featureOne;
private Boolean featureTwo;
private Boolean featureThree;
private Boolean featureFour;
/*The predicates based on the feature flags*/
private Predicate<FeatureFlag> isFeatureFlagOneEnabled = featureFlag.isFeatureOne();
private Predicate<FeatureFlag> isFeatureFlagTwoEnabled = featureFlag.isFeatureTwo();
private Predicate<FeatureFlag> isFeatureFlagThreeEnabled = featureFlag.isFeatureThree();
private Predicate<FeatureFlag> isFeatureFlagFourEnabled = featureFlag.isFeatureFour();
}
我想传递实际的谓词并迭代它们中的每一个,但我想知道我是否可以做一个通用函数来传递谓词列表及其要测试的值以及它们是否都为真函数 return 为 true 否则为 false
然后在这个 class 添加一些这样的代码,因为我想按需生成这个列表,例如我有一个客户 x 购买了 featureOne 和 featureTwo,在这个例子中我创建了一个这样的列表
Set<Predicate<FeatureFlag>> rulesForClientX = new HashSet<>();
rulesForClientX.add(isFeatureFlagOneEnabled);
rulesForClientX.add(isFeatureFlagTwoEnabled);
然后我想为该客户端创建一个特定的逻辑并将之前创建的谓词列表传递给它,但我想我需要这样的东西
Function<List<Predicate<FeatureFlag>>, Boolean> iteratePredicates = (predicates) -> {
//test each predicate and return true if all of them are true otherwise return false
}
您可以创建一个接受 Set<Predicate<FeatureFlag>>
和值的方法,然后您可以流式传输谓词集并使用 allMatch
public boolean testPredicates(Set<Predicate<FeatureFlag>> predicates, Integer value) {
return predicates.stream().allMatch(pre->pre.test(value));
}
连锁Predicate
s
您应该期待的是链接您拥有的谓词。
I pass the list of Predicates with its value to be tested and if all of them are true the function return me a true otherwise false
根据您的要求,这应该是这样的;
public Predicate<FeatureFlag> chainPredicates(Set<Predicate<FeatureFlag>> predicates) {
return predicates.stream()
.reduce(Predicate::and) // all true
.orElse(p -> false); // or false
}
消费Predicate
此外,您现在可以像
一样轻松地消费这个Predicate
boolean testFeatureFlag(Set<Predicate<FeatureFlag>> predicates, FeatureFlag value) {
return chainPredicates(predicates).test(value);
}
或在集合上说 List<FeatureFlag>
以过滤掉特定的 FeatureFlag
为:
List<FeatureFlag> selectiveFeatures(Set<Predicate<FeatureFlag>> predicates, List<FeatureFlag> featureFlags) {
Predicate<FeatureFlag> flagPredicate = chainPredicates(predicates);
return featureFlags.stream()
.filter(flagPredicate)
.collect(Collectors.toList());
}