php - Check if a string is solely composed of characters found in another string

Solution:

Use the first string (whitelist string) as the mask fortrim(). This will remove all matched characters from the second string. If there are any characters remaining after the trim, thenfalse.

since all characters of str2 are also in str1, so it should return true

Code: (Demo)

$str1 = "apple";
$str2 = "pal";
$str3 = "pole";
var_export(!strlen(trim($str2, $str1)));
echo "\n---\n";
var_export(!strlen(trim($str3, $str1)));
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Output:

true
---
false
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This answer is clean and direct because it does not need to generate temporary arrays for subsequent comparison.trim()'s character mask affords the same action thatstr_replace() is famous for, but without needing to split the whitelist string into an array first. I don't know ifltrim() orrtrim() would be any faster (on a microscopic level) thantrim(), but any of these functions will deliver the same output.

p.s. If you are guaranteed to only be working with letters, then you can use a falsey check instead ofstrlen().

!trim($str2, $str1)
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I say this because if you allowed numbers, then a trimmed string containing a zero would be considered falsey and return an incorrect result.

Solution:

You can use the array_intersect function after splitting each characters from a word using the str_split function as:

<?php

$a = "Apple";
$b = "pal";

$matched = !empty(array_intersect(str_split($b), str_split($a))); 
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array_intersect will return [ p, l ] ignoring 'A' due to case sensitive

echo $matched; // true

?>
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In case if you want to match ignoring the case of the characters then you can first lowercase the word using the strtolower function: and then split as:

$a = "Apple";

$a = strtolower($a);
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Solution:

You can use count_chars() to get array of present chars in each string and then compare those two arrays:

$s1 = 'apple';
$s2 = 'pal';

$c1 = count_chars($s1, 1);
$c2 = count_chars($s2, 1);

$ok = empty(array_diff_key($c2, $c1));
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