string 如何获取给定字符串的子字符串，直到第一次出现指定字符？
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/8256382/
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
How to take substring of a given string until the first appearance of specified character?
提问by Mihran Hovsepyan
The question is for perl.
问题是针对perl的。
For example if I have "hello.world"and the specified character is '.'then the result I want is "hello".
例如，如果我有"hello.world"并且指定的字符是，'.'那么我想要的结果是"hello".
回答by Jim Davis
回答by yko
回答by canavanin
use strict;
use warnings;
my $string = "hello.world";
my $dot = index($string, '.');
my $word = substr($string, 0, $dot);
print "$word\n";
gives you hello
给你 hello
回答by TLP
In the spirit of TIMTOWTDI, and introducing new features: Using the non-destructive option/r
本着 TIMTOWTDI 的精神，并引入新功能：使用非破坏性选项/r
my $partial = $string =~ s/\..*//sr;
The greedy .*end will chop off everything after the first period, including possible newline characters (/soption), but keep the original string intact and remove the need for parens to impose list context (/roption).
贪婪的.*结尾将砍掉第一个句点之后的所有内容，包括可能的换行符（/s选项），但保持原始字符串完好无损，并且不需要括号来强加列表上下文（/r选项）。
Quote from perlop:
来自 perlop 的引用：
If the /r (non-destructive) option is used then it runs the substitution on a copy of the string and instead of returning the number of substitutions, it returns the copy whether or not a substitution occurred. The original string is never changed when /r is used. The copy will always be a plain string, even if the input is an object or a tied variable.
如果使用 /r（非破坏性）选项，则它会在字符串的副本上运行替换，而不是返回替换次数，无论是否发生替换，它都会返回副本。使用 /r 时，原始字符串永远不会改变。即使输入是对象或绑定变量，副本也将始终是纯字符串。
回答by choroba
Another possibility:
另一种可能：
my $string = 'hello.world';
print ((split /\./, $string)[0], "\n");
