发出返回JSON值的问题 [英] Issue returning JSON value
问题描述
我没有从服务器获得JSON类型数据的响应。
I'm not getting response as JSON type data from server.
我正在使用JSON插件。
I'm using JSON plugin.
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 dataType ，我将获得该页面的HTML格式。
If dataType = 'json' I am not getting any response, but if I don't mention any dataType, I'm getting the HTML format of the page.
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：
<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 - 找不到操作或结果
没有为action action.Part
和结果{col1定义结果： col1，col2：col2}
org.apache.struts2.dispatcher.Dispatcher - Could not find action or result No result defined for action action.Part and result {"col1":"col1","col2":"col2"}
web.xml ：
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。
请更正我，这里有什么问题？
I'm not getting data to jQuery success. Please correct me, whats wrong here?
推荐答案
A dataType：'json'< jQuery Ajax 使用/ code>指定预期的数据类型在执行操作和结果时从 success 回调函数返回，并从服务器返回响应。
A dataType : 'json' is used by jQuery Ajax to specify a data type that is expected to return by the success callback function when the action and result is executed, and a response returned from the server.
dataType（默认值：智能猜测（xml，json，script，或html））
dataType(default: Intelligent Guess (xml,json,script, orhtml))
类型：字符串
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。）
The type of data that you're expecting back from the server. If none is specified, jQuery will try to infer it based on the MIME type of the response (an XML MIME type will yield XML, in 1.4 JSON will yield a JavaScript object, in 1.4 script will execute the script, and anything else will be returned as a string).
URL应正确指向操作映射。假设它将在默认命名空间中，否则你应该修改URL和映射以添加命名空间属性。
The URL should correctly point to the action mapping. Assume it will be in the default namespace, otherwise you should modify URL and mapping to add the namespaceattribute.
<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>
返回 json 如果结果类型不需要您手动构建 JSONObject 。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。
Returning json result type is not needed if you build the JSONObject manually. You can return text as stream result then convert a string to JSON if needed.
struts.xml ：
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中添加了不同的结果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，以使其工作。
I have placed different results with result type and content type to better describe the idea. You could return any of these results and return JSON object either stringified or not. The stringified version requires to parse returned data to get the JSON object. You can also choose which result type better serializes to suit your needs but my goal was to show that if you need to serialize the simple object then json plugin is not necessary to get it working.
参考文献：
- How can we return a text string as the response
- How to convert
JSONObjectto string
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