页面排序不适用于零大小

发布于 2025-01-10 15:11:50 字数 185 浏览 0 评论 0原文

我有以下代码：

PageRequest pageRequest = PageRequest.of(0, count, page.getSort());

当 count = 0 时，它不起作用，因为据我所知，排序不适用于 null。有没有办法用零大小来解决这个问题，或者我应该申请计算其他数字？

原文

I have a following code:

PageRequest pageRequest = PageRequest.of(0, count, page.getSort());

With count = 0 it isn't working, because as I see the sorting won't work with null. Is there a way to solve this with a zero size or I should apply to count other number?

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猛虎独行 2025-01-17 15:11:50

使用 count=0，您的 PageRequest 会转换为请求不带任何元素的排序集合。
排序，事实上完整的请求没有任何意义，因此你不应该传递 0，这实际上是由代码验证的：

https://github.com/spring-projects/spring-data-commons/blob/8e5010c490459785316e8dc55817a9ff61227290/src/main/java/org/springframework/data/domain/AbstractPageRequest.java#L27

    public AbstractPageRequest(int page, int size) {

        if (page < 0) {
            throw new IllegalArgumentException("Page index must not be less than zero!");
        }

        if (size < 1) {
            throw new IllegalArgumentException("Page size must not be less than one!");
        }

        this.page = page;
        this.size = size;
    }

With count=0 your PageRequest translate to requesting a sorted collection without any elements.
Sorting and in fact the complete request doesn't make any sense, therefore you should not pass 0 and that is actually verified by the code:

https://github.com/spring-projects/spring-data-commons/blob/8e5010c490459785316e8dc55817a9ff61227290/src/main/java/org/springframework/data/domain/AbstractPageRequest.java#L27

    public AbstractPageRequest(int page, int size) {

        if (page < 0) {
            throw new IllegalArgumentException("Page index must not be less than zero!");
        }

        if (size < 1) {
            throw new IllegalArgumentException("Page size must not be less than one!");
        }

        this.page = page;
        this.size = size;
    }

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