分解字符串，仅在分解匹配的最后一次出现时

Question

我正在尝试分解一个字符串，但我需要它只在最后一个 'and' 而不是每个 'and' 分解。有没有办法做到这一点？

<?php

$string = "one and two and three and four and five";
$string = explode(" and ", $string);

print_r($string);

?>

结果：

数组([0]=>一[1]=>二[2]=>三[3]=>四[4]=>五)

需要结果：

数组（[0] => 一二三四 [1] => 五）

Answer 1

我不知道是否有任何其他快速解决该需求的方法，但您可以尝试下面的代码；

$string = "one and two and three and four and five";
$stringArray = explode(" and ", $string);
$stringArrayItemCount = count($stringArray);
//Keep your last item
$stringArrayLastItem = $stringArray[$stringArrayItemCount-1];
//Remove last item from your array
unset($stringArray[$stringArrayItemCount-1]);
//Create new array imploding existing one + last item of your old array
$stringArray = array(implode(" and ",$stringArray),$stringArrayLastItem);

print_r($stringArray);

此示例的工作版本：http://ideone.com/qdZFtk

希望对您有所帮助。

Answer 2

写了一个函数来做到这一点：

<?php

$string = "one and two and three and four and five";
$string = explodeLast(" and ", $string);

echo $string;

function explodeLast($explodeAt, $string) {

$explode = explode($explodeAt, $string);
$count = count($explode);
$counter = 0;
$string = null;

while ($counter < $count-1) {

if ($counter < $count-2) {
$string .= $explode[$counter].$explodeAt;
} //end of if ($counter < $count-2)
else {
$string .= $explode[$counter];
} //end of else not ($counter < $count-2)

$counter++;
} //end of while ($counter < $count)

return $string;

} //end of function explodeLast($explode, $explodeAt)

?>

Answer 3

这应该有效：

$str = 'one and two and three and four and five';
$breakWord = ' and ';

$output = str_split($str, strrpos($str, $breakWord) + strlen($breakWord));

var_dump($output);

http://sandbox.onlinephpfunctions.com/code/0149b3ff973485befe97a1c6b241a6764bd2f289

编辑 - 正则表达式使用：

<?php
$str = 'one and two and three and four and the last part is actually longer than the first part';
$pattern = "/(.+) and (?!.* and )(.+)/";
preg_match($pattern, $str, $matches);

var_dump($matches);

http://sandbox.onlinephpfunctions.com/code/0bc68f46fe594e360134bdca256e5916a2f42f74

Answer 4

我的方法更加简单，使用 preg_match_all() 和一个简单的正则表达式代替：

$string = "one and two and three and four and five";
$pattern = '/(\w+\s)+/';
preg_match_all($pattern, $string, $matches);

$length = strlen($matches[0][0]);

echo $matches[0][0]; // 'one and two and three and four and'
echo substr($string, $length); // 'five'

这里唯一的问题是第一个匹配项仍然有尾随 'and'，如果需要，可以通过一些简单的编码将其删除。如果您想要更复杂的正则表达式，您可以使用正向前瞻和负向后视。

Answer 5

这似乎很容易，只使用基本的字符串函数就可以做到。

$x = strrpos($string, ' and ');
$s2 = array(substr($string, 0, $x), substr($string, $x + 5));

Answer 6

你也可以使用preg_split:

单字符定界符

$string = "one,two,three,four,five";
$delimiter = ",";

// The regexp will look like this:  /,([^,]+)$/   
$array = preg_split("/".$delimiter."([^".$delimiter."]+)$/", $string,
  -1, PREG_SPLIT_DELIM_CAPTURE | PREG_SPLIT_NO_EMPTY);

print_r($array);

正则表达式将最后一个定界符与最后一个项目匹配，但仅捕获该项目以便它保留在结果中，这要归功于 PREG_SPLIT_DELIM_CAPTURE。

PREG_SPLIT_NO_EMPTY 在那里是因为，我不确定为什么，拆分最后给出了一个空字符串。

多字符分隔符

此处必须调整正则表达式，使用负环视（如 this answer 中所述）：

$string = "one and two and three and four and five";
$delimiter = " and ";

$array = preg_split("/$delimiter((?(?!$delimiter).)*$)/", $string,
  -1, PREG_SPLIT_DELIM_CAPTURE | PREG_SPLIT_NO_EMPTY);

print_r($array);

(?(?!$delimiter).)*表示：只匹配不以$delimiter开头的字符。第一个 ? 阻止捕获额外的组。

分解字符串，仅在分解匹配的最后一次出现时

Exploding a string, only at the last occurrence of the explode match

php

explode

单字符定界符

多字符分隔符