Replace strings with variable in bash script - bash

I want to create a bash script to change the configuration file of an application. The script will read a variable and replace the value of a setting that I want to change with the variable value.
#!/bin/bash
mysqlurlenv1="jdbc:mysl://xx.xx.xx.xx:3306
mysqlurlenv2="jdbc:mysl://xx.xx.xx.xz:3306
configFile=app.conf
pattern=$1
case "$pattern" in
1)
sed -i.bak "s/^\(mysql.db.url =\).*/\1 ${mysqldburl_env1}/"
;;
2)
sed -i.bak "s/^\(mysql.db.url =\).*/\1 ${mysqldburl_env2}/"
;;
*)
echo "Wrong input...try again"
esac
I have a config file with the following entry
mysql.db.driver = com.mysql.jdbc.Driver
mysql.db.url = jdbc:mysql....
mysql.db.uname = admin
When I run the script I got this error
sed: -e expression #1, char 40: unknown option to `s'
How to get those defined variable in the config file ? Any help will appreciated
Rgds
Din

You are using this sed pattern s/FIND_STRING/REPLACE_STRING, that is the common way to do it.
But your REPLACE_STRING contains /, so your pattern will become s/FIND_STRING/jdbc:mysl://.... and that is invalid, because it contains too much /.
Sed accepts any character as field separator, so in your case you need change sed pattern field separator from / to something else, like |.
Example:
$ mysqlurlenv1="jdbc:mysl://xx.xx.xx.xx:3306"
$ echo "mysql.db.url = jdbc:mysql" | sed "s|^\(mysql.db.url =\).*|\1 ${mysqlurlenv1}|"
mysql.db.url = jdbc:mysl://xx.xx.xx.xx:3306

Related

String manipulation via script

I am trying to get a substring between &DEST= and the next & or a line break.
For example :
MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546
In this I need to extract "SFO"
MYREQUESTISTO8764GETTHIS&DEST=SANFRANSISCO&ORIG=6546
In this I need to extract "SANFRANSISCO"
MYREQUESTISTO8764GETTHISWITH&DEST=SANJOSE
In this I need to extract "SANJOSE"
I am reading a file line by line, and I need to update the text after &DEST= and put it back in the file. The modification of the text is to mask the dest value with X character.
So, SFO should be replaced with XXX.
SANJOSE should be replaced with XXXXXXX.
Output :
MYREQUESTISTO8764GETTHIS&DEST=XXX&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=XXXXXXXXXXXX&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=XXXXXXX
Please let me know how to achieve this in script (Preferably shell or bash script).
Thanks.
$ cat file
MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=PORTORICA
MYREQUESTISTO8764GETTHIS&DEST=SANFRANSISCO&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=SANJOSE
$ sed -E 's/^.*&DEST=([^&]*)[&]*.*$/\1/' file
SFO
PORTORICA
SANFRANSISCO
SANJOSE
should do it
Replacing airports with an equal number of Xs
Let's consider this test file:
$ cat file
MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=SANFRANSISCO&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=SANJOSE
To replace the strings after &DEST= with an equal length of X and using GNU sed:
$ sed -E ':a; s/(&DEST=X*)[^X&]/\1X/; ta' file
MYREQUESTISTO8764GETTHIS&DEST=XXX&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=XXXXXXXXXXXX&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=XXXXXXX
To replace the file in-place:
sed -i -E ':a; s/(&DEST=X*)[^X&]/\1X/; ta' file
The above was tested with GNU sed. For BSD (OSX) sed, try:
sed -Ee :a -e 's/(&DEST=X*)[^X&]/\1X/' -e ta file
Or, to change in-place with BSD(OSX) sed, try:
sed -i '' -Ee :a -e 's/(&DEST=X*)[^X&]/\1X/' -e ta file
If there is some reason why it is important to use the shell to read the file line-by-line:
while IFS= read -r line
do
echo "$line" | sed -Ee :a -e 's/(&DEST=X*)[^X&]/\1X/' -e ta
done <file
How it works
Let's consider this code:
search_str="&DEST="
newfile=chart.txt
sed -E ':a; s/('"$search_str"'X*)[^X&]/\1X/; ta' "$newfile"
-E
This tells sed to use Extended Regular Expressions (ERE). This has the advantage of requiring fewer backslashes to escape things.
:a
This creates a label a.
s/('"$search_str"'X*)[^X&]/\1X/
This looks for $search_str followed by any number of X followed by any character that is not X or &. Because of the parens, everything except that last character is saved into group 1. This string is replaced by group 1, denoted \1 and an X.
ta
In sed, t is a test command. If the substitution was made (meaning that some character needed to be replaced by X), then the test evaluates to true and, in that case, ta tells sed to jump to label a.
This test-and-jump causes the substitution to be repeated as many times as necessary.
Replacing multiple tags with one sed command
$ name='DEST|ORIG'; sed -E ':a; s/(&('"$name"')=X*)[^X&]/\1X/; ta' file
MYREQUESTISTO8764GETTHIS&DEST=XXX&ORIG=XXXX
MYREQUESTISTO8764GETTHIS&DEST=XXXXXXXXXXXX&ORIG=XXXX
MYREQUESTISTO8764GETTHISWITH&DEST=XXXXXXX
Answer for original question
Using shell
$ s='MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546'
$ s=${s#*&DEST=}
$ echo ${s%%&*}
SFO
How it works:
${s#*&DEST=} is prefix removal. This removes all text up to and including the first occurrence of &DEST=.
${s%%&*} is suffix removal_. It removes all text from the first & to the end of the string.
Using awk
$ echo 'MYREQUESTISTO8764GETTHIS&DEST=SFO&ORIG=6546' | awk -F'[=\n]' '$1=="DEST"{print $2}' RS='&'
SFO
How it works:
-F'[=\n]'
This tells awk to treat either an equal sign or a newline as the field separator
$1=="DEST"{print $2}
If the first field is DEST, then print the second field.
RS='&'
This sets the record separator to &.
With GNU bash:
while IFS= read -r line; do
[[ $line =~ (.*&DEST=)(.*)((&.*|$)) ]] && echo "${BASH_REMATCH[1]}fooooo${BASH_REMATCH[3]}"
done < file
Output:
MYREQUESTISTO8764GETTHIS&DEST=fooooo&ORIG=6546
MYREQUESTISTO8764GETTHIS&DEST=fooooo&ORIG=6546
MYREQUESTISTO8764GETTHISWITH&DEST=fooooo
Replace the characters between &DEST and & (or EOL) with x's:
awk -F'&DEST=' '{
printf("%s&DEST=", $1);
xlen=index($2,"&");
if ( xlen == 0) xlen=length($2)+1;
for (i=0;i<xlen;i++) printf("%s", "X");
endstr=substr($2,xlen);
printf("%s\n", endstr);
}' file

