我想使用C＃来检查字符串值是否包含字符串数组中的单词 . 例如，

string stringToCheck = "text1text2text3";

string[] stringArray = { "text1", "someothertext", etc... };

if(stringToCheck.contains stringArray) //one of the items?
{

}

如何检查'stringToCheck'的字符串值是否包含数组中的单词？

26 回答

展示了三种选择 . 我更愿意发现第三个是最简洁的 .

class Program {
    static void Main(string[] args) {
    string req = "PUT";
    if ((new string[] {"PUT", "POST"}).Any(s => req.Contains(s))) {
        Console.WriteLine("one.1.A");  // IS TRUE
    }
    req = "XPUT";
    if ((new string[] {"PUT", "POST"}).Any(s => req.Contains(s))) {
        Console.WriteLine("one.1.B"); // IS TRUE
    }
    req = "PUTX";
    if ((new string[] {"PUT", "POST"}).Any(s => req.Contains(s))) {
        Console.WriteLine("one.1.C");  // IS TRUE
    }
    req = "UT";
    if ((new string[] {"PUT", "POST"}).Any(s => req.Contains(s))) {
        Console.WriteLine("one.1.D"); // false
    }
    req = "PU";
    if ((new string[] {"PUT", "POST"}).Any(s => req.Contains(s))) {
        Console.WriteLine("one.1.E"); // false
    }
    req = "POST";
    if ((new string[] {"PUT", "POST"}).Any(s => req.Contains(s))) {
        Console.WriteLine("two.1.A"); // IS TRUE
    }
    req = "ASD";
    if ((new string[] {"PUT", "POST"}).Any(s => req.Contains(s))) {
        Console.WriteLine("three.1.A");  // false
    }


    Console.WriteLine("-----");
    req = "PUT";
    if (Array.IndexOf((new string[] {"PUT", "POST"}), req) >= 0)  {
        Console.WriteLine("one.2.A"); // IS TRUE
    }
    req = "XPUT";
    if (Array.IndexOf((new string[] {"PUT", "POST"}), req) >= 0)  {
        Console.WriteLine("one.2.B"); // false
    }
    req = "PUTX";
    if (Array.IndexOf((new string[] {"PUT", "POST"}), req) >= 0)  {
        Console.WriteLine("one.2.C"); // false
    }
    req = "UT";
    if (Array.IndexOf((new string[] {"PUT", "POST"}), req) >= 0)  {
        Console.WriteLine("one.2.D"); // false
    }
    req = "PU";
    if (Array.IndexOf((new string[] {"PUT", "POST"}), req) >= 0)  {
        Console.WriteLine("one.2.E"); // false
    }
    req = "POST";
    if (Array.IndexOf((new string[] {"PUT", "POST"}), req) >= 0)  {
        Console.WriteLine("two.2.A");  // IS TRUE
    }
    req = "ASD";
    if (Array.IndexOf((new string[] {"PUT", "POST"}), req) >= 0)  {
        Console.WriteLine("three.2.A");  // false
    }

    Console.WriteLine("-----");
    req = "PUT";
    if ((new string[] {"PUT", "POST"}.Contains(req)))  {
        Console.WriteLine("one.3.A"); // IS TRUE
    }
    req = "XPUT";
    if ((new string[] {"PUT", "POST"}.Contains(req)))  {
        Console.WriteLine("one.3.B");  // false
    }
    req = "PUTX";
    if ((new string[] {"PUT", "POST"}.Contains(req)))  {
        Console.WriteLine("one.3.C");  // false
    }
    req = "UT";
    if ((new string[] {"PUT", "POST"}.Contains(req)))  {
        Console.WriteLine("one.3.D");  // false
    }
    req = "PU";
    if ((new string[] {"PUT", "POST"}.Contains(req)))  {
        Console.WriteLine("one.3.E");  // false
    }
    req = "POST";
    if ((new string[] {"PUT", "POST"}.Contains(req)))  {
        Console.WriteLine("two.3.A");  // IS TRUE
    }
    req = "ASD";
    if ((new string[] {"PUT", "POST"}.Contains(req)))  {
        Console.WriteLine("three.3.A");  // false
    }

    Console.ReadKey();
    }
}

回复于 2024-04-20T06:39:19+08:00

这是你如何做到的：

string stringToCheck = "text1";
string[] stringArray = { "text1", "testtest", "test1test2", "test2text1" };
foreach (string x in stringArray)
{
    if (stringToCheck.Contains(x))
    {
        // Process...
    }
}

UPDATE: 您可能正在寻找更好的解决方案..请参阅下面的@Anton Gogolev的答案，其中使用了LINQ .

