Twilio “需要一个‘收件人’电话号码。” Java HttpRequest

Question

我正在尝试将请求从 Java 发送到 Twilio SMS API。 我正在使用java.net.http包：

var url = UriBuilder.fromUri(
                "https://api.twilio.com/2010-04-01/Accounts/MyAccount/Messages.json").build();

var urlEncodedBody = URLEncoder.encode(String.format("To=%s&From=%s&Body=%s",
                                                         "+1123456789",
                                                         "+1223456789",
                                                         "Hello"),
                                           StandardCharsets.UTF_8);
var request = HttpRequest.newBuilder(url)
                             .headers("Authorization", "Basic " + base64,
                                      "Content-Type", "application/x-www-form-urlencoded")
                             .method("POST", HttpRequest.BodyPublishers.ofString(
                                     urlEncodedBody))
                             .build();

HttpClient httpClient = HttpClient.newBuilder().build();

try {
            var response = httpClient.send(request, HttpResponse.BodyHandlers.ofString());
} catch (IOException |
                InterruptedException e) {
            throw new RuntimeException(e);
}

我不断收到错误响应：

{"code": 21604, "message": "A 'To' phone number is required.", "more_info": "https://www.twilio.com/docs/errors/21604", "status": 400}

知道我错过了什么吗？

Answer 1

我认为，当您对请求的正文进行编码时，您也在对& s 和= s 进行编码，因此它似乎只是一个字符串。

This answer on a different question建议使用Map并依次编码每个值。 以此为基础，您可以将字符串格式替换为：

Map<String, String> parameters = new HashMap<>();
parameters.put("To", "+1123456789");
parameters.put("From", "+1223456789");
parameters.put("Body", "Hello");

String urlEncodedBody = parameters.entrySet()
    .stream()
    .map(e -> e.getKey() + "=" + URLEncoder.encode(e.getValue(), StandardCharsets.UTF_8))
    .collect(Collectors.joining("&"));