Grep multi-line
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From: thdyoung on 13 Jul 2010 18:11 On Jul 13, 2:13 pm, Ben Bacarisse <ben.use...(a)bsb.me.uk> wrote: > Ed Morton <mortons...(a)gmail.com> writes: > > On 7/13/2010 1:32 AM, moonhkt wrote: > >> How to grep multi-line, when input particlar date ? > > >> Text file > >> 2010/06/21 Canon Scanner 9000F Could a kind person speak what this code means ? I see: announcement of a variable and a value assigned to it Then a 'pattern' - posix digits at the start of the line Then the 'action' which I can't read. It seems to announce another variable f which, I think, is conditional on the first field being the date Then f outside the last curly brace followed by the input Sorry to be thick but this seems to say ' print f (a line of input) if the first field of f (guessing really) is the date d ' Probably this is wrong coz I've haven't given an account of what the regex pattern is doing except note that it demands a number at the start of the line... thanks Tom > >> 2010/06/30 MacBook Pro Delivery, Original 2010/07/02 > >> S/N > >> etc... > >> 2010/07/06 Mail Rebate Form to MAC Apple,http://www.apple.com/hk/promo/rebate/bts.html > <snip> > > Try this (untested): > > > awk -v d="2010/07/06" '$0 ~ /^[[:digit:]]/ {f = ($1 == d)} f' file > > Is there a reason to prefer "$0 ~ /^[[:digit:]]/" over the simpler > "/^[[:digit:]]/" as the pattern? > > -- > Ben.
From: Ben Bacarisse on 13 Jul 2010 18:45 [I've re-ordered an snipped.] "thdyoung(a)googlemail.com" <thdyoung(a)gmail.com> writes: > On Jul 13, 2:13 pm, Ben Bacarisse <ben.use...(a)bsb.me.uk> wrote: >> Ed Morton <mortons...(a)gmail.com> writes: >> > On 7/13/2010 1:32 AM, moonhkt wrote: >> >> How to grep multi-line, when input particlar date ? >> >> >> Text file >> >> 2010/06/21 Canon Scanner 9000F > >> 2010/06/30 MacBook Pro Delivery, Original 2010/07/02 >> >> S/N >> >> etc... >> >> 2010/07/06 Mail Rebate Form to MAC Apple,http://www.apple.com/hk/promo/rebate/bts.html >> <snip> >> > Try this (untested): >> >> > awk -v d="2010/07/06" '$0 ~ /^[[:digit:]]/ {f = ($1 == d)} f' file >> > Could a kind person speak what this code means ? Let's start with my suggestion of getting rid of "$0 ~" -- it's less to explain: awk -v d="2010/07/06" '/^[[:digit:]]/ {f = ($1 == d)} f' file awk is passed four arguments: The first two set the variable d to the desired date. The third is the awk program and the fourth is the file to process. awk programs are a sequence of <test> {<action>} pairs (functions are different but this program has none). You can leave out either the <test> part or the {<action>} part. The <test> is evaluated against every line: if the line produces a true result from the test the actions are executed. If there is no {<action>} part awk assumes {print}, and if there is no <test> part awk assumes 1 (a test that is always true). This program has two such pairs which are (after inserting the default print action): /^[[:digit:]]/ {f = ($1 == d)} f {print} The first matches lines that start with a digit. Such a line sets f to the value of the test $1 == d. $1 is the first field on the line, so f gets set to true or false (actually 1 or 0) depending on whether the line starts with the date that we want (the value of d set on the command line). All lines (including this one) are tested against the test f. This is true (non-zero) or false (zero) depending on what has happened before. Once f has been set to 1, lines are printed until f gets set to false. This can happen only when a line is processed that starts with a digit but has $1 not equal to the date we want. Ed's original $0 ~ /.../ version is the same. ~ is the match operator and $0 is the whole line. A pattern on its own is equivalent to a match against $0. <snip> -- Ben.
