Spring 数据规范 - 如何按多个属性过滤对象

Question

我正在尝试通过多个属性实现对用户的搜索。 比方说，我的人有名字和姓氏。我想输入"Michael Doe"并列出所有具有此姓名和姓氏的人。当我只输入"迈克尔"时，我想要所有的迈克尔，依此类推。

我试图通过使用谓词来实现它，例如：

private static Specification<UserEntity> containsTextInAttributes(List<String> text, List<String> attributes) {
List<String> finalTextArray = text.stream().map(e -> "%"+e+"%").collect(Collectors.toList());
return (root, query, builder) -> builder.or(root.getModel().getDeclaredSingularAttributes().stream()
.filter(a -> attributes.contains(a.getName()))
.map(a -> builder.or(finalTextArray.stream().map(e -> builder.like(root.get(a.getName()), e)).toArray(Predicate[]::new)))
.toArray(Predicate[]::new)
);
}
public static Specification<UserEntity> containsTextInAllAttributes(List<String> text) {
return containsTextInAttributes(text, Arrays.asList("lastname", "firstname", "email", "phoneNumber"));
}

但是，当我搜索"Michael Doe"时，我有所有的Michaels和Lastname Doe。如何解决我的问题？

提前感谢您的帮助。

Answer 1

I."Michael D", "firstName", "lastName"或"mdoe@gmail", "email"

public <T> Specification<T> dynamicLike(String likeValue, String... properties) {
return (Root<T> root, CriteriaQuery<?> query, CriteriaBuilder cb) -> {
Expression<String> concat = null;
for (String property : properties) {
if (concat == null) {
concat = cb.concat("", root.get(property));
} else {
concat = cb.concat(concat, cb.concat(" ", root.get(property)));
}
}
return cb.like(cb.lower(concat), "%" + likeValue.toLowerCase() + "%");
};
}

第二。"Mic Do @gmail", "firstName", "lastName", "email"

public <T> Specification<T> dynamicLike2(String value, String... properties) {
return (Root<T> root, CriteriaQuery<?> query, CriteriaBuilder cb) -> {
String[] values = value.split("\s");
int minNumber = Integer.min(values.length, properties.length);
Predicate[] likes = new Predicate[minNumber];
for (int i = 0; i < minNumber; i++) {
likes[i] = cb.like(cb.lower(root.get(properties[i])), "%" + values[i].toLowerCase() + "%");
}
return cb.and(likes);
};
}