我没有从服务器获得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'我没有得到任何回应,但是如果我没有任何回应,那么dataType我正在获取页面的HTML格式。
dataType = 'json'
dataType
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-找不到动作或结果没有为动作action.Part和结果定义结果。{“ col1”:“ col1”,“ col2”:“ col2”}
org.apache.struts2.dispatcher.Dispatcher
{“ 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>
我没有获得有关jQuery成功的数据。请纠正我,这是怎么了?
jQuery AjaxdataType : 'json'使用A 来指定在执行操作和结果时从回调函数返回的数据类型,以及从服务器返回的响应。success
jQuery AjaxdataType : 'json'
success
dataType(默认值:智能猜测(xml,json,script,或html))
xml
json
script
html
类型: String
String
你期望从服务器返回的数据类型。如果未指定,则jQuery将尝试根据响应的MIME类型来推断它(XML MIME类型将产生XML,在1.4中,JSON将产生JavaScript对象,在1.4中,脚本将执行该脚本,而其他所有内容将以字符串形式返回)。
该URL应该正确指向动作映射。假设它将在默认名称空间中,否则你应该修改URL和映射以添加namespace属性。
namespace
<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如果你JSONObject手动构建,则不需要返回结果类型。你可以返回文本作为流结果,然后根据需要将字符串转换为JSON。
JSONObject
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"; } }
我对结果类型和内容类型放置了不同的结果,以更好地描述这个想法。你可以返回任何这些结果,并返回带或不带字符串的JSON对象。字符串化版本需要解析返回的数据以获得JSON对象。你还可以选择哪种序列类型更好地进行序列化以满足你的需要,但我的目标是表明,如果你需要序列化简单对象,则无需json插件即可使其工作。