string 如何替换perl中的空格
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How to Replace white space in perl
提问by DarRay
chomp($myString);
$myString =~ s/\///g;
can i replace those two with
我可以用这两个代替吗
$myString =~ s/\s//g;
are there any difference? Please explain.
有什么区别吗？请解释。
回答by mattexx
Your first code will take a newline off the end of $myString if it exists and then remove all "/" characters. The second line of code will remove all whitespace characters. Is there a typo?
如果 $myString 存在，您的第一个代码将从 $myString 的末尾取一个换行符，然后删除所有“/”字符。第二行代码将删除所有空白字符。有错别字吗？
Maybe you want to know you can replace this:
也许你想知道你可以替换这个：
chomp($myString);
$myString =~ s/\s//g;
with this:
有了这个：
$myString =~ s/\s//g;
If that's the question, then yes. Since a newline counts as whitespace, the second code example do the job of both lines above.
如果这是问题，那么是的。由于换行符算作空格，第二个代码示例完成了上面两行的工作。
回答by Nikhil Jain
From perldoc chomp:
从 perldoc chomp：
chompremove the newline from the end of an input record when you're worried that the final record may be missing its newline.
当您担心最终记录可能会丢失其换行符时，chomp从输入记录的末尾删除换行符。
When in paragraph mode ($/ = "" ), it removes all trailing newlines from the string.
When in slurp mode ($/ = undef) or fixed-length record mode ($/is a reference to an integer or the like, see perlvar) chomp()won't remove anything.
在段落模式下($/ = "" )，它会从字符串中删除所有尾随换行符。当处于 slurp 模式 ( $/ = undef) 或固定长度记录模式（$/是对整数等的引用，请参阅 perlvar）时，chomp()不会删除任何内容。
you can removeleading and trailing whitespace from strings like,
您可以从字符串中删除前导和尾随空格，例如，
$string =~ s{^\s+|\s+$}{}g
回答by Ben
Chomp will get rid of newlines at the end of your string, but won't remove whitespace. A typical trim function uses the following two substitution lines:
Chomp 将删除字符串末尾的换行符，但不会删除空格。典型的修剪函数使用以下两条替换行：
$string =~ s/^\s+//;
$string =~ s/\s+$//;
The first line removes any spaces at the beginning of your string, and the second removes spaces after the end of the string.
第一行删除字符串开头的所有空格，第二行删除字符串结尾之后的空格。
