Question

如果以下方法中的条件具有相似的模式，则所有这些都是有没有提出一种减少这种方法重复的常用方法的想法？

    public override Func<JObject, dynamic, string> version => (jobject, parameters) =>
    {
        bool hasValidObject = false;

        if (jobject["Version1"] != null)
        {
            _radio.Version1 = new VersionInfo(jobject["Version1"].Value<string>());
            hasValidObject = true;
        }

        if (jobject["Version2"] != null)
        {
            _radio.Version2 = new VersionInfo(jobject["Version2"].Value<string>());
            hasValidObject = true;
        }

        if (jobject["Version3"] != null)
        {
            _radio.Version3 = new VersionInfo(jobject["Version3"].Value<string>());
            hasValidObject = true;
        }

        if (jobject["Version4"] != null)
        {
            _radio.Version4 = new VersionInfo(jobject["Version4"].Value<string>());
            hasValidObject = true;
        }

        if (jobject["Version6"] != null)
        {
            _radio.Version6 = new VersionInfo(jobject["Version6"].Value<string>());
            hasValidObject = true;
        }

        if (hasValidObject)
        {
            return GenerateSuccess();
        }

        return GenerateUnsuccessful( try again.");
    };

Answer 1

一种方法是使用反射和循环

public override Func version => (jobject, parameters) => 
{ 
    bool hasValidObject = false;
    for (int i = 1; i<7;i++) 
    {
        hasValidObject = this.SetVersionInfo(i) || hasValidObject;
    }

    if (hasValidObject)
    {
        return GenerateSuccess();
    }

    return GenerateUnsuccessful( "try again.");
};


private bool SetVersionInfo(int i)
{
    if (jobject["Version" + i] == null) return false;

    _radio.GetType().GetProperty(propName)
        .SetValue(_radio, new VersionInfo(jobject["Version" + i].Value<string>()));
    return true;
}

另一种方法是在Dictionary<int,VersionInfo> Versions课程中创建_radio，然后您不需要反思：

private bool SetVersionInfo(int i)
{
    if (jobject["Version" + i] == null) return false;

    _radio.Versions[i] = new VersionInfo(jobject["Version" + i].Value<string>());
    return true;
}

Answer 2

另一种选择可以是switch语句：

public override Func<JObject, dynamic, string> version => (jobject, parameters) =>
{
    bool hasValidObject = false;

    foreach (char n in "12346")
    {
        var jObj = jobject["Version" + n];
        if (jObj != null)
        {
            var versionInfo = new VersionInfo(jObj.Value<string>());
            switch (n)
            {
                case '1': _radio.Version1 = versionInfo; break;
                case '2': _radio.Version2 = versionInfo; break;
                case '3': _radio.Version3 = versionInfo; break;
                case '4': _radio.Version4 = versionInfo; break;
                case '6': _radio.Version6 = versionInfo; break;
            }
            hasValidObject = true;
        }
    }

    return hasValidObject ? GenerateSuccess() : GenerateUnsuccessful(" try again.");
};

或带有委托数组的更高级版本..因为我喜欢垂直空间：]（也未经过测试）：

public override Func<JObject, dynamic, string> version => (jobject, parameters) =>
{
    Func<VersionInfo, VersionInfo>[] a = { null, _radio.Version1 = v, 
                v => _radio.Version2 = v, v => _radio.Version3 = v,
                v => _radio.Version4 = v, null, v => _radio.Version6 = v };

    var q = from n in new[] { 1, 2, 3, 4, 6 }
            let j = jobject["Version" + n] where j != null
            select a[n](new VersionInfo(j.Value<string>()));

    return q.Count() > 0 ? GenerateSuccess() : GenerateUnsuccessful(" try again.");
};

Answer 3

如果 jobject 是Dictionary类型，您可以使用

运行它

foreach(string version in jobject.keys)
{
     if(jobject[version]!=null){
     //do something
     }
}

问题在于收音机，但如果是你的班级，你可以在班级中使用一个版本的数组而不是那么多，并使用计数器来访问它们，你将使用它来计算循环的数量。编号

常见模式，这里所以我试图简化它以减少重复

3 个答案: