I tried this but it does not seem to work.
Please help thanks
TEST_STRING= test
echo Starting awk command
awk -v testString=$TEST_STRING'
BEGIN {
}
{
print testString
}
END {}
' file2
There are two problems here: You aren't actually assigning to TEST_STRING, and you're passing the program code in the same argument as the variable value. Both of these are caused by whitespace and quoting being in the wrong places.
TEST_STRING= test
...does not assign a value to TEST_STRING. Instead, it runs the command test, with an environment variable named TEST_STRING set to an empty value.
Perhaps instead you want
TEST_STRING=test
or
TEST_STRING=' test'
...if the whitespace is intentional.
Second, passing a variable to awk with -v, the right-hand side should be double-quoted, and there must be unquoted whitespace between that value and the program to be passed to awk (or other values). That is to say:
awk -v testString=$TEST_STRING' BEGIN
...will, if TEST_STRING contains no whitespace, pass the BEGIN as part of the value of testString, not as a separate argument!
awk -v testString="$TEST_STRING" 'BEGIN
...on, the other hand, ensures that the value of TEST_STRING is passed as part of the same argument as testString=, even if it contains whitespace -- and ensures that the BEGIN is passed as part of a separate argument.
Related
If I use the zsh shell and execute the following command I get
$zsh
$echo '$_GET["test"]'
preexec: bad math expression: operand expected at `"test"'
$echo '$_GET[]'
preexec: invalid subscript
In bash I get what I expect:
$bash
$echo '$_GET["test"]'
$_GET["test"]
I assume that zsh is trying to expand the $_GET variable. How can I avoid this? I always expected this to only happen within double quotes anyhow.
[update]
I found the following three lines in the .zshrc:
# Display last command interminal
echo -en "\e]2;Parrot Terminal\a"
preexec () { print -Pn "\e]0;$1 - Parrot Terminal\a" }
After commenting them out everything seems to work as expected.
What I understand is that preexec is executed after a command in the terminal has been submitted but before it is executed. The $1 is the command that one submitted.
I still do not understand the purpose of the two lines but is it because of the double quotes in the preexec print statement that the variables are expanded?
The combination of print -P together with the expansion of $1 is killing you. With this, you first get a "normal" expansion of $1, yielding something like "\e]0;echo '$_GET["test"]'...". Now -P causes print to do a prompt expansion on this string, which means that it has to expand $_GET["test"] as well. This causes the error.
I suggest to remove the -P, in particular since you don't have any characters in your string which would benefit from prompt expansion.
I am attempting to use the sqrt function from awk command in my script, but all it returns is 0. Is there anything wrong with my script below?
echo "enter number"
read root
awk 'BEGIN{ print sqrt($root) }'
This is my first time using the awk command, are there any mistakes that I am not understanding here?
Maybe you can try this.
echo "enter number"
read root
echo "$root" | awk '{print sqrt($0)}'
You have to give a data input to awk. So, you can pipe 'echo'.
The BEGIN statement is to do things, like print a header...etc before
awk starts reading the input.
$ echo "enter number"
enter number
$ read root
3
$ awk -v root="$root" 'BEGIN{ print sqrt(root) }'
1.73205
See the comp.unix.shell FAQ for the 2 correct ways to pass the value of a shell variable to an awk script.
UPDATE : My proposed solution turns out to be potentially dangerous. See Ed Morton's answer for a better solution. I'll leave this answer here as a warning.
Because of the single quotes, $root is interpreted by awk, not by the shell. awk treats root as an uninitialized variable, whose value is the empty string, treated as 0 in a numeric context. $root is the root'th field of the current line -- in this case, as $0, which is the entire line. Since it's in a BEGIN block, there is no current line, so $root is the empty string -- which again is treated as 0 when passed to sqrt().
You can see this by changing your command line a bit:
$ awk 'BEGIN { print sqrt("") }'
0
$ echo 2 | awk '{ print sqrt($root) }'
1.41421
NOTE: The above is merely to show what's wrong with the original command, and how it's interpreted by the shell and by awk.
One solution is to use double quotes rather than single quotes. The shell expands variable references within double quotes:
$ echo "enter number"
enter number
$ read x
2
$ awk "BEGIN { print sqrt($x) }" # DANGEROUS
1.41421
You'll need to be careful when doing this kind of thing. The interaction between quoting and variable expansion in the shell vs. awk can be complicated.
UPDATE: In fact, you need to be extremely careful. As Ed Morton points out in a comment, this method can result in arbitrary code execution given a malicious value for $x, which is always a risk for a value read from user input. His answer avoids that problem.
(Note that I've changed the name of your shell variable from $root to $x, since it's the number whose square root you want, not the root itself.)
My team gets Teradata DDL files generated through a front end tool. These files need to be corrected before executing.
A step in this is getting the DDL command on a single line
E.g.
create table ABC
(column A varchar2(100),
column B number(10)
);
replace view ABC_v as
select columnA, column B from
ABC;
should change to
create table ABC (column A varchar2(100),column B number(10));
replace view ABC_v as select columnA, column B from ABC;
In short, I am looking to replace every new line character with single space in a multi-line string.
