javascript - 如何获取store.sync()的返回消息；

Question

function onButtonClick(){
           var grid = Ext.getCmp('mygridpanel')
           var row = grid.getSelectionModel().getSelection()[0];
           var txtVehicleID = Ext.getCmp('txtVehicleID').getValue();
           var txtPlat_No = Ext.getCmp('txtPlat_No').getValue();
           console.log(txtVehicleID);

            var record = UserStore.findRecord('_id', txtVehicleID);
            record.set('_id', txtVehicleID);
            record.set('Plat_No',txtPlat_No);

            UserStore.sync({
                success: function(response) {
                            Ext.MessageBox.show({
                                title: "Information",
                                msg: "Update Success !",
                                icon: Ext.MessageBox.INFO,
                                buttons: Ext.MessageBox.OK,
                                fn: function(buttonId) {
                                    if (buttonId === "ok") {
                                        EditWin.close();
                                    }
                                }
                            });  
                },
                    failure: function(action){
                                   Ext.MessageBox.show({
                                title: "Information",
                                msg: "Update Failed !",
                                icon: Ext.MessageBox.ERROR,
                                buttons: Ext.MessageBox.OK,
                                fn: function(buttonId) {
                                    if (buttonId === "ok") {
                                        EditWin.close();
                                    }
                                }
                    });  

                }
            });
            //console.log("clicked");
}

php返回成功消息字符串

<?php
$data = file_get_contents("php://input");
//echo $data;
//$obj = var_dump(json_decode($data));

$obj = json_decode($data);
$_id = $obj->{'_id'};
$Plat_No = $obj->{'Plat_No'};

mysql_connect("localhost", "root", "Apacheah64") or die("Could not connect");
mysql_select_db("db_shuttlebus") or die("Could not select database");

$query = "UPDATE tbl_vehicle SET Plat_No ='". $Plat_No ."' WHERE _id=".$_id;

if (mysql_query($query)){
    echo '{"success":true,"message":"Update Success !"}';
}else{
    echo '{"success":false,"message":"Update Failed !"}';
}

?>

如何获取消息并显示在此处 msg: "Update Success !" ?

Answer 1

在您的 UserStore.sync() 配置对象中尝试定义 success 和 failure 处理程序，如下所示:

UserStore.sync({
    success: function(batch, options) {
        var message = batch.proxy.getReader().jsonData.message;
        // display success message...

    },
    failure: function(batch, options) {
        var message = batch.proxy.getReader().jsonData.message;
        // display error message...
    }
});