该方法必须返回类型为Specification< Employee>的结果.在Java 8中 [英] This method must return a result of type Specification<Employee> in Java 8

问题描述

如何使用Lambda表达式返回返回值?

How to return using if else using Lambda expression ?

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));
    }
}

我遇到以下错误:

此方法必须返回规范类型的结果

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);
}

推荐答案

您可以执行以下操作；

You can do something like this;

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;
    }
}

在map()方法中隐藏if逻辑...

hiding the if logic in the map() method...

将端点更新为；

findEmployees(@RequestParam String searchValue) { }

因为所有输入都可以接受为String类型；

Since all input can be accepted as type String;

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

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