How can I remove text from between square brackets and the brackets themselves?
如何从方括号和方括号之间删除文本?
For example, I need hello [quote="im sneaky"] world
例如，我需要一个“你好”的世界
to become
成为
hello world
你好，世界
Here's what I'm trying to use, but it's not doing the trick
这是我想用的，但没有用
preg_replace("/[\[(.)\]]/", '', $str);
preg_replace(" /[\[()\]]/“,”str美元);
I just end up with hello quote="im sneaky" world
我只是以“你好”结束了我的世界
Thanks
谢谢
3 个解决方案
#1
28
[ and ] are special characters in a regex. They are used to list characters of a match. [a-z] matches any lowercase letter between "a" and "z". [03b] matches a "0", "3", or "b". To match the characters "[" and "]", you have to escape them with a preceding \.
是正则表达式中的特殊字符。它们用于列出匹配的字符。[a-z]匹配“a”和“z”之间的任何小写字母。[03b]匹配a "0"、"3"或"b"。要匹配字符“[”和“]”，您必须使用前面的\来转义它们。
Your code currently says "replace any character of [](). with an empty string" (reordered from the order in which you typed them for clarity).
您的代码当前说的是“替换[]()的任何字符”。用一个空字符串“(从您输入它们的顺序中重新排序)。”
Greedy match:
贪婪匹配:
preg_replace('/\[.*\]/', '', $str); // Replace from one [ to the last ]
A greedy match could match multiple [s and ]s. That expression would take an example [of "sneaky"] text [with more "sneaky"] here and turn it into an example here.
一个贪婪的匹配可以匹配多个[s和]s。这个表达式将在这里举一个“卑鄙的”文本的例子，并把它变成这里的一个例子。
Perl has a syntax for a non-greedy match (you most likely don't want to be greedy):
Perl有一个非贪婪匹配的语法(您很可能不想变得贪婪):
preg_replace('/\[.*?\]/', '', $str);
Non-greedy matches try to catch as few characters as possible. Using the same example: an example [of "sneaky"] text [with more "sneaky"] here becomes an example text here.
非贪婪匹配试图捕获尽可能少的字符。使用相同的示例:这里有一个示例文本(带有更多的“sneaky”)。
Only up to the first following ]:
[只适用于下列第一项]:
preg_replace('/\[[^\]]*\]/', '', $str); // Find a [, look for non-] characters, and then a ]
This is more explicit, but harder to read. Using the same example text, you'd get the output of the non-greedy expression.
这更明确，但更难读。使用相同的示例文本，您将得到非贪婪表达式的输出。
Note that none of these deal explicitly with white space. The spaces on either side of [ and ] will remain.
注意，这些都没有显式地处理空白。[和]两边的空间将保持不变。
Also note that all of these can fail for malformed input. Multiple [s and ]s without matches could cause a surprising result.
还需要注意的是，对于格式错误的输入，所有这些都可能失败。没有匹配的多个s会导致令人惊讶的结果。
#2
7
Just in case you are looking for a recursive removal:
如果你在寻找递归删除:
$str = preg_replace("/\[([^\[\]]++|(?R))*+\]/", "", $str);
That will convert this:
这将转换:
This [text [more text]] is cool
这个[文本]很酷
to this:
:
This is cool
这是很酷的
#3
1
I think you actually want parens for your outer brackets since it's a group. square brackets are a range of expressions. Not sure how to type it in SO.
我认为你需要外括号的括号因为它是一个组。方括号是表达式的范围。不知道怎么把它打出来。
/(\\[.*\\])/
#1
28
[ and ] are special characters in a regex. They are used to list characters of a match. [a-z] matches any lowercase letter between "a" and "z". [03b] matches a "0", "3", or "b". To match the characters "[" and "]", you have to escape them with a preceding \.
是正则表达式中的特殊字符。它们用于列出匹配的字符。[a-z]匹配“a”和“z”之间的任何小写字母。[03b]匹配a "0"、"3"或"b"。要匹配字符“[”和“]”，您必须使用前面的\来转义它们。
Your code currently says "replace any character of [](). with an empty string" (reordered from the order in which you typed them for clarity).
您的代码当前说的是“替换[]()的任何字符”。用一个空字符串“(从您输入它们的顺序中重新排序)。”
Greedy match:
贪婪匹配:
preg_replace('/\[.*\]/', '', $str); // Replace from one [ to the last ]
A greedy match could match multiple [s and ]s. That expression would take an example [of "sneaky"] text [with more "sneaky"] here and turn it into an example here.
一个贪婪的匹配可以匹配多个[s和]s。这个表达式将在这里举一个“卑鄙的”文本的例子，并把它变成这里的一个例子。
Perl has a syntax for a non-greedy match (you most likely don't want to be greedy):
Perl有一个非贪婪匹配的语法(您很可能不想变得贪婪):
preg_replace('/\[.*?\]/', '', $str);
Non-greedy matches try to catch as few characters as possible. Using the same example: an example [of "sneaky"] text [with more "sneaky"] here becomes an example text here.
非贪婪匹配试图捕获尽可能少的字符。使用相同的示例:这里有一个示例文本(带有更多的“sneaky”)。
Only up to the first following ]:
[只适用于下列第一项]:
preg_replace('/\[[^\]]*\]/', '', $str); // Find a [, look for non-] characters, and then a ]
This is more explicit, but harder to read. Using the same example text, you'd get the output of the non-greedy expression.
这更明确，但更难读。使用相同的示例文本，您将得到非贪婪表达式的输出。
Note that none of these deal explicitly with white space. The spaces on either side of [ and ] will remain.
注意，这些都没有显式地处理空白。[和]两边的空间将保持不变。
Also note that all of these can fail for malformed input. Multiple [s and ]s without matches could cause a surprising result.
还需要注意的是，对于格式错误的输入，所有这些都可能失败。没有匹配的多个s会导致令人惊讶的结果。
#2
7
Just in case you are looking for a recursive removal:
如果你在寻找递归删除:
$str = preg_replace("/\[([^\[\]]++|(?R))*+\]/", "", $str);
That will convert this:
这将转换:
This [text [more text]] is cool
这个[文本]很酷
to this:
:
This is cool
这是很酷的
#3
1
I think you actually want parens for your outer brackets since it's a group. square brackets are a range of expressions. Not sure how to type it in SO.
我认为你需要外括号的括号因为它是一个组。方括号是表达式的范围。不知道怎么把它打出来。
/(\\[.*\\])/