让我们把你的优秀和最喜欢的扩展方法列一个列表。

要求是必须发布完整的代码,以及如何使用它的示例和解释。

基于对这个主题的高度兴趣,我在Codeplex上建立了一个名为extensionoverflow的开源项目。

请将您的回答标记为接受,以便将代码放入Codeplex项目。

请张贴完整的源代码,而不是一个链接。

Codeplex上新闻:

24.08.2010 Codeplex页面现在在这里:http://extensionoverflow.codeplex.com/

11.11.2008 XmlSerialize / XmlDeserialize现在是实现和单元测试。

11.11.2008仍有发展空间。;-)现在就加入!

11.11.2008第三位贡献者加入了ExtensionOverflow,欢迎加入BKristensen

11.11.2008 FormatWith现在是实现和单元测试。

09.11.2008第二个贡献者加入ExtensionOverflow。欢迎来到chakrit。

我们需要更多的开发人员。: -)

09.11.2008 ThrowIfArgumentIsNull现已在Codeplex上实现和单元测试。


当前回答

我发现这个方法很有用

public static IEnumerable<T> EmptyIfNull<T>(this IEnumerable<T> pSeq)
{
    return pSeq ?? Enumerable.Empty<T>();
}

它在调用代码中删除空检查。现在你可以

MyList.EmptyIfNull().Where(....)

其他回答

// This file contains extension methods for generic List<> class to operate on sorted lists.
// Duplicate values are OK.
// O(ln(n)) is still much faster then the O(n) of LINQ's searches/filters.
static partial class SortedList
{
    // Return the index of the first element with the key greater then provided.
    // If there's no such element within the provided range, it returns iAfterLast.
    public static int sortedFirstGreaterIndex<tElt, tKey>( this IList<tElt> list, Func<tElt, tKey, int> comparer, tKey key, int iFirst, int iAfterLast )
    {
        if( iFirst < 0 || iAfterLast < 0 || iFirst > list.Count || iAfterLast > list.Count )
            throw new IndexOutOfRangeException();
        if( iFirst > iAfterLast )
            throw new ArgumentException();
        if( iFirst == iAfterLast )
            return iAfterLast;

        int low = iFirst, high = iAfterLast;
        // The code below is inspired by the following article:
        // http://en.wikipedia.org/wiki/Binary_search#Single_comparison_per_iteration
        while( low < high )
        {
            int mid = ( high + low ) / 2;
            // 'mid' might be 'iFirst' in case 'iFirst+1 == iAfterLast'.
            // 'mid' will never be 'iAfterLast'.
            if( comparer( list[ mid ], key ) <= 0 ) // "<=" since we gonna find the first "greater" element
                low = mid + 1;
            else
                high = mid;
        }
        return low;
    }

    // Return the index of the first element with the key greater then the provided key.
    // If there's no such element, returns list.Count.
    public static int sortedFirstGreaterIndex<tElt, tKey>( this IList<tElt> list, Func<tElt, tKey, int> comparer, tKey key )
    {
        return list.sortedFirstGreaterIndex( comparer, key, 0, list.Count );
    }

    // Add an element to the sorted array.
    // This could be an expensive operation if frequently adding elements that sort firstly.
    // This is cheap operation when adding elements that sort near the tail of the list.
    public static int sortedAdd<tElt>( this List<tElt> list, Func<tElt, tElt, int> comparer, tElt elt )
    {
        if( list.Count == 0 || comparer( list[ list.Count - 1 ], elt ) <= 0 )
        {
            // either the list is empty, or the item is greater then all elements already in the collection.
            list.Add( elt );
            return list.Count - 1;
        }
        int ind = list.sortedFirstGreaterIndex( comparer, elt );
        list.Insert( ind, elt );
        return ind;
    }

    // Find first exactly equal element, return -1 if not found.
    public static int sortedFindFirstIndex<tElt, tKey>( this List<tElt> list, Func<tElt, tKey, int> comparer, tKey elt )
    {
        int low = 0, high = list.Count - 1;

        while( low < high )
        {
            int mid = ( high + low ) / 2;
            if( comparer( list[ mid ], elt ) < 0 )
                low = mid + 1;
            else
                high = mid; // this includes the case when we've found an element exactly matching the key
        }
        if( high >= 0 && 0 == comparer( list[ high ], elt ) )
            return high;
        return -1;
    }

    // Return the IEnumerable that returns array elements in the reverse order.
    public static IEnumerable<tElt> sortedReverse<tElt>( this List<tElt> list )
    {
        for( int i=list.Count - 1; i >= 0; i-- )
            yield return list[ i ];
    }
}

内置强制转换的FindControl:

public static T FindControl<T>(this Control control, string id) where T : Control
{
    return (T)control.FindControl(id);
}

这没什么了不起的,但我觉得这样可以写出更简洁的代码。

// With extension method
container.FindControl<TextBox>("myTextBox").SelectedValue = "Hello world!";

// Without extension method
((TextBox)container.FindControl("myTextBox")).SelectedValue = "Hello world!";

这可以放在codeplex项目中,如果需要的话

也许我写过和用过的最有用的扩展方法是:

http://www.codeproject.com/KB/cs/fun-with-cs-extensions.aspx?msg=2838918#xx2838918xx

我在所有项目中用到的两个小技巧(有些人会觉得很傻)是:

public static bool IsNull(this object o){
  return o == null;
}

and

public static bool IsNullOrEmpty(this string s){
  return string.IsNullOrEmpty(s);
}

它使我的代码更流畅。

if (myClassInstance.IsNull()) //... do something

if (myString.IsNullOrEmpty()) //... do something

我认为这些是很好的扩展属性;如果我们能得到这些。

public static class StringExtensions {

    /// <summary>
    /// Parses a string into an Enum
    /// </summary>
    /// <typeparam name="T">The type of the Enum</typeparam>
    /// <param name="value">String value to parse</param>
    /// <returns>The Enum corresponding to the stringExtensions</returns>
    public static T EnumParse<T>(this string value) {
        return StringExtensions.EnumParse<T>(value, false);
    }

    public static T EnumParse<T>(this string value, bool ignorecase) {

        if (value == null) {
            throw new ArgumentNullException("value");
        }

        value = value.Trim();

        if (value.Length == 0) {
            throw new ArgumentException("Must specify valid information for parsing in the string.", "value");
        }

        Type t = typeof(T);

        if (!t.IsEnum) {
            throw new ArgumentException("Type provided must be an Enum.", "T");
        }

        return (T)Enum.Parse(t, value, ignorecase);
    }
}

将字符串解析为Enum很有用。

public enum TestEnum
{
    Bar,
    Test
}

public class Test
{
    public void Test()
    {
        TestEnum foo = "Test".EnumParse<TestEnum>();
    }
 }

这要归功于斯科特·多尔曼

——编辑Codeplex项目——

我问过Scott Dorman,他是否介意我们在Codeplex项目中发布他的代码。我从他那里得到的回答是:

感谢你对SO帖子和CodePlex项目的提醒。我赞成你对这个问题的回答。是的,代码目前在CodeProject开放许可证(http://www.codeproject.com/info/cpol10.aspx)下有效地处于公共领域。 我没有问题,这被包括在CodePlex项目,如果你想把我添加到项目(用户名是sdorman),我会添加该方法加上一些额外的枚举助手方法。