42

有没有办法使用爆炸功能仅在最后一个分隔符出现时爆炸?

$string = "one_two_  ... _three_four";

$explodeResultArray = explode("_", $string);

结果应该是:

echo $explodeResultArray[0]; // "one_two_three ...";
echo $explodeResultArray[1]; // "four";
4

11 回答 11

96

直截了当:

$parts = explode('_', $string);
$last = array_pop($parts);
$parts = array(implode('_', $parts), $last);
echo $parts[0]; // outputs "one_two_three"

常用表达:

$parts = preg_split('~_(?=[^_]*$)~', $string);
echo $parts[0]; // outputs "one_two_three"

字符串反转:

$reversedParts = explode('_', strrev($string), 2);
echo strrev($reversedParts[0]); // outputs "four"
于 2013-05-17T13:58:12.317 回答
58

无需解决方法。explode()接受负限制。

$string = "one_two_three_four";
$part   = implode('_', explode('_', $string, -1));
echo $part;

结果是

one_two_three
于 2013-05-17T23:30:13.467 回答
13

I chose to use substring becasue you want a string up to a particular point:

$string = "one_two_three_four_five_six_seven";
$part1 = substr("$string",0, strrpos($string,'_'));
$part2 = substr("$string", (strrpos($string,'_') + 1));
var_dump($part1,$part2);

RESULTS:

string(27) "one_two_three_four_five_six"
string(5) "seven"
于 2013-05-17T14:09:35.660 回答
9

您可以执行以下操作:

$string = "one_two_three_four";
$explode = explode('_', $string); // split all parts

$end = '';
$begin = '';

if(count($explode) > 0){
    $end = array_pop($explode); // removes the last element, and returns it

    if(count($explode) > 0){
        $begin = implode('_', $explode); // glue the remaining pieces back together
    }
}

编辑: array_shift 应该是 array_pop

于 2013-05-17T13:57:49.863 回答
5
<?php
$lastPos = strrpos($string, '_');
if ($lastPos !== false) {
    $start = substr($string, 0, $lastPos);
    $end = substr($string, $lastPos+1);
} else {
    // no delimeter found!
}

如果只关心最后一部分,那就更简单了。

<?php
$end = substr(strrchr($string, '_'), 1);
于 2013-05-17T14:03:33.360 回答
3

使用 preg_match()

$string = "one_two_three_four";

$arr = array();
preg_match("/(^.*)_(.*?)$/", $string, $arr);

print_r($arr);

输出: Array ( [0] => one_two_three_four [1] => one_two_three [2] => four )

于 2013-05-17T13:59:31.027 回答
3

使用结束+爆炸

$explodeResultArray = end(explode("_", $string));

$explodeResultArray will = 四

于 2018-05-14T12:14:24.270 回答
2
// reverse $string right after definition
$string = "one_two_three_four_five_six";
$string = implode("_",array_reverse(explode("_",$string)));

// chop off the first part
list($result, $string) = explode("_", $string, 2);

echo "$result --- $string";

输出:

six --- five_four_three_two_one 
于 2013-05-17T15:14:41.773 回答
1
$explodeResultArray = explode("_", $string);
$last_item = end($explodeResultArray);
$key = count($explodeResultArray) - 1;
unset($explodeResultArray[$key]);
$arr[] = (implode($explodeResultArray,'_'));
$arr[] = $last_item;
print_r($arr);

Output

Array
(
    [0] => one_two_  ... _three
    [1] => four
)
于 2013-05-20T11:45:33.877 回答
1

我有类似的需求,并受到@NLZ 回答的启发,我制作了一个可重用的函数,其功能与 regular 相同explode(),但向后(尽管我添加了一个选项来反转数组顺序 contra regular explode()):

function backward_explode($delimiter, $string, $limit = null, $keep_order = true) {
    if ((string)$delimiter === "") {
        return false;
    }

    if ($limit === 0 || $limit === 1) {
        return array($string);
    }

    $explode = explode($delimiter, $string);

    if ($limit === null || $limit === count($explode)) {
        return $keep_order? $explode : array_reverse($explode);
    }

    $parts = array();

    if ($limit > 0) {
        for ($i = 1; $i < $limit; $i++) {
            $parts[] = array_pop($explode);
        }
        $remainder = implode($delimiter, $explode);
        $parts[] = $remainder;
        if ($keep_order) {
            $parts = array_reverse($parts);
        }
    } else {
        if (strpos($string, $delimiter) === false) {
            return array();
        }
        $parts = $explode;
        array_splice($parts, 0, abs($limit));
        if (!$keep_order) {
            $parts = array_reverse($parts);
        }
    }

    return $parts;
}

(也作为要点。)

所以:

$string = 'one two three four';
var_dump(backward_explode(' ', $string));
var_dump(backward_explode(' ', $string, 2));
var_dump(backward_explode(' ', $string, 3));
var_dump(backward_explode(' ', $string, 2, false));
var_dump(backward_explode(' ', $string, -1));
var_dump(backward_explode(' ', $string, 1)); // same as with $limit = 0
var_dump(backward_explode('#', $string, -2));
var_dump(backward_explode('', $string, 3));

我们得到:

array (size=4)
  0 => string 'one' (length=3)
  1 => string 'two' (length=3)
  2 => string 'three' (length=5)
  3 => string 'four' (length=4)
array (size=2)
  0 => string 'one two three' (length=13)
  1 => string 'four' (length=4)
array (size=3)
  0 => string 'one two' (length=7)
  1 => string 'three' (length=5)
  2 => string 'four' (length=4)
array (size=2)
  0 => string 'four' (length=4)
  1 => string 'one two three' (length=13)
array (size=3)
  0 => string 'two' (length=3)
  1 => string 'three' (length=5)
  2 => string 'four' (length=4)
array (size=1)
  0 => string 'one two three four' (length=18)
array (size=0)
  empty
boolean false
于 2014-05-14T08:50:57.093 回答
0

对于这样的任务,您可以只使用strstrstrrchr

$valueBeforeLastUnderscore = rtrim(strrev(strstr(strrev($value), '_')), '_');
$valueAfterLastUnderscore = ltrim(strrchr($value, '_'), '_');

话虽如此,我更喜欢正则表达式的答案。

于 2018-09-06T08:11:53.370 回答