如何檢查兩個字符串是否包含相同的字母？

Question

$textone = "pate"; //$_GET 
$texttwo = "tape"; 
$texttre = "tapp"; 

if ($textone ??? $texttwo) { 
echo "The two strings contain the same letters"; 
} 
if ($textone ??? $texttre) { 
echo "The two strings NOT contain the same letters"; 
} 

什麼if聲明我在找什麼？

Answer 1

我想一個解決辦法是，考慮到以下兩個變量：

$textone = "pate"; 
$texttwo = "tape"; 

1.首先，分割字符串，得到的字母兩個數組：

$arr1 = preg_split('//', $textone, -1, PREG_SPLIT_NO_EMPTY); 
$arr2 = preg_split('//', $texttwo, -1, PREG_SPLIT_NO_EMPTY); 

需要注意的是，在他的評論中指出@Mike，而不是使用preg_split()像我第一次做，對於這樣的情況，人們將使用str_split()會更好：

$arr1 = str_split($textone); 
$arr2 = str_split($texttwo); 

2.然後，排序的那些陣列，所以字母按字母順序排列：

sort($arr1); 
sort($arr2); 

之後，破滅的陣列，打造話所有的字母按字母順序排列：

$text1Sorted = implode('', $arr1); 
$text2Sorted = implode('', $arr2); 

4.最後，比較這些兩個單詞：

if ($text1Sorted == $text2Sorted) { 
    echo "$text1Sorted == $text2Sorted"; 
} 
else { 
    echo "$text1Sorted != $text2Sorted"; 
} 

---

談到這個想法變成一個比較函數會給你的代碼如下部分：

function compare($textone, $texttwo) { 
    $arr1 = str_split($textone); 
    $arr2 = str_split($texttwo); 

    sort($arr1); 
    sort($arr2); 

    $text1Sorted = implode('', $arr1); 
    $text2Sorted = implode('', $arr2); 

    if ($text1Sorted == $text2Sorted) { 
     echo "$text1Sorted == $text2Sorted<br />"; 
    } 
    else { 
     echo "$text1Sorted != $text2Sorted<br />"; 
    } 
} 

並呼籲你的兩個話該功能：

compare("pate", "tape"); 
compare("pate", "tapp"); 

會得到你以下結果：

aept == aept 
aept != appt 

Answer 2

使用===和!==

if ($textone === $texttwo) { 
    echo "The two strings contain the same letters"; 
}else{ 
    echo "The two strings NOT contain the same letters"; 
} 

或

if ($textone === $texttwo) { 
    echo "The two strings contain the same letters"; 
} 

if ($textone !== $texttwo) { 
    echo "The two strings NOT contain the same letters"; 
}