使用ActionScript拆分数组

某人的快速数组问题...所以我有这个数组:
var bodyText:Array = ["1||^^1a::##1b||1", "2||^^2a::##2b||2"]; 
我需要像这样解析和格式化它: 问题:1 正确答案:1a 答案错误:1b 反馈:1 问题2 正确答案:2a 答案不正确:2b 反馈:2 我非常接近解决方案,但无论出于何种原因,我遇到了正确/错误密钥的问题,例如,这是我的代码跟踪:
Question: 1
Correct Answer: 1a
^^1a
##1b
Incorrect Answer: 1b
Feedback: 1

Question: 2
Correct Answer: 2a
^^2a
##2b
Incorrect Answer: 2b
Feedback: 2
这是我的剧本,非常感谢任何帮助!谢谢!
var bodyText:Array = ["1||^^1a::##1b||1", "2||^^2a::##2b||2"];
var _txt:String;
for (var i:Number = 0; i < bodyText.length; i++) {
    var _tb:Array = bodyText[i].split("||");
    for (var j:Number = 0; j < _tb.length; j++) {
        //question
        _txt = "Question: " + _tb[0] + "n";

        //answers
        var _kb:Array = _tb[1].split("::");
        for (var k:Number = 0; k < _kb.length; k++) {
            _txt += _kb[k].split("^^").join("Correct Answer: ") + "n";
            _txt += _kb[k].split("##").join("Incorrect Answer: ") + "n";
        }

        //feedback
        _txt += "Feedback: " + _tb[2] + "n";
    }
    trace(_txt);
}
    
已邀请:
我知道了...
var bodyText:Array = ["1||^^1a::##1b||1", "2||^^2a::##2b::##2c||2"];
var _txt:String;
for (var i:Number = 0; i < bodyText.length; i++) {
    var _tb:Array = bodyText[i].split("||");
    for (var j:Number = 0; j < _tb.length; j++) {
        //Question:
        _txt = "Question: " + _tb[0] + "n";

        //Answers:
        var _kb:Array = _tb[1].split("::");
        for (var k:Number = 0; k < _kb.length; k++) {
            if (_kb[k].indexOf("^^") != -1) {
                _txt += _kb[k].split("^^").join("Correct Answer: ") + "n";
            } else {
                _txt += _kb[k].split("##").join("Incorrect Answer: ") + "n";
            }
        }
        //Feedback:
        _txt += "Feedback: " + _tb[2] + "n";
    }
    trace(_txt);
}
    
_txt += _kb[0].split("^^").join("Correct Answer: ") + "n";
_txt += _kb[1].split("##").join("Incorrect Answer: ") + "n";
如果每个问题只有一个正确答案和一个错误答案,则上述内容就足够了。 你获得
^^1a
的原因是因为你的分裂方式
for (var k:Number = 0; k < _kb.length; k++) {
            _txt += _kb[k].split("^^").join("Correct Answer: ") + "n";
            _txt += _kb[k].split("##").join("Incorrect Answer: ") + "n";
        }
这条线
_kb[k].split("^^").join("Correct Answer: ") + "n";
取代
^^
为kb [0] 但第二行
_txt += _kb[k].split("##").join("Incorrect Answer: ") + "n";
找不到
##
并返回
^^1a
在下一次迭代中它将是相反的,k现在是1:
##1b
    
你的第二个for循环是不必要的,请参阅下面的代码(试过,工作):
var bodyText:Array = ["1||^^1a::##1b||1", "2||^^2a::##2b||2"];
            var _txt:String;
            for (var i:Number = 0; i < bodyText.length; i++) {
                var _tb:Array = bodyText[i].split("||");
                for (var j:Number = 0; j < _tb.length; j++) {
                    //question
                    _txt = "Question: " + _tb[0] + "n";

                    //answers
                    var _kb:Array = _tb[1].split("::");
                    _txt += _kb[0].split("^^").join("Correct Answer: ") + "n";
                    _txt += _kb[1].split("##").join("Incorrect Answer: ") + "n";

                    //feedback
                    _txt += "Feedback: " + _tb[2] + "n";
                }
                trace(_txt);
            }
干杯, 抢     

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