我嘗試發送POST帶有這樣的正文的請求{ "method": "getAreas", "methodProperties": { "prop1" : "value1", "prop2" : "value2", }}這是我的代碼static final String HOST = "https://somehost.com"; public String sendPost(String method, Map<String, String> methodProperties) throws ClientProtocolException, IOException { HttpPost post = new HttpPost(HOST); List<NameValuePair> urlParameters = new ArrayList<>(); urlParameters.add(new BasicNameValuePair("method", method)); List<NameValuePair> methodPropertiesList = methodProperties.entrySet().stream() .map(entry -> new BasicNameValuePair(entry.getKey(), entry.getValue())) .collect(Collectors.toList()); // ??? urlParameters.add(new BasicNameValuePair("methodProperties", methodPropertiesList)); post.setEntity(new UrlEncodedFormEntity(urlParameters)); try (CloseableHttpClient httpClient = HttpClients.createDefault(); CloseableHttpResponse response = httpClient.execute(post)) { return EntityUtils.toString(response.getEntity()); } }但 的構造函數BasicNameValuePair適用(String, String)。所以我需要另一堂課。有什么辦法可以添加methodPropertiesList嗎urlParameters？
3 回答
長風秋雁
TA貢獻1757條經驗 獲得超7個贊
您的請求看起來像 json 結構，因此發布如下數據：
class pojo1{
String method;
Map<String,String> methodProperties;
}
String postUrl = "www.site.com";// put in your url
Gson gson = new Gson();
HttpClient httpClient = HttpClientBuilder.create().build();
HttpPost post = new HttpPost(postUrl);
StringEntity postingString = new StringEntity(gson.toJson(pojo1));//gson.tojson() converts your pojo to json
post.setEntity(postingString);
post.setHeader("Content-type", "application/json");
HttpResponse response = httpClient.execute(post);
慕無忌1623718
TA貢獻1744條經驗 獲得超4個贊
對于這個問題有一個眾所周知的方法。在大多數情況下，您將創建自己的對象來描述要在 HttpPost 中發送的內容。所以你會得到類似的東西：
static final String HOST = "https://somehost.com";
? public String sendPost(String method,
? ? ? Map<String, String> methodProperties) throws ClientProtocolException, IOException {
? ? HttpPost post = new HttpPost(HOST);
? ? MyResource resource = new MyResource();
? ? resource.setMethod(method);
? ? MyNestedResource nestedResource = new MyNestedResource();
? ? nestedResource.setMethodProperties(methodProperties);
? ? resource.setNestedResourceMethodProperties(nestedResource);
? ? StringEntity strEntity = new StringEntity(gson.toJson(resource));
? ? post.setEntity(strEntity);
? ? try (CloseableHttpClient httpClient = HttpClients.createDefault();
? ? ? ? CloseableHttpResponse response = httpClient.execute(post)) {
? ? ? return EntityUtils.toString(response.getEntity());
? ? }
? }
這通常是您使用嵌套結構處理更復雜的 json 對象的方式。您必須為要發送的資源創建類（在您的示例中，它可能是一個類并在其中使用映射，但通常您也為嵌套對象創建一個類，如果它具有特定的結構）。
慕妹3146593
TA貢獻1820條經驗 獲得超9個贊
static final String HOST = "https://somehost.com";
? public String sendPost(String method, Map<String, String> methodProperties) throws ClientProtocolException, IOException {
? ? HttpPost post = new HttpPost(HOST);??
? ? Gson gson = new Gson();
? ? Params params = new Params(method, methodProperties);
? ? StringEntity entity = new StringEntity(gson.toJson(params));? ?
? ? post.setEntity(entity);
? ? try (CloseableHttpClient httpClient = HttpClients.createDefault();
? ? ? ? CloseableHttpResponse response = httpClient.execute(post)) {
? ? ? return EntityUtils.toString(response.getEntity());
? ? }
? }
? class Params {
? ? String method;? ?
? ? Map<String, String> methodProperties;
? ? public Params(String method, Map<String, String> methodProperties) {
? ? ? this.method = method;
? ? ? this.methodProperties = methodProperties;
? ? }
? ? //getters
? }
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