发出返回JSON值的问题

Question

我没有收到来自服务器的JSON类型数据的响应.

我正在使用JSON插件.

jQuery( "#dialog-form" ).dialog({ 
    autoOpen: false,
    height: 500,
    width: 750,
    modal: true,
    buttons :{
        "Search" : function(){
            jQuery.ajax({type : 'POST',
            dataType : 'json',
             url : '<s:url action="part" method="finder" />',
         success : handledata})
        }
    }
});
var handledata = function(data)
{
    alert(data);
}

如果dataType = 'json'我没有得到任何回复,但如果我没有提及任何回复,dataType我将获得该页面的HTML格式.

public String list(){
    JSONObject jo = new JSONObject();
    try {
        Iterator it = findList.iterator();
        while(it.hasNext()){
             SearchResult part = (SearchResult) it.next();
             jo.put("col1",part.getcol1());
             jo.put("col2",part.getcol2());
        }
        log.debug("--------->:"+jo.toString());
    } catch (Exception e) {
        log.error(e);
    }
    return jo.toString();
}

struts.xml中:

<package name="default" namespace="/ajax" extends="json-default">
  <action name="finder" 
       class="action.Part" method="finder" name="finder">
       <result type="json" />
  </action>
</package>

JSP页面:

<div id="dialog-form" >
    <form action="" id="channelfinder">
        <textarea id="products" name="prodnbr"<s:property value='prodNbr'/>   
    </form>
</div>

控制台错误:

org.apache.struts2.dispatcher.Dispatcher - 无法找到操作或结果没有为操作动作定义结果.部分和结果{"col1":"col1","col2":"col2"}

web.xml:

    <?xml version="1.0" encoding="ISO-8859-1"?>
     <web-app version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee"      xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
            xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
      <display-name>/parts</display-name>
      <description>Parts List Web App</description>

    <filter>
          <filter-name>struts-cleanup</filter-name>
          <filter-class>org.apache.struts2.dispatcher.ActionContextCleanUp</filter-class>
        </filter>

        <filter>
        <filter-name>sitemesh</filter-name>
        <filter-class>com.opensymphony.module.sitemesh.filter.PageFilter</filter-class>
        </filter>

       <filter>
        <filter-name>struts2</filter-name>
        <filter-class>org.apache.struts2.dispatcher.ng.filter.StrutsPrepareAndExecuteFilter</filter-class>
        <init-param>
            <param-name>actionPackages</param-name>
            <param-value>com.action</param-value>
        </init-param>
    </filter>


    <filter-mapping>
        <filter-name>struts-cleanup</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>

    <filter-mapping>
        <filter-name>sitemesh</filter-name>
        <url-pattern>/*</url-pattern>
        <dispatcher>REQUEST</dispatcher>
        <dispatcher>FORWARD</dispatcher>
        <dispatcher>INCLUDE</dispatcher>
    </filter-mapping>

    <filter-mapping>
        <filter-name>struts2</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>
    <error-page>
        <exception-type>java.lang.Throwable</exception-type>
        <location>/errorPage.jsp</location>
    </error-page>
    <error-page>
        <error-code>404</error-code>
        <location>/errorPage.jsp</location>
    </error-page>

  <!-- Spring -->
  <context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/applicationContext.xml</param-value>
  </context-param>
  <listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
  </listener>

  </web-app>

我没有获得jQuery成功的数据.请纠正我,这里有什么不对吗？

Answer 1

关于Struts2-JSON插件

Struts2 JSON插件以特定方式工作:

JSON插件提供了一个"json"结果类型,可将操作序列化为JSON.

它将整个Action序列化为JSON,除外

• 瞬态特性
• 没有Getter的属性

如果您不希望序列化整个Action,但只需要选择一个对象,则可以指定根对象:

使用"root"属性(OGNL表达式)指定要序列化的根对象.

它可以struts.xml像这样完成:

<result type="json">
    <param name="root">
        objectToBeSerialized
    </param>
</result>

行动应该有:

private CustomObject objectToBeSerialized;

public CustomObject getObjectToBeSerialized(){
    return this.objectToBeSerialized;
}

CustomObject可以是Primitive,String,Array等...

以这种方式使用它(它的构建方式),你可以返回SUCCESS并ERROR喜欢任何其他AJAX Struts2 Action,而不破坏框架约定,并从AJAX jQuery调用的回调函数访问序列化JSON对象,就像任何其他字段一样(如果使用rootObject,则"数据" var handledata = function(data)将是您的对象,否则它将是您的Action).