From: Janis Papanagnou on 13 Jul 2010 18:46 On 14/07/10 00:11, thdyoung(a)googlemail.com wrote: > On Jul 13, 2:13 pm, Ben Bacarisse <ben.use...(a)bsb.me.uk> wrote: >> Ed Morton <mortons...(a)gmail.com> writes: >>> On 7/13/2010 1:32 AM, moonhkt wrote: >>>> How to grep multi-line, when input particlar date ? >> >>>> Text file >>>> 2010/06/21 Canon Scanner 9000F > Could a kind person speak what this code means ? > > I see: announcement of a variable and a value assigned to it > Then a 'pattern' - posix digits at the start of the line > Then the 'action' which I can't read. It seems to announce another > variable f which, I think, is conditional on the first field being the > date > Then f outside the last curly brace followed by the input > > Sorry to be thick but this seems to say ' print f (a line of input) if > the first field of f (guessing really) is the date d ' > > Probably this is wrong coz I've haven't given an account of what the > regex pattern is doing except note that it demands a number at the > start of the line... Reformatted and added the default action... $0 ~ /^[[:digit:]]/ {f = ($1 == d)} f { print $0 } First line: whenever a line starts with a digit, evaluate whether the first field equals the requested date, if so then the flag is set to true (1) otherwise false (0). Second line: if flag is set print the line Actually Ed's code is a variant of the one I posted earlier this morning, which I explain as well for comparison. The code was (!/^ / && f=($1~d)) || (f && /^ /) Reformatted, slightly rearranged, and default action added gives !/^ / && f=($1~d) { print $0 } /^ / && f { print $0 } First line: whenever a line starts not with white space, and the flag is set, which depends on the first field matches the date, print the line Second line: whenever a line starts with white space, and the flag is set, print the line Janis > > thanks > > Tom > >> > > >> 2010/06/30 MacBook Pro Delivery, Original 2010/07/02 >>>> S/N >>>> etc... >>>> 2010/07/06 Mail Rebate Form to MAC Apple,http://www.apple.com/hk/promo/rebate/bts.html >> <snip> >>> Try this (untested): >> >>> awk -v d="2010/07/06" '$0 ~ /^[[:digit:]]/ {f = ($1 == d)} f' file >> >> Is there a reason to prefer "$0 ~ /^[[:digit:]]/" over the simpler >> "/^[[:digit:]]/" as the pattern? >> >> -- >> Ben. >
From: moonhkt on 13 Jul 2010 21:02 On Jul 14, 6:46 am, Janis Papanagnou <janis_papanag...(a)hotmail.com> wrote: > On 14/07/10 00:11, thdyo...(a)googlemail.com wrote: > > > > > > > On Jul 13, 2:13 pm, Ben Bacarisse <ben.use...(a)bsb.me.uk> wrote: > >> Ed Morton <mortons...(a)gmail.com> writes: > >>> On 7/13/2010 1:32 AM, moonhkt wrote: > >>>> How to grep multi-line, when input particlar date ? > > >>>> Text file > >>>> 2010/06/21 Canon Scanner 9000F > > Could a kind person speak what this code means ? > > > I see: announcement of a variable and a value assigned to it > > Then a 'pattern' - posix digits at the start of the line > > Then the 'action' which I can't read. It seems to announce another > > variable f which, I think, is conditional on the first field being the > > date > > Then f outside the last curly brace followed by the input > > > Sorry to be thick but this seems to say ' print f (a line of input) if > > the first field of f (guessing really) is the date d ' > > > Probably this is wrong coz I've haven't given an account of what the > > regex pattern is doing except note that it demands a number at the > > start of the line... > > Reformatted and added the default action... > > $0 ~ /^[[:digit:]]/ {f = ($1 == d)} > f { print $0 } > > First line: > whenever a line starts with a digit, > evaluate whether the first field equals the requested date, > if so then the flag is set to true (1) otherwise false (0). > Second line: > if flag is set print the line > > Actually Ed's code is a variant of the one I posted earlier this morning, > which I explain as well for comparison. The code was > > (!/^ / && f=($1~d)) || (f && /^ /) > > Reformatted, slightly rearranged, and default action added gives > > !/^ / && f=($1~d) { print $0 } > /^ / && f { print $0 } > > First line: > whenever a line starts not with white space, > and the flag is set, which depends on > the first field matches the date, > print the line > Second line: > whenever a line starts with white space, > and the flag is set, > print the line > > Janis > > > > > > > thanks > > > Tom > > > >> 2010/06/30 MacBook Pro Delivery, Original 2010/07/02 > >>>> S/N > >>>> etc... > >>>> 2010/07/06 Mail Rebate Form to MAC Apple,http://www.apple.com/hk/promo/rebate/bts.html > >> <snip> > >>> Try this (untested): > > >>> awk -v d="2010/07/06" '$0 ~ /^[[:digit:]]/ {f = ($1 == d)} f' file > > >> Is there a reason to prefer "$0 ~ /^[[:digit:]]/" over the simpler > >> "/^[[:digit:]]/" as the pattern? > > >> -- > >> Ben.- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - Thank a lot Following coding tested OK. #!/bin/ksh file=sch_mail.sch #awk -v d="2010/06/30" '$0 ~ /^[[:digit:]]/ {f = ($1 == d)} f' $file #awk -v d="2010/06/30" '(!/^ / && f=($1~d)) || (f && /^ /)' $file awk -v d="2010/06/30" ' !/^ / && f=($1~d) { print $0 } /^ / && f { print $0 } ' $file
From: thdyoung on 14 Jul 2010 14:38 Thank you Janis for a clear explanation. I get your first line. Second line: how is 'f' set ? Does it have a default when called ? Is ' f ' some kind of standard usage in awk ? Tom > Reformatted, slightly rearranged, and default action added gives > > !/^ / && f=($1~d) { print $0 } > /^ / && f { print $0 } > > First line: > whenever a line starts not with white space, > and the flag is set, which depends on > the first field matches the date, > print the line > Second line: > whenever a line starts with white space, > and the flag is set, > print the line > > Janis
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