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
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