Select string between two groups with bash regex

Question

I have a string like this:

zone "example.com" {
    type slave;
    file "db.example.com";
    masters { x.x.x.x; };
};

zone "foo.com" {
    type slave;
    file "db.foo.com";
    masters { x.x.x.x; };
};

...

I would like to delete a specific zone.

I know how to delete it once I have the regex:

sudo sed -i '$REGEX' /file/path

But I am not quite sure of the regex itself. I know the beginning would be (zone "example.com" {) and the end would be something like (};\s};) but how do i also include the content in between?

Nathan Wilson · Accepted Answer · 2015-06-18 14:51:26Z

4

sed -i '/example\.com/,/^$/d' filename

This will delete lines in the range starting with the first the line containing "example.com" and ending with a blank line.

answered Jun 18, 2015 at 14:51

Nathan Wilson

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3 Comments

123 Over a year ago

I've well overcomplicated this haha

maxisme Over a year ago

is there any need for the '\' before the .com I tried without and still working.

Nathan Wilson Over a year ago

The period matches any character, so the target /example.com/ without the \ will also match example7com or example com.

123 · Accepted Answer · 2015-06-18 14:47:13Z

2

You could do something like this in sed

REGEX=example
sed -n "/$REGEX/{h};/^$/h;x;/$REGEX/{x;d};/$REGEX/"'!{x;p}' file

zone "foo.com" {
    type slave;
    file "db.foo.com";
    masters { x.x.x.x; };
};

answered Jun 18, 2015 at 14:47

123

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2 Comments

maxisme Over a year ago

This is duplicating everything else in the file

123 Over a year ago

@Maximilian shouldn't be.

anubhava · Accepted Answer · 2015-06-18 14:32:06Z

0

Using awk you can handle this using a null record separator i.e. -v RS=:

awk -v name='example.com' -v RS= '!($0 ~ name)' zones
zone "foo.com" {
    type slave;
    file "db.foo.com";
    masters { x.x.x.x; };
};

answered Jun 18, 2015 at 14:32

anubhava

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