我如何编写两个函数,如果它以指定的字符/字符串开头或以指定的字符串结尾,那么它们将接受字符串并返回?

例如:

$str = '|apples}';

echo startsWith($str, '|'); //Returns true
echo endsWith($str, '}'); //Returns true

当前回答

这可能奏效

function startsWith($haystack, $needle) {
     return substr($haystack, 0, strlen($needle)) == $needle;
}

资料来源:https://stackoverflow.com/a/4419658

其他回答

专注于startswith,如果您确定字符串不为空,那么在比较之前,在第一个字符、strlen等上添加一个测试会加快速度:

function startswith5b($haystack, $needle) {
    return ($haystack{0}==$needle{0})?strncmp($haystack, $needle, strlen($needle)) === 0:FALSE;
}

不知怎么的,速度快了(20%-30%)。添加另一个字符测试,如$haystack{1}==$needle{1},似乎并不能加快速度,甚至可能会减慢速度。

==似乎比==条件运算符(a)?b: c似乎比如果(a)b更快;否则c;


对于那些问“为什么不使用strpos?”并称其他解决方案为“不必要的工作”的人


strpos速度很快,但它不是这项工作的正确工具。

为了理解,这里有一个小模拟作为示例:

Search a12345678c inside bcdefga12345678xbbbbb.....bbbbba12345678c

电脑在里面做什么?

    With strccmp, etc...

    is a===b? NO
    return false



    With strpos

    is a===b? NO -- iterating in haysack
    is a===c? NO
    is a===d? NO
    ....
    is a===g? NO
    is a===g? NO
    is a===a? YES
    is 1===1? YES -- iterating in needle
    is 2===3? YES
    is 4===4? YES
    ....
    is 8===8? YES
    is c===x? NO: oh God,
    is a===1? NO -- iterating in haysack again
    is a===2? NO
    is a===3? NO
    is a===4? NO
    ....
    is a===x? NO
    is a===b? NO
    is a===b? NO
    is a===b? NO
    is a===b? NO
    is a===b? NO
    is a===b? NO
    is a===b? NO
    ...
    ... may many times...
    ...
    is a===b? NO
    is a===a? YES -- iterating in needle again
    is 1===1? YES
    is 2===3? YES
    is 4===4? YES
    is 8===8? YES
    is c===c? YES YES YES I have found the same string! yay!
    was it at position 0? NOPE
    What you mean NO? So the string I found is useless? YEs.
    Damn.
    return false

假设strlen不迭代整个字符串(但即使在这种情况下),这一点都不方便。

无副本,无实习循环:

function startsWith(string $string, string $start): bool
{
    return strrpos($string, $start, - strlen($string)) !== false;
}

function endsWith(string $string, string $end): bool
{
    return ($offset = strlen($string) - strlen($end)) >= 0 
    && strpos($string, $end, $offset) !== false;
}

拉里维尔9.0

如果你正在使用Laravel,那么你可以执行以下操作(如果你没有使用Laraver,那么你真的应该这样做)。

Str::of('a long string')->startsWith('a');
Str::of('a long string')->endsWith('string');

//true
//true

这是一个接受答案的多字节安全版本,它适用于UTF-8字符串:

function startsWith($haystack, $needle)
{
    $length = mb_strlen($needle, 'UTF-8');
    return (mb_substr($haystack, 0, $length, 'UTF-8') === $needle);
}

function endsWith($haystack, $needle)
{
    $length = mb_strlen($needle, 'UTF-8');
    return $length === 0 ||
        (mb_substr($haystack, -$length, $length, 'UTF-8') === $needle);
}

如果速度对你很重要,试试这个。(我认为这是最快的方法)

仅适用于字符串,如果$haystack只有1个字符

function startsWithChar($needle, $haystack)
{
   return ($needle === $haystack[0]);
}

function endsWithChar($needle, $haystack)
{
   return ($needle === $haystack[strlen($haystack) - 1]);
}

$str='|apples}';
echo startsWithChar('|',$str); //Returns true
echo endsWithChar('}',$str); //Returns true
echo startsWithChar('=',$str); //Returns false
echo endsWithChar('#',$str); //Returns false