Servlets 在ExtJs中将参数传递给服务器

Servlets 在ExtJs中将参数传递给服务器,servlets,extjs,Servlets,Extjs,我是ExtJs新手 我想在单击按钮时将文本框中的值传递给服务器(Servlet)。但由于我是新手,我不知道怎么做 请有人帮我了解这一点,或者给我推荐一些教程或例子,可以帮助你 代码如下所示: new Ext.Button({ text: "Send to server", handler: function () { Ext.Ajax.request({ url: 'myPage.php', success: fun

我是ExtJs新手

我想在单击按钮时将文本框中的值传递给服务器(Servlet)。但由于我是新手,我不知道怎么做

请有人帮我了解这一点,或者给我推荐一些教程或例子,可以帮助你

代码如下所示:

new Ext.Button({
    text: "Send to server",
    handler: function () {
        Ext.Ajax.request({
            url: 'myPage.php',
            success: function (){alert('Value has been sent!');},
            failure: function (){alert('Failure of sending...');},
            headers: {
                'my-header': 'foo'
            },
            params: { foo: myTextField.getValue() }
        });
    }
})

在url中输入Servlet类名。 如果您有一个具有多个按钮(例如保存、更新、删除)的表单,则可以执行以下操作:

// Your form fields ...
var buttonAdd = new Ext.Button({text:'Add', handler:addFunction});
var deleteAdd = new Ext.Button({text:'Delete', handler:deleteFunction});
function addFunction(){
    Ext.Ajax.Request({
        url: 'MyServlet', // you can fix a parameter like this : MyServlet?action=add
        method: 'POST',
        params: {
            myField1: myField1.getValue()
            // all your params.... 
        }
        success: function (result, request){
            alert('Succesfully added ' + result.responseText);
        },
        failure: function (result, request){
            alert('Error in server' + result.responseText);
        }
});

function deleteFunction(){
    Ext.Ajax.Request({
        url: 'MyServlet', // you can fix a parameter like this : MyServlet?action=delete
        method: 'POST',
        params: {
            myField1: myField1.getValue()
            // all your params.... 
        }
        success: function (result, request){
            alert('Succesfully added ' + result.responseText);
        },
        failure: function (result, request){
            alert('Error in server' + result.responseText);
        }
});
}
在Servlet中,您可以执行以下操作:

public void doPost(HttpServletRequest request, HttpServletResponse response){
    String action = request.getParameter("action");
    if(action.equals("add")){
        // Your code for add method goes here
    } else if(action.equals("delete")){
        // Your code for delete method goes here
    }
}