如何在Linq中执行SQL Like％？

我有一个试图转换为Linq的SQL过程：

SELECT O.Id, O.Name as Organization
FROM Organizations O
JOIN OrganizationsHierarchy OH ON O.Id=OH.OrganizationsId
where OH.Hierarchy like '%/12/%'

我最关心的行是：

where OH.Hierarchy like '%/12/%'

我有一列存储例如/ 1/3/12 /之类的层次结构，因此我只使用％/ 12 /％进行搜索。

我的问题是，Linq或.NET与使用百分号等效吗？

.Where(oh => oh.Hierarchy.Contains("/12/"))

您也可以使用.StartsWith()或.EndsWith()。

用这个：

from c in dc.Organization
where SqlMethods.Like(c.Hierarchy, "%/12/%")
select *;

我假设您正在使用Linq-to-SQL *（请参阅下面的注释）。如果是这样，请使用string.Contains，string.StartsWith和string.EndsWith生成使用SQL LIKE运算符的SQL。

from o in dc.Organization
join oh in dc.OrganizationsHierarchy on o.Id equals oh.OrganizationsId
where oh.Hierarchy.Contains(@"/12/")
select new { o.Id, o.Name }

要么

from o in dc.Organization
where o.OrganizationsHierarchy.Hierarchy.Contains(@"/12/")
select new { o.Id, o.Name }

注意： * =如果在.net 3.5中使用ADO.Net实体框架（EF / L2E），请注意它将不会执行与Linq-to-SQL相同的转换。尽管L2S进行了适当的转换，但L2E v1（3.5）会转换为t-sql表达式，该表达式将强制对您查询的表进行全表扫描，除非where子句或联接过滤器中有另一个更好的区分符。
更新：这已在EF / L2E v4（.net 4.0）中修复，因此它将像L2S一样生成SQL LIKE。

如果您使用的是VB.NET，则答案为“ *”。这是您的where子句的样子……

Where OH.Hierarchy Like '*/12/*'

注意：“ *”匹配零个或多个字符。这是Like运算符的msdn文章。

好吧indexOf也对我有用

var result = from c in SampleList
where c.LongName.IndexOf(SearchQuery) >= 0
select c;

使用这样的代码

try
{
    using (DatosDataContext dtc = new DatosDataContext())
    {
        var query = from pe in dtc.Personal_Hgo
                    where SqlMethods.Like(pe.nombre, "%" + txtNombre.Text + "%")
                    select new
                    {
                        pe.numero
                        ,
                        pe.nombre
                    };
        dgvDatos.DataSource = query.ToList();
    }
}
catch (Exception ex)
{
    string mensaje = ex.Message;
}

.NET核心现在拥有 EF.Functions.Like

如果您不匹配数字字符串，通常情况下总是很好：

.Where(oh => oh.Hierarchy.ToUpper().Contains(mySearchString.ToUpper()))

我总是这样做：

from h in OH
where h.Hierarchy.Contains("/12/")
select h

我知道我不使用like语句，但是在后台将它翻译成带有like语句的查询是可行的。

试试这个，对我来说很好

from record in context.Organization where record.Hierarchy.Contains(12) select record;

System.Data.Linq.SqlClient.SqlMethods.Like("mystring", "%string")

在Linq中使用Contains，就像在SQL中使用Like一样。

string _search="/12/";

。。。

.Where(s => s.Hierarchy.Contains(_search))

您可以在Linq中编写SQL脚本，如下所示：

 var result= Organizations.Join(OrganizationsHierarchy.Where(s=>s.Hierarchy.Contains("/12/")),s=>s.Id,s=>s.OrganizationsId,(org,orgH)=>new {org,orgH});

对于那些像我这样在寻找如何在LINQ中实现“ SQL Like”方法的人来说，我的工作非常好。

我无法以任何方式更改数据库来更改列排序规则。因此，我必须在LINQ中找到一种方法来做到这一点。

我使用的辅助方法SqlFunctions.PatIndex类似于实际的SQL LIKE运算符。

首先，我需要在搜索值中列举所有可能的变音符号（我刚刚学到的一个单词），以获得类似以下内容：

déjà     => d[éèêëeÉÈÊËE]j[aàâäAÀÂÄ]
montreal => montr[éèêëeÉÈÊËE][aàâäAÀÂÄ]l
montréal => montr[éèêëeÉÈÊËE][aàâäAÀÂÄ]l

然后以LINQ为例：

var city = "montr[éèêëeÉÈÊËE][aàâäAÀÂÄ]l";
var data = (from loc in _context.Locations
                     where SqlFunctions.PatIndex(city, loc.City) > 0
                     select loc.City).ToList();

