Question

我知道Extjs有异步行为，但我仍然需要一种方法来打印console.log（'x'+ x的值）;作为“x 1的值”

Ext.onReady(function() {
    var x=0;
    Ext.MessageBox.confirm('Confirm', 'Are you sure you want to do that?', showResult );

    function showResult(btn){

            Ext.Msg.show({
                   title: 'number',
                   msg: 'Please enter your number:',
                   width: 300,
                   buttons: Ext.MessageBox.OKCANCEL,
                   multiline: true,
                   fn: saveNumber,
                   icon: Ext.MessageBox.INFO
                });
        function saveNumber(){
            x=1;
        }


    }
    console.log('value of x'+x); //want this value to be 1 without changing the line of this statement
});

Answer 1

这是不可能的，你必须使用一个函数来完成你的回调任务。一个想法是：

Ext.Msg.show({
        title:'Number?',
        message: 'Please enter your number: ',
        buttons: Ext.MessageBox.OKCANCEL,
        multiline: true,
        icon: Ext.MessageBox.INFO,
        fn: function(btn,txt) {
            if(btn==='ok') {
                var number = parseInt(txt,10);
                console.log('Your number is: ' + number);
            }
        }
    });

甚至更好：

Ext.Msg.prompt('Number?', 'Please enter your number: ', function(btn,txt){  
        if(btn==='ok') {
            var number = parseInt(txt,10);
            console.log(number);
        }
    });

需要绕过ExtJS的异步性质

1 个答案: