Bash 正则表达式字符串变量匹配
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Bash regex string variable match
提问by The Georgia
I have the following script i wrote in perl that works just fine. But i am trying to achieve the same thing using bash.
我有以下用 perl 编写的脚本，效果很好。但我正在尝试使用 bash 来实现同样的目的。
#!/usr/bin/perl
use 5.010;
use strict;
INIT {
my $string = 'Seconds_Behind_Master: 1';
my ($s) = ($string =~ /Seconds_Behind_Master: ([\d]+)/);
if ($s > 10) {
print "Too long... ${s}";
} else {
print "It's ok";
}
}
__END__
How can i achieve this using a bash script? Basically, i want to be able to read and match the value at the end of the string "Seconds_Behind_Master: N" where N can be any value.
我如何使用 bash 脚本实现这一目标？基本上，我希望能够读取和匹配字符串“Seconds_Behind_Master: N”末尾的值，其中 N 可以是任何值。
回答by Krzysztof Krasoń
You can use a tool for it e.g. sedif you want to stay with regexps:
你可以使用一个工具，例如，sed如果你想继续使用正则表达式：
#!/bin/sh
string="Seconds_Behind_Master: 1"
s=`echo $string | sed -r 's/Seconds_Behind_Master: ([0-9]+)//g'`
if [ $s -gt 10 ]
then
echo "Too long... $s"
else
echo "It's OK"
fi
回答by xxfelixxx
You can use regular expression in bash, just like in perl.
您可以在 bash 中使用正则表达式，就像在 perl 中一样。
#!/bin/bash
STRING="Seconds_Behind_Master: "
REGEX="Seconds_Behind_Master: ([0-9]+)"
RANGE=$( seq 8 12 )
for i in $RANGE; do
NEW_STRING="${STRING}${i}"
echo $NEW_STRING;
[[ $NEW_STRING =~ $REGEX ]]
SECONDS="${BASH_REMATCH[1]}"
if [ -n "$SECONDS" ]; then
if [[ "$SECONDS" -gt 10 ]]; then
echo "Too Long...$SECONDS"
else
echo "OK"
fi
else
echo "ERROR: Failed to match '$NEW_STRING' with REGEX '$REGEX'"
fi
done
Output
输出
Seconds_Behind_Master: 8
OK
Seconds_Behind_Master: 9
OK
Seconds_Behind_Master: 10
OK
Seconds_Behind_Master: 11
Too Long...11
Seconds_Behind_Master: 12
Too Long...12
回答by tripleee
The specific case of "more than a single digit" is particularly easy with just a pattern match:
“多于一位数”的特定情况仅通过模式匹配就特别容易：
case $string in
*Seconds_Behind_Master: [1-9][0-9]*) echo Too long;;
*) echo OK;;
esac
To emulate what your Perl code is doing more closely, you can extract the number with simple string substitutions.
要更接近地模拟 Perl 代码的作用，您可以使用简单的字符串替换来提取数字。
s=${string##*Seconds_Behind_Master: }
s=${s%%[!0-9]*}
[ $s -gt 10 ] && echo "Too long: $s" || echo OK.
These are glob patterns, not regular expressions; *matches any string, [!0-9]matches a single character which is not a digit. All of this is Bourne-compatible, i.e. not strictly Bash only (you can use /bin/shinstead of /bin/bash).
这些是 glob 模式，而不是正则表达式；*匹配任何字符串，[!0-9]匹配不是数字的单个字符。所有这些都是与 Bourne 兼容的，即严格来说不仅仅是 Bash（您可以使用/bin/sh代替/bin/bash）。
