blocks|key|2145240|text|一个更快速的解决方案将停止搜索它发现了一个不存在的词：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2145241|var+text+=+"Hello+Swift-world"
var+textArray+=+["Hello",+"world"]

let+match+=+textArray.reduce(true)+{+!$0+?+false+:+(text.range(of:+$1)+!=+nil+)+}|code-block|syntax|javascript|2145242|另一种在发现不匹配后停止的方法：|2145243|let+match+=+textArray.first(where:+{+!text.contains($0)+})+==+nil|2145244|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

A more Swifty solution that will stop searching after it found a non-existent word:

<pre><code>var text = "Hello Swift-world"
var textArray = ["Hello", "world"]

let match = textArray.reduce(true) { !$0 ? false : (text.range(of: $1) != nil ) }
</code></pre>

Another way to do it which stops after it found a non-match:

<pre><code>let match = textArray.first(where: { !text.contains($0) }) == nil
</code></pre>

blocks|key|526766|text|如果您正在寻找更少的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|526767|let+text+=+"Hello+Swift-world"
let+wordList+=+["Hello",+"world"]

let+success+=+!wordList.contains(where:+{+!text.contains($0)+})
print(success)|code-block|syntax|javascript|526768|它也比您的解决方案更有效率，因为contains方法一找到“不包含”的单词就会返回。|offset|length|style|CODE|526769|在Swift+4或更高版本中，您可以使用allSatisfy|BOLD|526770|let+success+=+wordList.allSatisfy(text.contains)|526771|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/documentation/swift/array/2994715-allsatisfy^0|0|0|G|8|0|1|7|K|A|K|A|0|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|11|8|@$I|12|J|13|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|14|8|@$I|15|J|16|K|O]|$I|17|J|18|K|L]]|9|@$I|19|J|1A|1|1B]]|A|$]]|$1|P|3|Q|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|R|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|S|$T|$5|U|V|W|A|$X|Y]]]]

If you are looking for less code:
<pre><code>let text = &quot;Hello Swift-world&quot;
let wordList = [&quot;Hello&quot;, &quot;world&quot;]

let success = !wordList.contains(where: { !text.contains($0) })
print(success)
</code></pre>
It is also a little more efficient than your solution because
the <code>contains</code> method returns as soon as a &quot;not contained&quot; word
is found.
As of Swift 4 or later, you can use <a href="https://developer.apple.com/documentation/swift/array/2994715-allsatisfy" rel="nofollow noreferrer"><code>allSatisfy</code></a>:
<pre><code>let success = wordList.allSatisfy(text.contains)
</code></pre>

blocks|key|2624655|text|is+regular+expressions的另一种可能性|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2624656|//+*'s+are+wildcards
let+regexp+=+"(?=.*Hello*)(?=.*world*)"

if+let+range+=+Text.range(of:regexp,+options:+.regularExpression)+{
++++print("this+string+contains+Hello+world")
}+else+{
++++print("this+string+doesn't+have+the+words+we+want")
}|code-block|syntax|javascript|2624657|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/39676876/search-for-a-regular-expression-in-swift-3^0|0|M|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Another possibility <a href="https://stackoverflow.com/questions/39676876/search-for-a-regular-expression-in-swift-3">is regular expressions</a>:

<pre><code>// *'s are wildcards
let regexp = "(?=.*Hello*)(?=.*world*)"

if let range = Text.range(of:regexp, options: .regularExpression) {
 print("this string contains Hello world")
} else {
 print("this string doesn't have the words we want")
}
</code></pre>

I have a long string in swift3 and want to check if it contains word1 and word2. It could also be more than 2 search words. I found out following solution:

<pre><code>var Text = "Hello Swift-world"
var TextArray = ["Hello", "world"]
var count = 0

for n in 0..&lt;TextArray.count {
 if (Text.contains(TextArray[n])) {
 count += 1
 }
}

if (count == TextArray.count) {
 print ("success")
}
</code></pre>

But this seems very complicated, is there not an easier way to solve this? (Xcode8 and swift3)

Search multiple words in one string in swift

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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swift3中有一个长字符串，希望检查它是否包含word1、和 word2。它也可能是两个以上的搜索词。我发现了以下解决方案：var Text = "Hello Swift-world"var TextArray = ["Hello", "world"]var count = 0for n in 0..<TextArr...

问快速搜索一个字符串中的多个单词
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问快速搜索一个字符串中的多个单词EN