5

我有以下控制器方法:

@RequestMapping(method = GET, produces = APPLICATION_JSON_VALUE)
@Transactional(readOnly = true)
public ResponseEntity list(Pageable pageable, PagedResourcesAssembler pagedResourcesAssembler) {
    Page<Customer> customers = customerRepository.findAll(pageable);
    return ResponseEntity.ok().body(pagedResourcesAssembler.toResource(customers, customerResourceAssembler));
}

@RequestMapping(value = "/search", method = GET, produces = APPLICATION_JSON_VALUE)
@Transactional(readOnly = true)
public ResponseEntity search(@RequestParam("q") String q, Pageable pageable, PagedResourcesAssembler pagedResourcesAssembler) {
    Specification spec = where(..some specs..);
    Page<Customer> customers = customerRepository.findAll(spec, pageable);
    return ResponseEntity.ok().body(pagedResourcesAssembler.toResource(customers, customerResourceAssembler));
}

第一种方法将所有客户资源作为页面返回。

第二个也返回分页结果,但可以选择提供q查询参数进行过滤。

来自搜索方法的 JSON HATEOAS 响应包含一个下一页链接,例如:

{
     "rel": "next",
     "href": "http://localhost:8080/api/customers/search?page=1&size=10{&sort}"
}

问题是这里的q查询参数丢失了。

我应该在这里使用PagedResourcesAssembler不同的吗?

4

1 回答 1

8

我想我通过手动创建链接找到了解决方案;请参见下面的示例。

@RequestMapping(value = "/search", method = GET, produces = APPLICATION_JSON_VALUE)
@Transactional(readOnly = true)
public ResponseEntity search(@RequestParam("q") String q, Pageable pageable, PagedResourcesAssembler pagedResourcesAssembler) {

    // create link to search method with q; pass link as 3th param to paged resource assembler's toResource method
    Link link = linkTo(methodOn(CustomerController.class).search(q, pageable, pagedResourcesAssembler)).withSelfRel();

    Specification spec = where(..some specs..);
    Page<Customer> customers = customerRepository.findAll(spec, pageable);
    return ResponseEntity.ok().body(pagedResourcesAssembler.toResource(customers, customerResourceAssembler, link));
}
于 2014-11-14T18:55:59.457 回答