Warning: file_get_contents(/data/phpspider/zhask/data//catemap/2/node.js/34.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Java ExtJSservlet工作不正常_Java_Javascript_Servlets_Extjs - Fatal编程技术网

Java ExtJSservlet工作不正常

Java ExtJSservlet工作不正常,java,javascript,servlets,extjs,Java,Javascript,Servlets,Extjs,我似乎不明白为什么servlet不能正常工作。 当我在extjs上按save时,它没有拾取任何内容。我有一个控制器peopleTemplate,它充当servlet并执行查询 下面是我的servlet public class peopleTemplate extends HttpServlet { private static final long serialVersionUID = 1L; protected void doPost(HttpServletRequest request

我似乎不明白为什么servlet不能正常工作。 当我在extjs上按save时,它没有拾取任何内容。我有一个控制器peopleTemplate,它充当servlet并执行查询

下面是我的servlet

public class peopleTemplate extends HttpServlet {
private static final long serialVersionUID = 1L;


protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException,
        IOException {
    PrintWriter out = response.getWriter();
    String action = request.getParameter("action");
    String firstname = request.getParameter("firstname").trim();
    if(action.equals("add")) {
        ExecuteQuery sq = new ExecuteQuery();


    }

    ExecuteQuery sq = new ExecuteQuery();

    String confirmUpdate = null;
    confirmUpdate = sq.peopleTemplate(firstname);

    // return delete confirmation
    response.setContentType("text/html");
    out.print(confirmUpdate);


}
}

下面是我的查询执行

public ExecuteQuery() {
    try {
        conn = DBConnect.getInstance().dbOracleConnect();
    } catch (Exception e) {
        e.printStackTrace();
        logger.error("Database Connection error: " + e);
    }
}


public String peopleTemplate(String firstname) {

    String result = null;
    String query = "INSERT INTO ts_people SET firstname =?" ;

    try {
        ps = conn.prepareStatement(query);
        ps.setString(1, "0");
        ps.setString(2, firstname);

        int numberOfRowsAffected = ps.executeUpdate();
        if (numberOfRowsAffected == 1) {
            result = "Bid Number " + firstname + " activated.";
        } else {
            result = "Update failed: Bid Number " + firstname + " not found.";
        }

    } catch (Exception e) {


}
    return result;
}
}

这是javascript文件中的按钮

   buttons: [{
        text: 'Save',
           handler: function(){
    Ext.Ajax.request({
        url : 'src/javas/peopleTemplate',
        method: 'POST',                   
        params :{'firstname': firstname},       
        success: function ( result, request ) {
            resultData = result.responseText;
            Ext.MessageBox.show({title:'Delete Status ', msg: resultData, buttons: Ext.MessageBox.OK, icon: Ext.MessageBox.INFO});              

        },
        failure: function ( result, request ) {
            Ext.MessageBox.show({title:'Delete Status ', msg: 'Request failed.', buttons: Ext.MessageBox.OK, icon: Ext.MessageBox.INFO});               

        }   
    });

为什么按save时没有出现任何问题?

如果您没有解释它是如何失败的(“工作不正常”、“它没有拾取任何东西”、“没有出现任何问题”不够精确),那么您是唯一可以进一步调查此问题的人:验证请求是否离开浏览器、到达服务器、成功处理,以预期的格式发送响应,等等。通过这种方式,您将发现哪个部分出现故障。一个突出的问题是
peopleTemplate
中的空
catch(异常e)
块:如果出现任何问题,它将自动失败并返回
null
。您仍然没有回答试图帮助您的人的其他问题。对于同一件事,现在有3个问题,但你需要一步一步地做。确定是否发送了ajax请求,检查服务器是否接收到它,然后检查servlet是否正在生成结果