C＃遍历类属性

我目前正在设置类对象的所有值Record。

这是我现在用来逐个属性填充记录的代码。

// Loop through each field in the result set
for (int i = 0; i <= resultItems.Length; i++)
{

    Record newRecord = new Record()
    {
            itemtype =   resultItems[i - (fieldCount - 0)],
            itemdesc =   resultItems[i - (fieldCount - 1)],
            prodcode =   resultItems[i - (fieldCount - 2)],
            proddesc =   resultItems[i - (fieldCount - 3)],
            curstat =    resultItems[i - (fieldCount -4)],
            totfree =    resultItems[i - (fieldCount -5)],
            totphys =    resultItems[i - (fieldCount -6)],
            pcolgroup =  resultItems[i - (fieldCount -7)],
            scolgroup =  resultItems[i - (fieldCount -8)],
            totpo =      resultItems[i - (fieldCount - 9)],
            totso =      resultItems[i - (fieldCount - 10)],
            quality =    resultItems[i - (fieldCount - 11)],
            statusdesc = resultItems[i - (fieldCount - 12)],
            groupcode =  resultItems[i - (fieldCount - 13)],
            qualitydes = resultItems[i - (fieldCount - 14)],
            pcoldesc =   resultItems[i - (fieldCount - 15)],
            scoldesc =   resultItems[i - (fieldCount - 16)],
            pgroupdesc = resultItems[i - (fieldCount - 17)],
    };
}

是否可以在不对所有属性名称进行硬编码的情况下动态地遍历每个属性？

像这样：

// Create new Record instance
Record newRecord = new Record();

for (int e = 0; e < propertyCount.Length - 1; e++)
{
    newRecord[fieldname] = resultItems[i - (fieldCount - e)];
}

您可以使用反射来做到这一点。据我了解，您可以枚举类的属性并设置值。您必须尝试一下，并确保您了解属性的顺序。有关此方法的更多信息，请参考此MSDN文档。

作为提示，您可以执行以下操作：

Record record = new Record();

PropertyInfo[] properties = typeof(Record).GetProperties();
foreach (PropertyInfo property in properties)
{
    property.SetValue(record, value);
}

value您要写入的值在哪里（因此来自resultItems数组）。

// the index of each item in fieldNames must correspond to 
// the correct index in resultItems
var fieldnames = new []{"itemtype", "etc etc "};

for (int e = 0; e < fieldNames.Length - 1; e++)
{
    newRecord
       .GetType()
       .GetProperty(fieldNames[e])
       .SetValue(newRecord, resultItems[e]);
}

是的，您可以在Record类上创建一个索引器，该索引器从属性名称映射到正确的属性。这会将所有从属性名称到属性的绑定都放在一个位置，例如：

public class Record
{
    public string ItemType { get; set; }

    public string this[string propertyName]
    {
        set
        {
            switch (propertyName)
            {
                case "itemType":
                    ItemType = value;
                    break;
                    // etc
            }   
        }
    }
}

另外，正如其他人提到的那样，请使用反射。

我尝试了塞缪尔·斯莱德的建议。没为我工作。该PropertyInfo 名单来为空。因此，我尝试了以下方法，并且对我有用。

    Type type = typeof(Record);
    FieldInfo[] properties = type.GetFields();
    foreach (FieldInfo property in properties) {
       Debug.LogError(property.Name);
    }