回复于 2024-04-20T06:39:19+08:00

0
这是如何做：
```
if(stringArray.Any(stringToCheck.Contains))
/* or a bit longer: (stringArray.Any(s => stringToCheck.Contains(s))) */
```
这将检查 stringToCheck 是否包含 stringArray 中的任何一个子字符串 . 如果要确保它包含所有子字符串，请将 Any 更改为 All ：
```
if(stringArray.All(stringToCheck.Contains))
```
回复于 2024-04-20T06:39:19+08:00

711

试试这个：

无需使用LINQ

if (Array.IndexOf(array, Value) >= 0)
{
    //Your stuff goes here
}

回复于 2024-04-20T06:39:19+08:00

0
只需使用linq方法：
```
stringArray.Contains(stringToCheck)
```
（抱歉无法在现有答案上添加评论，因为我的声誉<50）
回复于 2024-04-20T06:39:19+08:00

最简单和样本方式 .

bool bol=Array.Exists(stringarray,E => E == stringtocheck);

回复于 2024-04-20T06:39:19+08:00

也许这样的东西：

string stringToCheck = "text1text2text3";
string[] stringArray = new string[] { "text1" };
if (Array.Exists<string>(stringArray, (Predicate<string>)delegate(string s) { 
    return stringToCheck.IndexOf(s, StringComparison.OrdinalIgnoreCase) > -1; })) {
    Console.WriteLine("Found!");
}

回复于 2024-04-20T06:39:19+08:00

115

使用Linq和方法组将是最快速，更紧凑的方法 .

var arrayA = new[] {"element1", "element2"};
var arrayB = new[] {"element2", "element3"};
if (arrayB.Any(arrayA.Contains)) return true;

回复于 2024-04-20T06:39:19+08:00

string strName = "vernie";
string[] strNamesArray = { "roger", "vernie", "joel" };

if (strNamesArray.Any(x => x == strName))
{
   // do some action here if true...
}

回复于 2024-04-20T06:39:19+08:00

尝试：

String[] val = { "helloword1", "orange", "grape", "pear" };
String sep = "";
string stringToCheck = "word1";

bool match = String.Join(sep,val).Contains(stringToCheck);
bool anothermatch = val.Any(s => s.Contains(stringToCheck));

回复于 2024-04-20T06:39:19+08:00

1
你也可以像Anton Gogolev建议的那样做，检查 stringArray1 中的 any item 是否与 stringArray2 中的 any item 匹配：
```
if(stringArray1.Any(stringArray2.Contains))
```
同样，stringArray1中的 all items 与stringArray2中的 all items 匹配：
```
if(stringArray1.All(stringArray2.Contains))
```
回复于 2024-04-20T06:39:19+08:00
-1
我在控制台应用程序中使用以下内容来检查参数
```
var sendmail = args.Any( o => o.ToLower() == "/sendmail=true");
```
回复于 2024-04-20T06:39:19+08:00
3
我会使用Linq但它仍然可以通过以下方式完成：
```
new[] {"text1", "text2", "etc"}.Contains(ItemToFind);
```
回复于 2024-04-20T06:39:19+08:00

您可以定义自己的 string.ContainsAny() 和 string.ContainsAll() 方法 . 作为奖励，我甚至抛出了一个允许不区分大小写的比较等的 string.Contains() 方法 .

public static class Extensions
{
    public static bool Contains(this string source, string value, StringComparison comp)
    {
        return source.IndexOf(value, comp) > -1;
    }

    public static bool ContainsAny(this string source, IEnumerable<string> values, StringComparison comp = StringComparison.CurrentCulture)
    {
        return values.Any(value => source.Contains(value, comp));
    }

    public static bool ContainsAll(this string source, IEnumerable<string> values, StringComparison comp = StringComparison.CurrentCulture)
    {
        return values.All(value => source.Contains(value, comp));
    }
}

您可以使用以下代码测试这些：

public static void TestExtensions()
    {
        string[] searchTerms = { "FOO", "BAR" };
        string[] documents = {
            "Hello foo bar",
            "Hello foo",
            "Hello"
        };

        foreach (var document in documents)
        {
            Console.WriteLine("Testing: {0}", document);
            Console.WriteLine("ContainsAny: {0}", document.ContainsAny(searchTerms, StringComparison.OrdinalIgnoreCase));
            Console.WriteLine("ContainsAll: {0}", document.ContainsAll(searchTerms, StringComparison.OrdinalIgnoreCase));
            Console.WriteLine();
        }
    }

回复于 2024-04-20T06:39:19+08:00

试试这个，这里的示例：检查字段是否包含数组中的任何单词 . 检查字段（someField）是否包含数组中的任何单词 .

String[] val = { "helloword1", "orange", "grape", "pear" };   

Expression<Func<Item, bool>> someFieldFilter = i => true;

someFieldFilter = i => val.Any(s => i.someField.Contains(s));

回复于 2024-04-20T06:39:19+08:00

public bool ContainAnyOf(string word, string[] array) 
    {
        for (int i = 0; i < array.Length; i++)
        {
            if (word.Contains(array[i]))
            {
                return true;
            }
        }
        return false;
    }

回复于 2024-04-20T06:39:19+08:00

0
我使用了与Maitrey684的IndexOf类似的方法和Theomax的foreach循环来创建它 . （注意：前3个“字符串”行只是一个如何创建数组并使其成为正确格式的示例） .

