有这样的字符串:
$str = "dateto:'2015-10-07 15:05' xxxx datefrom:'2015-10-09 15:05' yyyy asdf"
期望的结果是:
[0] => Array (
[0] => dateto:'2015-10-07 15:05'
[1] => xxxx
[2] => datefrom:'2015-10-09 15:05'
[3] => yyyy
[4] => asdf
)
我得到的东西:
preg_match_all("/\'(?:[^()]|(?R))+\'|'[^']*'|[^(),\s]+/", $str, $m);
是:
[0] => Array (
[0] => dateto:'2015-10-07
[1] => 15:05'
[2] => xxxx
[3] => datefrom:'2015-10-09
[4] => 15:05'
[5] => yyyy
[6] => asdf
)
还尝试使用preg_split("/[\s]+/", $str)
,但如果值介于引号之间,则无法知道如何转义。任何人都可以告诉我如何,也请解释正则表达式。谢谢!
答案 0 :(得分:1)
通常情况下,当你想要分割一个字符串时,使用createTemplateFromFile()
并不是最好的方法(这似乎有点反直觉,但大部分时间都是如此)。更有效的方法是使用描述所有不是分隔符的模式(使用preg_split
)查找所有项目(白色空格):
preg_match_all
模式细节:
$pattern = <<<'EOD'
~(?=\S)[^'"\s]*(?:'[^']*'[^'"\s]*|"[^"]*"[^'"\s]*)*~
EOD;
if (preg_match_all($pattern, $str, $m))
$result = $m[0];
请注意,使用这个稍长的模式,您的示例字符串的五个项目只能找到60个步骤。您也可以使用这种更短/更简单的模式:
~ # pattern delimiter
(?=\S) # the lookahead assertion only succeeds if there is a non-
# white-space character at the current position.
# (This lookahead is useful for two reasons:
# - it allows the regex engine to quickly find the start of
# the next item without to have to test each branch of the
# following alternation at each position in the strings
# until one succeeds.
# - it ensures that there's at least one non-white-space.
# Without it, the pattern may match an empty string.
# )
[^'"\s]* #"'# all that is not a quote or a white-space
(?: # eventual quoted parts
'[^']*' [^'"\s]* #"# single quotes
|
"[^"]*" [^'"\s]* # double quotes
)*
~
但效率稍低。
答案 1 :(得分:0)
对于您的示例,您可以将preg_split与negative lookbehind (?<!\d)
一起使用,即:
<?php
$str = "dateto:'2015-10-07 15:05' xxxx datefrom:'2015-10-09 15:05' yyyy asdf";
$matches = preg_split('/(?<!\d)(\s)/', $str);
print_r($matches);
输出:
Array
(
[0] => dateto:'2015-10-07 15:05'
[1] => xxxx
[2] => datefrom:'2015-10-09 15:05'
[3] => yyyy
[4] => asdf
)
演示:
正则表达式解释:
(?<!\d)(\s)
Assert that it is impossible to match the regex below with the match ending at this position (negative lookbehind) «(?<!\d)»
Match a single character that is a “digit” «\d»
Match the regex below and capture its match into backreference number 1 «(\s)»
Match a single character that is a “whitespace character” «\s»