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

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;
}

其他回答

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。

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);
}

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';

我意识到这已经完成,但您可能需要查看strncmp,因为它允许您将字符串的长度与之进行比较,因此:

function startsWith($haystack, $needle, $case=true) {
    if ($case)
        return strncasecmp($haystack, $needle, strlen($needle)) == 0;
    else
        return strncmp($haystack, $needle, strlen($needle)) == 0;
}    

前面的许多答案也同样有效。然而,这可能是尽可能短的,你可以让它做你想做的事情。你只需要声明你希望它“回归真实”。因此,我包含了返回布尔值真/假和文本真/假的解决方案。

// boolean true/false
function startsWith($haystack, $needle)
{
    return strpos($haystack, $needle) === 0 ? 1 : 0;
}

function endsWith($haystack, $needle)
{
    return stripos($haystack, $needle) === 0 ? 1 : 0;
}


// textual true/false
function startsWith($haystack, $needle)
{
    return strpos($haystack, $needle) === 0 ? 'true' : 'false';
}

function endsWith($haystack, $needle)
{
    return stripos($haystack, $needle) === 0 ? 'true' : 'false';
}