如何与Spring RestTemplate发送阵列？(How to send array with

我如何使用Spring RestTemplate发送数组参数？

这是在服务器端执行：

@RequestMapping(value = "/train", method = RequestMethod.GET)
@ResponseBody
public TrainResponse train(Locale locale, Model model, HttpServletRequest request, 
    @RequestParam String category,
    @RequestParam(required = false, value = "positiveDocId[]") String[] positiveDocId,
    @RequestParam(required = false, value = "negativeDocId[]") String[] negativeDocId) 
{
    ...
}

这是我已经试过：

Map<String, Object> map = new HashMap<String, Object>();
map.put("category", parameters.getName());
map.put("positiveDocId[]", positiveDocs); // positiveDocs is String array
map.put("negativeDocId[]", negativeDocs); // negativeDocs is String array
TrainResponse response = restTemplate.getForObject("http://localhost:8080/admin/train?category={category}&positiveDocId[]={positiveDocId[]}&negativeDocId[]={negativeDocId[]}", TrainResponse.class, map);

以下是这显然是不正确的实际要求网址：

http://localhost:8080/admin/train?category=spam&positiveDocId%5B%5D=%5BLjava.lang.String;@4df2868&negativeDocId%5B%5D=%5BLjava.lang.String;@56d5c657`

一直在试图四处搜寻，但没有找到一个解决方案。 任何指针将不胜感激。

Answer 1:

Spring的UriComponentsBuilder的伎俩也允许变量扩展。 假设你想传递的字符串作为参数“ATTR”的资源，而您只需要一个URI与路径变量数组：

UriComponents comp = UriComponentsBuilder.fromHttpUrl(
    "http:/www.example.com/widgets/{widgetId}").queryParam("attr", "width", 
        "height").build();
UriComponents expanded = comp.expand(12);
assertEquals("http:/www.example.com/widgets/12?attr=width&attr=height", 
   expanded.toString());

否则，如果您需要定义应该是在运行时扩展一个URI，你不知道数组的大小提前，使用http://tools.ietf.org/html/rfc6570 UriTemplate与{？关键*}占位符，并与来自UriTemplate类展开https://github.com/damnhandy/Handy-URI-Templates 。

UriTemplate template = UriTemplate.fromTemplate(
    "http://example.com/widgets/{widgetId}{?attr*}");
template.set("attr", Arrays.asList(1, 2, 3));
String expanded = template.expand();
assertEquals("http://example.com/widgets/?attr=1&attr=2&attr=3", 
    expanded);

对于非Java语言看https://code.google.com/p/uri-templates/wiki/Implementations 。

Answer 2:

我最终通过收集循环构造URL。

Map<String, Object> map = new HashMap<String, Object>();
map.put("category", parameters.getName());

String url = "http://localhost:8080/admin/train?category={category}";
if (positiveDocs != null && positiveDocs.size() > 0) {
    for (String id : positiveDocs) {
        url += "&positiveDocId[]=" + id;
    }
}
if (negativeDocId != null && negativeDocId.size() > 0) {
    for (String id : negativeDocId) {
        url += "&negativeDocId[]=" + id;
    }
}

TrainResponse response = restTemplate.getForObject(url, TrainResponse.class, map);

Answer 3:

试试这个

改变从你的要求映射

@RequestMapping(value = "/train", method = RequestMethod.GET)

至

 @RequestMapping(value = "/train/{category}/{positiveDocId[]}/{negativeDocId[]}", method = RequestMethod.GET)

和你在restTemplate网址

变更网址如下格式

http://localhost:8080/admin/train/category/1,2,3,4,5/6,7,8,9

Answer 4:

这里是我是如何实现的吧：

其余的控制器：

@ResponseBody
@RequestMapping(value = "/test", method = RequestMethod.GET)
public ResponseEntity<Response> getPublicationSkus(
       @RequestParam(value = "skus[]", required = true) List<String> skus) {
    ...
}

请求：

List<String> skus = Arrays.asList("123","456","789");

Map<String, String> params = new HashMap<>();
params.put("skus", toPlainString(skus));

Response response = restTemplate.getForObject("http://localhost:8080/test?skus[]={skus}", 
                                      Response.class, params);

然后你只需要实现转换的方法List或String[]用逗号分隔中，例如一个普通字符串Java 8会是这样的：

private static String toPlainString(List<String> skus) {
    return skus.stream().collect(Collectors.joining(","));
}

文章来源: How to send array with Spring RestTemplate?