java - 在 Java 8 中，此方法必须返回 Specification<Employee> 类型的结果

Question

如何使用 Lambda 表达式使用 if else 返回？

public static Specification<Employee> textInAllColumns(Object value) {
    if (value instanceof String) {
        return (root, query, builder) -> builder
            .or(root.getModel().getDeclaredSingularAttributes().stream()
                .filter(a -> {
                    return a.getJavaType()
                        .getSimpleName()
                        .equalsIgnoreCase("String") ? true : false;
                })
                .map(a -> builder.like(root.get(a.getName()), getString((String) value)))
                .toArray(Predicate[]::new));
    } else if (value instanceof Integer) {
        return (root, query, builder) -> builder
            .or(root.getModel().getDeclaredSingularAttributes().stream()
                .filter(a -> {
                    return a.getJavaType()
                        .getSimpleName()
                        .equalsIgnoreCase("Integer") ? true : false;
                })
                .map(a -> builder.equal(root.get(a.getName()), value))
                .toArray(Predicate[]::new));
    }
}

我收到以下错误:

This method must return a result of type Specification

@GetMapping("/findEmployees")
public ResponseEntity<List<Employee>> findEmployees(@RequestParam Object searchValue) {
    List<Employee> employees = employeeService.searchGlobally(searchValue);
    return new ResponseEntity<>(employees, HttpStatus.OK);
}

Answer 1

你可以做这样的事情；

public static Specification<Employee> textInAllColumns(Object value) {
    return (root, query, builder) -> builder.or(root.getModel().getDeclaredSingularAttributes().stream()
            .filter(attr -> attr.getJavaType().equals(value.getClass()))
            .map(attr -> map(value, root, builder, attr))
            .toArray(Predicate[]::new));
}

private static Object map(Object value, Root root, CriteriaBuilder builder, SingularAttribute a) {
    switch (value.getClass().getSimpleName()) {
        case "String":
            return builder.like(root.get(a.getName()), getString((String) value));
        case "Integer":
            return builder.equal(root.get(a.getName()), value);
        case "Date":
            return //date mapping
        default:
            return //default;
    }
}

隐藏 map() 方法中的 if 逻辑...

---

将您的端点更新为；

findEmployees(@RequestParam String searchValue) { }

由于所有输入都可以作为 String 类型接受；

Object finalValue = searchValue;
try {
    finalValue = Integer.parseInt(searchValue);
} catch (Exception e) {
    // ignore
}
try {
    finalValue = parseDateFromStr(searchValue);
} catch (Exception e) {
    // ignore
}
// use finalValue as input to your logic, it will contain correct type