关于你的具体案例

在您的情况下,假设您的对象结构如下所示

row1 [col1, col2], 
row2 [col1, col2], 
rowN [col1, col2]

您可以创建一个包含两列的对象列表:

价值对象

public class MyRow implements Serializable {
    private static final long serialVersionUID = 1L;

    private String col1; 
    private String col2;

    // Getters
    public String getCol1(){ 
        return this.col1; 
    }
    public String getCol2(){ 
        return this.col2; 
    }
}

动作类

public class PartAction implements Serializable {
    private static final long serialVersionUID = 1L;

    private List<MyRow> rows;   

    // Getter
    public  List<MyRow> getRows() { 
        return this.rows; 
    } 

    public String finder() {
        String result = Action.SUCCESS;
        rows = new ArrayList<MyRow>();

        try {
            Iterator it = findList.iterator();
            while(it.hasNext()) {
                SearchResult part = (SearchResult) it.next();
                MyRow row = new MyRow();
                row.setCol1(part.getcol1());
                row.setCol2(part.getcol2());
                rows.add(row);
            }
        } catch (Exception e) {
            result = Action.ERROR;
            log.error(e);
        }
        return result;
    }  
} 

Struts.xml

<package name="default" namespace="/ajax" extends="json-default">
    <action name="finder" class="action.Part" method="finder" name="finder">
        <result type="json" >
            <param name="root">
                rows
            </param>
        </result>
  </action>
</package>

要在AJAX回调函数中测试它,只需使用$.each:

var handledata = function(data) {
    $.each(data, function(index) {
        alert(data[index].col1);
        alert(data[index].col2);
    });     
}

当然你可以使用一个List<List<String>>而不是一个Custom对象,或者你喜欢的任何其他对象结构:它只是为了让你明白.

Answer 2

jQuery AjaxdataType : 'json'使用A 来指定在执行操作和结果时期望由回调函数返回的数据类型,以及从服务器返回的响应.success

dataType(默认值:智能猜测(xml,json,script,或html))

类型: String

您期望从服务器返回的数据类型.如果没有指定,jQuery将尝试根据响应的MIME类型推断它(XML MIME类型将产生XML,在1.4 JSON中将产生一个JavaScript对象,在1.4脚本中将执行脚本,其他任何东西将是以字符串形式返回).

URL应正确指向操作映射.假设它将在默认命名空间中,否则您应该修改URL和映射以添加namespace属性.

<script type="text/javascript">
  $(function() {
    $("#dialog-form").dialog ({
      autoOpen: true,
      height: 500,
      width: 750,
      modal: true,
      buttons : {
        "Search" : function() {
          $.ajax({
            url : '<s:url action="part" />',
            success : function(data) {
              //var obj = $.parseJSON(data);
              var obj = data;
              alert(JSON.stringify(obj));
            }
          });
        }
      }
    });
  });
</script>

json如果JSONObject手动构建,则不需要返回结果类型.您可以将文本作为流结果返回,然后根据需要将字符串转换为JSON.

struts.xml:

<package name="default" extends="struts-default">
  <action name="part" class="action.PartAction" method="finder">    
    <result type="stream">
      <param name="contentType">text/html</param>
      <param name="inputName">stream</param>
    </result>
  </action>
</package>

行动:

public class PartAction extends ActionSupport {

  public class SearchResult {
    private String col1;
    private String col2;

    public String getCol1() {
      return col1;
    }

    public void setCol1(String col1) {
      this.col1 = col1;
    }

    public String getCol2() {
      return col2;
    }

    public void setCol2(String col2) {
      this.col2 = col2;
    }

    public SearchResult(String col1, String col2) {
      this.col1 = col1;
      this.col2 = col2;
    }
  }

  private InputStream stream;

  //getter here
  public InputStream getStream() {
    return stream;
  }

  private List<SearchResult> findList = new ArrayList<>();

  public List<SearchResult> getFindList() {
    return findList;
  }

  public void setFindList(List<SearchResult> findList) {
    this.findList = findList;
  }

  private String list() {
    JSONObject jo = new JSONObject();
    try {
      for (SearchResult part : findList) {
        jo.put("col1", part.getCol1());
        jo.put("col2", part.getCol2());
      }
      System.out.println("--------->:"+jo.toString());
    } catch (Exception e) {
      e.printStackTrace();
      System.out.println(e.getMessage());
    }
    return jo.toString();
  }

  @Action(value="part", results = {
    @Result(name="stream", type="stream", params = {"contentType", "text/html", "inputName", "stream"}),
    @Result(name="stream2", type="stream", params = {"contentType", "application/json", "inputName", "stream"}),
    @Result(name="json", type="json", params={"root", "findList"})
  })
  public String finder() {
    findList.add(new SearchResult("val1", "val2"));
    stream = new ByteArrayInputStream(list().getBytes());
    return "stream2";
  }
}

我在结果类型和内容类型上放置了不同的结果,以便更好地描述这个想法.您可以返回任何这些结果并返回JSON对象,无论是否为字符串.字符串化版本需要解析返回的数据以获取JSON对象.您还可以选择哪种结果类型更好地序列化以满足您的需求,但我的目标是表明如果您需要序列化简单对象,那么json插件不是必需的.

参考文献:

• 我们如何返回文本字符串作为响应
  
• 如何转换JSONObject为字符串