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’我没有得到任何响应,但如果我没有提到任何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 – Could not find action or result
No result defined for action action.Part
and result {“col1″:”col1″,”col2″:”col2”}
web.xml中：
<?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？
解决方法:
数据类型：jQuery Ajax使用’json’来指定在执行操作和结果时期望由成功回调函数返回的数据类型,以及从服务器返回的响应.
dataType(default: Intelligent Guess (xml,json,script, orhtml))Type:
StringThe 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,否则您应该修改URL和映射以添加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>
如果手动构建JSONObject,则不需要返回json结果类型.您可以将文本返回为stream result,然后根据需要将字符串转换为JSON.
struts.xml中：
<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：
> How can we return a text string as the response
> How to convert JSONObject to string
