AJAX请求中的JSON对象为Null或空

Null or empty JSON object from AJAX request

本文关键字：Null 或空 对象 JSON 请求 AJAX      更新时间：2023-09-26

最终删除了我之前发布的另一个与此相关的问题，因为它已经过时，还有另一件事需要解决。下面的答案非常清楚，比其他答案更清楚，我相信这对其他有类似困惑的人会有所帮助。

本质上，我想在Java中使用一个来自客户端JS的String。下面的JS警报工作正常，但是我的JSON对象在Servlet中为null或为空。我真的只需要把一个字符串传递给servlet，这样我就可以运行SQL查询，但我遇到了很多麻烦

谢谢！

html文件

       <div class="jumbotron" id="jumboframe">
            <h1>Enter your URL below</h1>
            <input class="txtUrl" type="text" id="txtUrl" value="...." onfocus="if(this.value == '....') { this.value = ''; }"/>
            <p><a class="btn btn-lg submitButton" href="testpage.html" onclick="getURL()" role="button">Start searching!</a></p>
        </div>
        <script>
            function getURL() {
                var urlString = document.getElementById("txtUrl").value;
                var params = {url: urlString};

            $.ajax({
                    type: "POST",
                    url: "testpage.html",
                    contentType: "application/json",
                    dataType: "text", // "JSON" here doesn't call alerts below
                    data: {
                        'url': urlString
                    },
                    success: function(data) {
                        alert("did it!");
                        alert(JSON.stringify(params));
                        alert(JSON.stringify(data));
                    }
                });
            }
        </script>

Java servlet

public class JavaServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
public void doGet(HttpServletRequest request, HttpServletResponse response) throws IOException {
    doPost(request, response);
}
public void doPost(HttpServletRequest request, HttpServletResponse response) throws IOException {
    PrintWriter out = response.getWriter();
    StringBuilder builder = new StringBuilder();
    BufferedReader reader = request.getReader();
    String line;
    while ((line = reader.readLine()) != null) {
        builder.append(line);
    }
    out.println(builder.toString());
    String url = request.getParameter("url");
    Map<String, String[]> map = request.getParameterMap();  // empty {}
    out.println(map.toString());  // null
    out.println(url);  // null

web.xml

  <servlet>
      <servlet-name>JavaServlet</servlet-name>
      <servlet-class>path.to.my.JavaServlet</servlet-class>
  </servlet>
  <servlet-mapping>
    <servlet-name>JavaServlet</servlet-name>
    <url-pattern>/testpage.html</url-pattern>
  </servlet-mapping>

再次感谢

如果您想从请求数据中获得Json，那么您应该将其作为字符串传递，然后，您的html可能看起来像这样（我已经将servlet路径更改为JavaServlet.do）

<div class="jumbotron" id="jumboframe">
            <h1>Enter your URL below</h1>
            <input class="txtUrl" type="text" id="txtUrl" value="...." onfocus="if(this.value == '....') { this.value = ''; }"/>
            <p><a class="btn btn-lg submitButton" href="#" onclick="getURL()" role="button">Start searching!</a></p>
        </div>
        <script>
            function getURL() {
                var urlString = document.getElementById("txtUrl").value;
                var params = {url: urlString};
                var jsonDataStr = JSON.stringify(params); // Here you get the json String
            $.ajax({
                    type: "POST",
                    url: "JavaServlet.do",
                    contentType: "application/json",
                    dataType: "text", // "JSON" here doesn't call alerts below * Because the servlet thrown an error, you should handle it *
                    data: jsonDataStr,
                    success: function(data) {
                        alert("Your URL is " + data); // **get the response text**
                        alert("Handle it");
                    },
                    error: function(data) { // Handle the error
                        alert(data.error().getResponseHeader());
                    }
                });
            }
        </script>

现在，JavaServlet将接收一个Json作为String，您必须解析它（使用像Gson这样的第三方库）。

public void doPost(HttpServletRequest request, HttpServletResponse response) throws IOException {
    PrintWriter out = response.getWriter();
    StringBuilder builder = new StringBuilder();
    BufferedReader reader = request.getReader();
    String line;
    while ((line = reader.readLine()) != null) {
        builder.append(line);
    }
    String jsonString = builder.toString(); // get the json from client
    UrlDto urlDto = new Gson().fromJson(UrlDto.class,jsonString); // parse you json to Java object
    String yourUrl = urlDto.getUrl();
    out.print(yourUrl); // **Set the response text**
    out.close();

WEB.XML和URLDto

class UrlDto {
    private String url;
    // G&S ...
}
    <servlet>
        <servlet-name>JavaServlet</servlet-name>
        <servlet-class>servlet.JavaServlet</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>JavaServlet</servlet-name>
        <url-pattern>/JavaServlet.do</url-pattern>
    </servlet-mapping>

我希望这能帮助你

相关文章：