从我的表单中发布参数数据工作得很好,但是我需要使用JSON发布相同的信息。这怎么能做到呢?我的表单看起来像这样。
<form enctype="multipart/form-data" method="post" action="http://myrestfulservice.com/">
<input type="text" name="name">
<input type="text" name="email">
<textarea name="about" rows="3"></textarea>
<input name="file" type="file">
</form>
我的JSON字符串如下所示。
{
"name":"Josh U",
"email":"xxx@gmail.com",
"about":"Get rich programming or die tryna be something else",
"file":"@/home/josh/image.png"
}
我尝试过使用JSON字符串,但得到以下结果
Exception in thread "main" java.lang.RuntimeException: Failed : HTTP error code : 500
at com.ijoshluisaac.restful.client.RESTFulClientPostUsingJavaHTTPClientLibrary.main(RESTFulClientPostUsingJavaHTTPClientLibrary.java:35)
我的JAVA代码如下所示
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;
public class RESTFulClientPostUsingJavaHTTPClientLibrary {
// JSON content to post
private static final String JSON_OBJECT = "{\"name\":\"Josh U\",\"email\":\"xxx@gmail.com\",\"about\":\"Get rich programming or die tryna be something else\",\"file\":\"/home/josh/image.png\"}";
//Web services URLs
private static final String URL_SR = "http://myrestfulservice.com/";
public static void main(String[] args) {
try {
URL url = new URL(URL_SR);
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setDoOutput(true);
conn.setRequestMethod("POST");
conn.setRequestProperty("Content-Type", "application/json");
OutputStream os = conn.getOutputStream();
os.write(JSON_OBJECT.getBytes());
os.flush();
if (conn.getResponseCode() != HttpURLConnection.HTTP_CREATED) {
throw new RuntimeException("Failed : HTTP error code : "
+ conn.getResponseCode());
}
BufferedReader br = new BufferedReader(new InputStreamReader(
(conn.getInputStream())));
String output;
System.out.println("Successfully Executed RESTFul POST .... \n");
while ((output = br.readLine()) != null) {
System.out.println(output);
}
conn.disconnect();
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
}
2条答案
按热度按时间kjthegm61#
我认为你必须做两个步骤:
1.第一步是将表单内容转换为JS对象(困难步骤)。
1.第二步,将JS对象转换为JSON格式(简单步骤)。
您只需要向jQuery serializeArray添加更多功能,就可以得到我们所需的JS对象:
然后使用
JSON.stringify
函数将JS对象转换为JSON格式。q9rjltbz2#
使用jquery(因为它提供了许多开箱即用的特性,如将表单参数序列化到json)并使用下面的代码
请参阅http://api.jquery.com/serializeArray/