The string can start with either create, replace or drop and it will always end with a ; (semicolon)
Thanks in advance for your help
Here's a simple solution in shell:
#!/bin/sh
while read first rest; do
case "$first" in
create|replace|drop) echo "" ;;
esac
printf "%s %s " "$first" "$rest"
done < inputfile
echo ""
This adds a blank line to the beginning of the output because I'm lazy. But you see the logic, I'm sure. To avoid the blank line, you can use a temporary variable to determine whether you've actually started pulling in input data yet.
You could do something sort-of similar using awk:
awk '
BEGIN {
a["create"];
a["replace"];
a["drop"];
}
$1 in a && h {
print substr(h,2);h="";
}
{
h=h" "$0;
}
END {
print substr(h,2);
}
' inputfile
Instead of simply prepending a newline before keywords, this solution builds lines of output in variables, then prints them when they're complete.
Alternately, you could use sed to implement the same idea:
sed -rne '/^(create|replace|drop) /{;x;s/\n/ /g;/./p;d;};H;${;x;s/\n/ /g;p;}' inputfile
In all three of these solutions, I haven't bothered to check whether the input string ends in a semicolon. You can add that check to each of them once you decide how you want to handle that failure. (Report an error? Send the command via email? Ignore it?)
Note also that DDL, like SQL, should be able to interpret commands provided on multiple lines. SQL is whitespace agnostic -- an unquoted newline should be the same as a space (though perhaps Teradata behaves differently).
i have once question, suppose i am using "=" as fiels seperator, in this case if my string contain for example
abc=def\=jkl
so if i use = as fields seperator, it will split into 3 as
abc def\ jkl
but as i have escaped 2nd "=" , my output should be as
abc def\=jkl
Can anyone please provide me any suggestion , if i can achieve this.
Thanks in advance
I find it simplest to just convert the offending string to some other string or character that doesn't appear in your input records (I tend to use RS if it's not a regexp* since that cannot appear within a record, or the awk builtin SUBSEP otherwise since if that appears in your input you have other problems) and then process as normal other than converting back within each field when necessary, e.g.:
$ cat file
abc=def\=jkl
$ awk -F= '{
gsub(/\\=/,RS)
for (i=1; i<=NF; i++) {
gsub(RS,"\\=",$i)
print i":"$i
}
}' file
1:abc
2:def\=jkl
* The issue with using RS if it is an RE (i.e. multiple characters) is that the gsub(RS...) within the loop could match a string that didn't get resolved to a record separator initially, e.g.
$ echo "aa" | gawk -v RS='a$' '{gsub(RS,"foo",$1); print "$1=<"$1">"}'
$1=<afoo>
When the RS is a single character, e.g. the default newline, that cannot happen so it's safe to use.
If it is like the example in your question, it could be done.
awk doesn't support look-around regex. So it would be a bit difficult to get what you want by setting FS.
If I were you, I would do some preprocessing, to make the data easier to be handled by awk. Or you could read the line, and using other functions by awk, e.g. gensub() to remove those = s you don't want to have in result, and split... But I guess you want to achieve the goal by playing field separator, so I just don't give those solutions.
However it could be done by FPAT variable.
awk -vFPAT='\\w*(\\\\=)?\\w*' '...' file
this will work for your example. I am not sure if it will work for your real data.
let's make an example, to split this string: "abc=def\=jkl=foo\=bar=baz"
kent$ echo "abc=def\=jkl=foo\=bar=baz"|awk -vFPAT='\\w*(\\\\=)?\\w*' '{for(i=1;i<=NF;i++)print $i}'
abc
def\=jkl
foo\=bar
baz
I think you want that result, don't you?
my awk version:
kent$ awk --version|head -1
GNU Awk 4.0.2
I want to check if a multiline text matches an input. grep comes close, but I couldn't find a way to make it interpret pattern as plain text, not regex.
How can I do this, using only Unix utilities?
Use grep -F:
-F, --fixed-strings
Interpret PATTERN as a list of fixed strings, separated by newlines, any of which is to be matched. (-F is specified by
POSIX.)
EDIT: Initially I didn't understand the question well enough. If the pattern itself contains newlines, use -z option:
-z, --null-data
Treat the input as a set of lines, each terminated by a zero
byte (the ASCII NUL character) instead of a newline. Like the
-Z or --null option, this option can be used with commands like
sort -z to process arbitrary file names.
I've tested it, multiline patterns worked.
From man grep
-F, --fixed-strings
Interpret PATTERN as a list of fixed strings, separated by
newlines, any of which is to be matched. (-F is specified by
POSIX.)
If the input string you are trying to match does not contain a blank line (eg, it does not have two consecutive newlines), you can do:
awk 'index( $0, "needle\nwith no consecutive newlines" ) { m=1 }
END{ exit !m }' RS= input-file && echo matched
If you need to find a string with consecutive newlines, set RS to some string that is not in the file. (Note that the results of awk are unspecified if you set RS to more than one character, but most awk will allow it to be a string.) If you are willing to make the sought string a regex, and if your awk supports setting RS to more than one character, you could do:
awk 'END{ exit NR == 1 }' RS='sought regex' input-file && echo matched