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
我希望以下答案既有效又简单:
$content = "The main string to search";
$search = "T";
//For compare the begining string with case insensitive.
if(stripos($content, $search) === 0) echo 'Yes';
else echo 'No';
//For compare the begining string with case sensitive.
if(strpos($content, $search) === 0) echo 'Yes';
else echo 'No';
//For compare the ending string with case insensitive.
if(stripos(strrev($content), strrev($search)) === 0) echo 'Yes';
else echo 'No';
//For compare the ending string with case sensitive.
if(strpos(strrev($content), strrev($search)) === 0) echo 'Yes';
else echo 'No';
其他回答
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;
}
PHP 8更新
PHP 8包含了新的str_starts_with和str_ends_with函数,它们最终为这个问题提供了一个高效便捷的解决方案:
$str = "beginningMiddleEnd";
if (str_starts_with($str, "beg")) echo "printed\n";
if (str_starts_with($str, "Beg")) echo "not printed\n";
if (str_ends_with($str, "End")) echo "printed\n";
if (str_ends_with($str, "end")) echo "not printed\n";
该特性的RFC提供了更多信息,同时也讨论了明显(但不那么明显)用户区域实现的优点和问题。
到目前为止,所有的答案似乎都做了大量不必要的工作、strlen计算、字符串分配(substr)等。“strpos”和“stripos”函数返回$haystack中$needle第一次出现的索引:
function startsWith($haystack,$needle,$case=true)
{
if ($case)
return strpos($haystack, $needle, 0) === 0;
return stripos($haystack, $needle, 0) === 0;
}
function endsWith($haystack,$needle,$case=true)
{
$expectedPosition = strlen($haystack) - strlen($needle);
if ($case)
return strrpos($haystack, $needle, 0) === $expectedPosition;
return strripos($haystack, $needle, 0) === $expectedPosition;
}
substr函数在许多特殊情况下都会返回false,所以这里是我的版本,它处理了这些问题:
function startsWith( $haystack, $needle ){
return $needle === ''.substr( $haystack, 0, strlen( $needle )); // substr's false => empty string
}
function endsWith( $haystack, $needle ){
$len = strlen( $needle );
return $needle === ''.substr( $haystack, -$len, $len ); // ! len=0
}
测试(真表示良好):
var_dump( startsWith('',''));
var_dump( startsWith('1',''));
var_dump(!startsWith('','1'));
var_dump( startsWith('1','1'));
var_dump( startsWith('1234','12'));
var_dump(!startsWith('1234','34'));
var_dump(!startsWith('12','1234'));
var_dump(!startsWith('34','1234'));
var_dump('---');
var_dump( endsWith('',''));
var_dump( endsWith('1',''));
var_dump(!endsWith('','1'));
var_dump( endsWith('1','1'));
var_dump(!endsWith('1234','12'));
var_dump( endsWith('1234','34'));
var_dump(!endsWith('12','1234'));
var_dump(!endsWith('34','1234'));
此外,substra_compare函数也值得一看。http://www.php.net/manual/en/function.substr-compare.php
我会这样做
function startWith($haystack,$needle){
if(substr($haystack,0, strlen($needle))===$needle)
return true;
}
function endWith($haystack,$needle){
if(substr($haystack, -strlen($needle))===$needle)
return true;
}