如果要比较2个数组，它们将以分号分隔，但最后一个值后面没有一个数组 . 如果在数组的字符串形式中附加一个分号（即a; b; c变为a; b; c;），则可以使用“x;”匹配无论它处于什么位置：
```
bool found = false;
string someString = "a-b-c";
string[] arrString = someString.Split('-');
string myStringArray = arrString.ToString() + ";";

foreach (string s in otherArray)
{
    if (myStringArray.IndexOf(s + ";") != -1) {
       found = true;
       break;
    }
}

if (found == true) { 
    // ....
}
```
回复于 2024-04-20T06:39:19+08:00
-1
```
string [] lines = {"text1", "text2", "etc"};

bool bFound = lines.Any(x => x == "Your string to be searched");
```
如果搜索的字符串与数组'lines'的任何元素匹配，则bFound设置为true .
回复于 2024-04-20T06:39:19+08:00

试试这个

string stringToCheck = "text1text2text3";
string[] stringArray = new string[] { "text1" };

var t = lines.ToList().Find(c => c.Contains(stringToCheck));

它将返回您正在查找的文本的第一个出处的行 .

回复于 2024-04-20T06:39:19+08:00

如果 stringArray 包含大量不同长度的字符串，请考虑使用Trie来存储和搜索字符串数组 .

public static class Extensions
{
    public static bool ContainsAny(this string stringToCheck, IEnumerable<string> stringArray)
    {
        Trie trie = new Trie(stringArray);
        for (int i = 0; i < stringToCheck.Length; ++i)
        {
            if (trie.MatchesPrefix(stringToCheck.Substring(i)))
            {
                return true;
            }
        }

        return false;
    }
}

这是 Trie 类的实现

public class Trie
{
    public Trie(IEnumerable<string> words)
    {
        Root = new Node { Letter = '\0' };
        foreach (string word in words)
        {
            this.Insert(word);
        }
    }

    public bool MatchesPrefix(string sentence)
    {
        if (sentence == null)
        {
            return false;
        }

        Node current = Root;
        foreach (char letter in sentence)
        {
            if (current.Links.ContainsKey(letter))
            {
                current = current.Links[letter];
                if (current.IsWord)
                {
                    return true;
                }
            }
            else
            {
                return false;
            }
        }

        return false;
    }

    private void Insert(string word)
    {
        if (word == null)
        {
            throw new ArgumentNullException();
        }

        Node current = Root;
        foreach (char letter in word)
        {
            if (current.Links.ContainsKey(letter))
            {
                current = current.Links[letter];
            }
            else
            {
                Node newNode = new Node { Letter = letter };
                current.Links.Add(letter, newNode);
                current = newNode;
            }
        }

        current.IsWord = true;
    }

    private class Node
    {
        public char Letter;
        public SortedList<char, Node> Links = new SortedList<char, Node>();
        public bool IsWord;
    }

    private Node Root;
}

如果 stringArray 中的所有字符串具有相同的长度，那么最好只使用 HashSet 而不是 Trie

public static bool ContainsAny(this string stringToCheck, IEnumerable<string> stringArray)
{
    int stringLength = stringArray.First().Length;
    HashSet<string> stringSet = new HashSet<string>(stringArray);
    for (int i = 0; i < stringToCheck.Length - stringLength; ++i)
    {
        if (stringSet.Contains(stringToCheck.Substring(i, stringLength)))
        {
            return true;
        }
    }

    return false;
}

回复于 2024-04-20T06:39:19+08:00

1

简单的解决方案，不需要linq任何

String.Join（“，”，array）.Contains（Value“，”）;

回复于 2024-04-20T06:39:19+08:00

int result = Array.BinarySearch(list.ToArray(), typedString, StringComparer.OrdinalIgnoreCase);

回复于 2024-04-20T06:39:19+08:00

试试这个，不需要循环..

string stringToCheck = "text1";
List<string> stringList = new List<string>() { "text1", "someothertext", "etc.." };
if (stringList.Exists(o => stringToCheck.Contains(o)))
{

}

回复于 2024-04-20T06:39:19+08:00

要完成上述答案，请 IgnoreCase 检查使用：

stringArray.Any(s => stringToCheck.IndexOf(s, StringComparison.CurrentCultureIgnoreCase) > -1)

回复于 2024-04-20T06:39:19+08:00

0
就我而言，上述答案无效 . 我正在检查数组中的字符串并将其分配给布尔值 . 我修改了@Anton Gogolev的答案并删除了 Any() 方法并将 stringToCheck 放在 Contains() 方法中 .
```
bool = stringArray.Contains(stringToCheck);
```
回复于 2024-04-20T06:39:19+08:00

我使用以下代码检查字符串是否包含字符串数组中的任何项：

foreach (string s in stringArray)
{
    if (s != "")
    {
        if (stringToCheck.Contains(s))
        {
            Text = "matched";
        }
    }
}

回复于 2024-04-20T06:39:19+08:00

使用C＃检查字符串是否包含字符串数组中的字符串

26 回答

试试这个：

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