php 在PHP中捕获方括号之间的文本
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/10104473/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Capturing text between square brackets in PHP
提问by Chuck Le Butt
I need some way of capturing the text between square brackets. So for example, the following string:
我需要某种方法来捕获方括号之间的文本。例如，以下字符串：
[This] is a [test] string, [eat] my [shorts].
[This] is a [test] string, [eat] my [shorts].
Could be used to create the following array:
可用于创建以下数组：
Array (
[0] => [This]
[1] => [test]
[2] => [eat]
[3] => [shorts]
)
I have the following regex, /\[.*?\]/but it only captures the first instance, so:
我有以下正则表达式，/\[.*?\]/但它只捕获第一个实例，所以：
Array ( [0] => [This] )
How can I get the output I need? Note that the square brackets are NEVER nested, so that's not a concern.
我怎样才能得到我需要的输出？请注意，方括号从不嵌套，因此不必担心。
回答by Naki
Matches all strings with brackets:
匹配所有带括号的字符串：
$text = '[This] is a [test] string, [eat] my [shorts].';
preg_match_all("/\[[^\]]*\]/", $text, $matches);
var_dump($matches[0]);
If You want strings without brackets:
如果你想要没有括号的字符串：
$text = '[This] is a [test] string, [eat] my [shorts].';
preg_match_all("/\[([^\]]*)\]/", $text, $matches);
var_dump($matches[1]);
Alternative, slower version of matching without brackets (using "*" instead of "[^]"):
替代的、较慢的不带括号匹配版本（使用“*”而不是“[^]”）：
$text = '[This] is a [test] string, [eat] my [shorts].';
preg_match_all("/\[(.*?)\]/", $text, $matches);
var_dump($matches[1]);
