Question

我需要从java代码本身调用控制器。控制器如下，

@RequestMapping(value = "temp", method = RequestMethod.POST)
@ResponseBody
public String uploadDataFromExcel(@RequestBody Map<String, String> colMapObj, @ModelAttribute ReqParam reqParam) {
}

我正在尝试使用http post调用上述控制器，如下所示，

String url ="http://localhost:8081/LeadM" + "/temp/?searchData="+ reqParam.getSearchData()+" &exportDiscardRec=" + reqParam.isExportDiscardRec() + "&fileName=" + reqParam.getFileName() + "&sheetName=" + reqParam.getSheetName() +  "&importDateFormat=" + reqParam.getImportDateFormat()  + "&selectedAddressTypes="+ reqParam.getSelectedAddressTypes() + "&duplicatesHandleOn=" + reqParam.getDuplicatesHandleOn() + colMapObj;

HttpPost httpPost = new HttpPost(url);
        CloseableHttpClient httpClient = HttpClients.createDefault();
        CloseableHttpResponse httpResponse = httpClient.execute(httpPost);
        HttpEntity entity = httpResponse.getEntity();
        InputStream rstream = entity.getContent();
        jsonObject = new JSONObject(new JSONTokener(rstream));

其中reqParam是一个类对象是类对象，colMapObj是我想传递给上面控制器的地图。但是当执行http post时，它会在url中给出异常。如果有人知道正确的方式，请建议，谢谢。

Answer 1

这应该有效

@RequestMapping(value = "/temp", method = RequestMethod.POST)
@ResponseBody
public String uploadDataFromExcel(@RequestBody Map<String, String> colMapObj, @ModelAttribute ReqParam reqParam) {
}

和url应该是

String url ="http://localhost:8081/LeadM" + "/temp?"+ reqParam.getSearchData()+" &exportDiscardRec=" + reqParam.isExportDiscardRec() + "&fileName=" + reqParam.getFileName() + "&sheetName=" + reqParam.getSheetName() +  "&importDateFormat=" + reqParam.getImportDateFormat() + "&selectedAddressTypes="+ reqParam.getSelectedAddressTypes() + "&duplicatesHandleOn=" + reqParam.getDuplicatesHandleOn() + colMapObj;

Answer 2

网址不适用于spaces.From上面的代码：＆＃34; ＆安培; exportDiscardRec =＆＃34; 要避免此类问题，请尽可能使用 URIBuilder 或类似内容。

现在对于请求，您没有正确构建请求，例如您没有提供正文。检查以下示例：

    Map<String, String> colMapObj = new HashMap<>();
    colMapObj.put("testKey", "testdata");

    CloseableHttpClient client = HttpClients.createDefault();
    HttpPost httpPost = new HttpPost(url);

    JSONObject body = new JSONObject(colMapObj);
    StringEntity entity = new StringEntity(body.toString());
    httpPost.setEntity(entity);
    httpPost.setHeader("Accept", "application/json");
    httpPost.setHeader("Content-type", "application/json");

    CloseableHttpResponse response = client.execute(httpPost);
    System.out.println(response.getEntity().toString());
    client.close();

更多示例只是google＆＃34; apache http客户端发布示例＆＃34; （例如http://www.baeldung.com/httpclient-post-http-request）

Answer 3

对您的查询字符串进行编码。

String endpoint = "http://localhost:8081/LeadM/tmp?";
String query = "searchData="+ reqParam.getSearchData()+" &exportDiscardRec=" + reqParam.isExportDiscardRec() + "&fileName=" + reqParam.getFileName() + "&sheetName=" + reqParam.getSheetName() +  "&importDateFormat=" + reqParam.getImportDateFormat()  + "&selectedAddressTypes="+ reqParam.getSelectedAddressTypes() + "&duplicatesHandleOn=" + reqParam.getDuplicatesHandleOn() + colMapObj;

String q = URLEncoder.encode(query, "UTF-8");

String finalUrl = endpoint + q;

如果这不起作用，则在连接之前编码各个参数。

旁注

如果你在同一个jvm中运行，那么你可以直接调用方法
如果您拥有上传方法，请考虑将查询字符串更改为表格参数

无法使用http post传递请求参数

3 个答案: