Can you help me with this script ?
#!/bin/bash
data=`date '+%Y %m %d'`
data2=" 7 0 0"
string=$data$data2
awk -v str=$string '{print mktime("str");}'
I´m getting this output:
awk: cmd. line:1: fatal: cannot open file `09' for reading (No such file or directory)
You should not quote str inside awk otherwise it is treated as a literal string. You must also quote the shell variables properly.
#!/bin/bash
data=$(date '+%Y %m %d')
data2=" 7 0 0"
string="$data$data2"
awk -v str="$string" 'BEGIN {print mktime(str)}'
#anubhava has the right answer. Here's why: after variable substitution, the shell sees this line in your code:
awk -v str=2014 01 09 7 0 0 '{print mktime("str");}'
So, you're passing "2014" as the "str" variable, your awk script is "01" (which is a valid awk program that will print every line), and passing the file arguments "09", "7", ... Hence the error "cannot open file `09' for reading"
Lesson: in shell scripting, unless you know when not to, always quote your "$variables".
You want this:
#!/bin/bash
data="`date '+%Y %m %d'`"
data2=" 7 0 0"
string="$data$data2"
awk -v str="$string" 'BEGIN {print mktime(str);exit;}'
Another alternative for the awk line is: (This has the advantage of working on older awk versions that lack the -v option, like Solaris 10 /usr/bin/awk (They tend to miss mktime though))
awk "BEGIN {print mktime(\"$string\");exit;}"
The BEGIN block is needed to make awk run the code at startup and not for every input line.
The exit is needed to make it quit before it waits for input. (It seems to work without it on GAWK)
The quotes are needed to pass the variable as a single entry... This overview of shell quoting is important reading for anyone attempting shell scripting...
The quotes in the awk code needs to be left out to get awk to use the variable str and not the string str.
You likely also want to replace the backticks (``) with $(), it is much easier to read in complicated, nested cases and supported in all modern shells (It seems like it might be in the POSIX standard, I haven't checked though...).. (It doesn't affect the code in this case though)
Related
Is it possible to store an awk script inside a shell variable; for example:
export script="'{printf(\$2); printf("\"\\n\"");}'"
echo $script
'{printf($2); printf("\n");}'
The script functions properly when I call it directly as such:
awk '{printf($2); printf("\n");}' testFile.txt
prints proper output
When I try and pass the script as a shell variable, I run into issues.
awk $script testFile.txt
awk: syntax error at source line 1
context is
>>> ' <<<
missing }
awk: bailing out at source line 1
I get a slightly different error when I wrap the variable in double quotes
awk "$script" testFile.txt
awk: syntax error at source line 1
context is
>>> ' <<<
awk: bailing out at source line 1
I'm still learning exactly how shell expansions work, I would appreciate any suggestions about what I am missing here.
Error in your quoting
export script='{printf($2); printf("\n");}'
awk "${script}" YourFile
I am not sure about the proper answer to this, but a very ugly (and probably unstable depending on the $script contents) workaround would be:
echo $script | awk '{print "awk "$0" testFile.txt"}' | bash
This is just printing the contents of $script in an awk statement that is then executed by bash. I am not particularly proud of this, but maybe it helps!
When you type
awk '{printf($2); printf("\n");}' testFile.txt
awk only sees {printf($2); printf("\n");} -- the shell removes the quotes
(see Quote Removal in the bash manual)
Heed #NeronLeVelu's answer.
I have written the following script that extracts a number from an rss file.
#!/bin/sh
wget -O selic https://conteudo.bcb.gov.br/api/feed/pt-br/PAINEL_INDICADORES/juros
line=$(grep 'dailyratevalue' selic)
index=$(awk -v var=$line 'BEGIN {print index(var, "dailyratevalue") }')
end=$((index+21))
echo $line | cut -c $index-$end | tail -c 4 | tr ',' '.' > selic
In zsh it works perfectly, but i need it to work in bash, too. I have tried running it on bash but i get the following error
awk: cmd. line:1: <content
awk: cmd. line:1: ^ syntax error
The error pattern <content comes from the line that is being fed as a parameter to awk, which makes no sense to me, since awk is just supposed to get me the position of the pattern i want.
What could this be?
index=$(awk -v var="$line" 'BEGIN {print index(var, "dailyratevalue") }')
should fix it.
awk can do all of the extra steps. You can just
wget -qO - https://conteudo.bcb.gov.br/api/feed/pt-br/PAINEL_INDICADORES/juros | \
awk -F '&[gl]t;' '/dailyratevalue/ {sub(",", ".", $25); print $25;}'
and obtain the value you want.
This is setting the FS and getting the field you want for the line that matches dailyratevalue.
#DiegoTorresMilano's answer is probably better overall, but if you want to do it in bash, the main thing you need to do is double-quote your variable references. Without double-quotes around them, bash (and most shells other than zsh) splits variables into "words", and also expands anything that looks like a wildcard expression into a list of matching filenames. You almost never want this, so use double-quotes. In your case, there are two places they're needed: around $line here:
index=$(awk -v var="$line" 'BEGIN {print index(var, "dailyratevalue") }')
and here:
echo "$line" | cut -c $index-$end | tail -c 4 | tr ',' '.' > selic
Note that you don't need double-quotes around the $( ) expressions, because they're on the right side of an assignment statement, and that isn't subject to word splitting and wildcard expansion. If they occurred elsewhere, you'd probably want double-quotes around them too.
BTW, shellcheck.net is really good at pointing out common mistakes like this, so I recommend running your scripts through it (even when they seem to be working correctly).
Does awk support using a different delimiter character than ' in print? e.g. Instead of awk '{print $1}', something like awk -d # #{print $1}#
I was actually looking at the C source code and it's a pretty short program; is there an alternative version that allow that?
It can't: The ' isn't passed to awk; instead, it's understood by the shell itself. Thus, when you run awk '{print $1}', what you're actually calling at the OS level is something like:
/* this is C syntax, so the double-quotes are C quotes; only their contents are literal */
execvp("awk", { "awk", "{print $1}", NUL });
Notably, the single-quotes aren't there at all any more -- they were parsed out by the shell when it understood them as instructions for how it should break the command into an argument list.
To see it in another way, consider if you put your script in a separate file and called
awk -f my.awk
The contents of my.awk would simply be
{print $1}
not
'{print $1}'
The single quotes are only used by the shell to ensure that the script is passed literally to awk, rather than being subject to any particular shell processing that could change the script before awk could read it.
I'm on a Mac, and I want to find a field in a CSV file adjacent to a search string
This is going to be a single file with a hard path; here's a sample of it:
84:a5:7e:6c:a6:b0, AP-ATC-151g84
84:a5:7e:6c:a6:b1, AP-A88-131g84
84:a5:7e:73:10:32, AP-AG7-133g56
84:a5:7e:73:10:30, AP-ADC-152g81
84:a5:7e:73:10:31, AP-D78-152e80
so if my search string is "84:a5:7e:73:10:32"
I want to get returned "AP-AG7-133g56"
I had been working within an Applescript, but maybe a shell script will do.
I just need the proper syntax for opening the file and having awk search it. Again, I'm weak conceptually on how shell commands run, how they must be executed, etc
This errors, gives me ("command not found"):
set the_file to "/Users/Paw/Desktop/AP-Decoder 3.app/Contents/Resources/BSSIDtable.csv"
set the_val to "70:56:81:cb:a2:dc"
do shell script "'awk $1 ~ the_val {print $2} the_file'"
Thank you for coddling me...
This is a relatively simple:
awk '$1 == "70:56:81:cb:a2:dc," {print "The answer is "$2}' 'BSSIDtable.csv'
(the "The answer is " text can be omitted if you only wish to see only the data, but this shows you how to get more user-friendly output if desired).
The comma is included since awk uses white space for separators so the comma becomes part of column 1.
If the thing you're looking for is in a shell variable, you can use -v to provide that to awk as an awk variable:
lookfor="70:56:81:cb:a2:dc,"
awk -v mac=$lookfor '$1 == mac {print "The answer is "$2}' 'BSSIDtable.csv'
As an aside, your AppleScript solution is probably not working because the $1/$2 are being interpreted as shell variable rather than awk variables. If you insist on using AppleScript, you will have to figure out how to construct a shell command that quotes the awk commands correctly.
My advice is to just use the shell directly, the number of people proficient in that almost certainly far outnumber those proficient in AppleScript :-)
if sed is available (normaly on mac, event if not tagged in OP)
simple but read all the file
sed -n 's/84:a5:7e:73:10:32,[[:blank:]]*//p' YourFile
quit after first occurence (so average of 50% faster on huge file)
sed -n -e '/84:a5:7e:73:10:32,[[:blank:]]*/!b' -e 's///p;q' YourFile
awk
awk '/^84:a5:7e:73:10:32/ {print $2}'
# OR using a variable for batch interaction
awk -v Src='84:a5:7e:73:10:32' '$1 == Src {print $2}'
# OR assuming that case is unknow
awk -v Src='84:a5:7e:73:10:32' 'BEGIN{IGNORECASE=1} $1 == Src {print $2}'
by default it take $0 as compare test if a regex is present, just add the ^ to take first field content
I have a shell script that constructs an awk program as a string then pass that string to awk. This is because I want to use values of shell variables in the awk program.
My code looks like this:
awk_prog="'{if (\$4~/$shell_var/) print \$1,\$2}'"
echo $awk_prog
awk $awk_prog $FILENAME
However, when I pass the string to awk, I always get the error:
'{if ($4~/regex/) print $1,$2}'
awk: '{if
awk: ^ invalid char ''' in expression
What does that error message mean? I tried the -F: switch but it does not help. How can I settle this issue?
Thank you.
This is caused by shell quoting. The following will work:
awk_prog="{ if (\$4 ~ /$shell_var/) print \$1, \$2 }"
echo "$awk_prog"
awk "$awk_prog" $FILENAME
When you run awk '{ print }' foo from the command line, the shell interprets and removes the quotes around the program so awk receives two arguments - the first is the program text and the second is the filename foo. Your example was sending awk the program text '{if ...}' which is invalid syntax as far as awk is concerned. The outer quotes should not be present.
In the snippet that I gave above, the shell uses the quotes in the awk_prog= line to group the contents of the string into a single value and then assigns it to the variable awk_prog. When it executes the awk "$awk_prog"... line, you have to quote the expansion of $awk_prog so awk receives the program text as a single argument.
There's another way to get your shell variable into awk -- use awk's -v option:
awk -v pattern="$shell_var" '$4 ~ pattern {print $1, $2}' "$FILENAME"
Use -v multiple times if you have several variables to pass to awk.
If you truly want to hold your awk program in a shell variable, build it up using printf:
awk_script="$( printf '$4 ~ /%s/ {print $1, $2}' "$shell_var" )"
awk "$awk_script" "$FILENAME"
Note the use of quotes in the printf command: single quotes around the template to protect the dollar signs you want awk to interpret, double quotes for shell variables.
Another (IMO simpler) solution which (I think) addresses what you are intuitively trying to do is simply to use eval. You want the shell to behave as if you had literally typed:
awk '{if ($4~/foo/) print $1,$2}' path
(where foo and path are the literal contents of $shell_var and $FILENAME). To make that happen, just slap an eval on the front of your last line (and perhaps quotes for good measure, but they aren't necessary in this case) so that your last line is:
eval "awk $awk_prog $FILENAME"