因此，为了我的需要，我编写了一个Helper / Extension方法

   public static class SqlServerHelper
    {

        private static readonly List<KeyValuePair<string, string>> Diacritics = new List<KeyValuePair<string, string>>()
        {
            new KeyValuePair<string, string>("A", "aàâäAÀÂÄ"),
            new KeyValuePair<string, string>("E", "éèêëeÉÈÊËE"),
            new KeyValuePair<string, string>("U", "uûüùUÛÜÙ"),
            new KeyValuePair<string, string>("C", "cçCÇ"),
            new KeyValuePair<string, string>("I", "iîïIÎÏ"),
            new KeyValuePair<string, string>("O", "ôöÔÖ"),
            new KeyValuePair<string, string>("Y", "YŸÝýyÿ")
        };

        public static string EnumarateDiacritics(this string stringToDiatritics)
        {
            if (string.IsNullOrEmpty(stringToDiatritics.Trim()))
                return stringToDiatritics;

            var diacriticChecked = string.Empty;

            foreach (var c in stringToDiatritics.ToCharArray())
            {
                var diac = Diacritics.FirstOrDefault(o => o.Value.ToCharArray().Contains(c));
                if (string.IsNullOrEmpty(diac.Key))
                    continue;

                //Prevent from doing same letter/Diacritic more than one time
                if (diacriticChecked.Contains(diac.Key))
                    continue;

                diacriticChecked += diac.Key;

                stringToDiatritics = stringToDiatritics.Replace(c.ToString(), "[" + diac.Value + "]");
            }

            stringToDiatritics = "%" + stringToDiatritics + "%";
            return stringToDiatritics;
        }
    }

如果您有任何建议要改进此方法，请您多多包涵。

太晚了，但是我将其结合在一起，以便能够使用SQL Like样式通配符进行字符串比较：

public static class StringLikeExtensions
{
    /// <summary>
    /// Tests a string to be Like another string containing SQL Like style wildcards
    /// </summary>
    /// <param name="value">string to be searched</param>
    /// <param name="searchString">the search string containing wildcards</param>
    /// <returns>value.Like(searchString)</returns>
    /// <example>value.Like("a")</example>
    /// <example>value.Like("a%")</example>
    /// <example>value.Like("%b")</example>
    /// <example>value.Like("a%b")</example>
    /// <example>value.Like("a%b%c")</example>
    /// <remarks>base author -- Ruard van Elburg from StackOverflow, modifications by dvn</remarks>
    /// <remarks>converted to a String extension by sja</remarks>
    /// <seealso cref="/programming/1040380/wildcard-search-for-linq"/>
    public static bool Like(this String value, string searchString)
    {
        bool result = false;

        var likeParts = searchString.Split(new char[] { '%' });

        for (int i = 0; i < likeParts.Length; i++)
        {
            if (likeParts[i] == String.Empty)
            {
                continue;   // "a%"
            }

            if (i == 0)
            {
                if (likeParts.Length == 1) // "a"
                {
                    result = value.Equals(likeParts[i], StringComparison.OrdinalIgnoreCase);
                }
                else // "a%" or "a%b"
                {
                    result = value.StartsWith(likeParts[i], StringComparison.OrdinalIgnoreCase);
                }
            }
            else if (i == likeParts.Length - 1) // "a%b" or "%b"
            {
                result &= value.EndsWith(likeParts[i], StringComparison.OrdinalIgnoreCase);
            }
            else // "a%b%c"
            {
                int current = value.IndexOf(likeParts[i], StringComparison.OrdinalIgnoreCase);
                int previous = value.IndexOf(likeParts[i - 1], StringComparison.OrdinalIgnoreCase);
                result &= previous < current;
            }
        }

        return result;
    }

    /// <summary>
    /// Tests a string containing SQL Like style wildcards to be ReverseLike another string 
    /// </summary>
    /// <param name="value">search string containing wildcards</param>
    /// <param name="compareString">string to be compared</param>
    /// <returns>value.ReverseLike(compareString)</returns>
    /// <example>value.ReverseLike("a")</example>
    /// <example>value.ReverseLike("abc")</example>
    /// <example>value.ReverseLike("ab")</example>
    /// <example>value.ReverseLike("axb")</example>
    /// <example>value.ReverseLike("axbyc")</example>
    /// <remarks>reversed logic of Like String extension</remarks>
    public static bool ReverseLike(this String value, string compareString)
    {
        bool result = false;

        var likeParts = value.Split(new char[] {'%'});

        for (int i = 0; i < likeParts.Length; i++)
        {
            if (likeParts[i] == String.Empty)
            {
                continue;   // "a%"
            }

            if (i == 0)
            {
                if (likeParts.Length == 1) // "a"
                {
                    result = compareString.Equals(likeParts[i], StringComparison.OrdinalIgnoreCase);
                }
                else // "a%" or "a%b"
                {
                    result = compareString.StartsWith(likeParts[i], StringComparison.OrdinalIgnoreCase);
                }
            }
            else if (i == likeParts.Length - 1) // "a%b" or "%b"
            {
                result &= compareString.EndsWith(likeParts[i], StringComparison.OrdinalIgnoreCase);
            }
            else // "a%b%c"
            {
                int current = compareString.IndexOf(likeParts[i], StringComparison.OrdinalIgnoreCase);
                int previous = compareString.IndexOf(likeParts[i - 1], StringComparison.OrdinalIgnoreCase);
                result &= previous < current;
            }
        }

        return result;
    }
}