Unable to define line number in sed by variable

I'm trying to use an array to define the lines to replace using sed; I can delete the lines using a variable for the line number but I can't get sed to use the variable to define the line number to write to. The problem seems to reside in the insert line. How do you pass the value of an array as a line number to sed?
#!/bin/bash
lineNum=$(sed -n '/max_allowed_packet/=' /etc/mysql/my.cnf)
IFS= #There's a space as the delimiter#
ary=($lineNum)
#for key in "${!ary[#]}";
# do
# sed -i '$ary[$key]'d /etc/mysql/my.cnf;
# #The folllowing line errors#
# sed -i "'$ary[$key]'imax_allowed_packet = 32M" /etc/mysql/my.cnf;
# #The above line errors#
#done
#for val in "${ary[#]}";
# do
# sed -i "${val}d" /etc/mysql/my.cnf;
# sed -i "${val}imax_allowed_packet = 32M" /etc/mysql/my.cnf;
# done
for val in "${ary[#]}";
do
sed -i "${val}s/.*/imax_allowed_packet = 32M" /etc/mysql/my.cnf";
done
For the first stanza of script I get the following output:
Error: sed: -e expression #1, char 1: unknown command: `''
For the second Stanza I get the following output:
sed: -e expression #1, char 3: unknown command:
'
sed: -e expression #1, char 3: unknown command:
'
For the third Stanza I get the following output:
./test.sh: line 22: unexpected EOF while looking for matching `"'
./test.sh: line 24: syntax error: unexpected end of file
Edit, rewriting the sed commands as sed -i "${ary[$key]}" generates the following error output: sed: -e expression #1, char 3: unknown command: `
I think you're over-complicating the issue. Your script can be reduced to this:
sed 's/\(max_allowed_packet\).*/\1 = 32M/' /etc/mysql.cnf
This performs a substitution on every occurrence of max_allowed_packet, setting the rest of the line to = 32M. Add the -i switch to overwrite the file when you're happy with the result.
Problems with your attempt
Shell parameters are not expanded within single quotes, so you would need to use double quotes, e.g. sed -i "${ary[$key]}d". You can use set -x to see what is happening here - at the moment, you will see the literal string $ary[$key], rather than the array value.
If I understand your intention correctly (you want to substitute the entire line), there's no need to call sed twice:
for val in "${ary[#]}"; do
sed -i.bak "${val}s/.*/imax_allowed_packet = 32M" /etc/mysql/my.cnf
done
I have chosen to loop through the values of the array, instead of the keys, in order to simplify things a little. When using the -i option, it is always a good idea to specify a backup file, as I have done.

ssh sed not changing variables correctly

I'm trying to use sed to change a variable in the site.js file on my server.
Here is the line: var url = "page.php"; I'm looking to just substitute page.php for whatever.php.
I thought this would be pretty simple and I figured this would work with no issues:
sed -i "s/\url = \".*\"/\url = \"page2.php\"/" /home/site.js
It works okay except instead of getting: var url = "page2.php"; I get: var R1 = "page2.php";
Why is the url value being changed to R1 when I use sed here?
You don't need \ before url.
sed -i -r 's#url\s*=\s*"[^"]+"#url = "page2.php"#' /home/site.js
Extra escaping of " can be eliminated by enclosing sed expression with ' instead of "
It's better to use different separator than / (here #) when the strings themselves may contain /
Try doing this :
sed -i -r 's#(var\s+url\s*=\s*")[^"]+"#\1whatever.php"#' file.js
/ is not mandatory as delimiter, I've picked up # there.
Here's another example: Took me while to figure that you change the / for delimiter and not the / in the directory path.
Use # instead of / for sed delimiter if you have dir path names.
First I tried this:
[root#ip-172-35-24-37 ec2-user]# egrep -q "^(\s*\S+\s+)/dev/shm(\s+\S+\s+\S+)(\s+\S+\s+\S+)(\s*#.*)?\s*$" /etc/fstab && sed -ri "s/^(\s*\S+\s+)/dev/shm(\s+\S+\s+\S+)(\s+\S+\s+\S+)(\s*#.*)?\s*$/\1/dev/shm\2nodev\3\4/" /etc/fstab
And got this error:
sed: -e expression #1, char 20: unknown option to `s'
So then I used # for the sed delimiter instead of /:
[root#ip-172-35-24-37 ec2-user]# egrep -q "^(\s*\S+\s+)/dev/shm(\s+\S+\s+\S+)(\s+\S+\s+\S+)(\s*#.*)?\s*$" /etc/fstab && sed -ri "s#^(\s*\S+\s+)/dev/shm(\s+\S+\s+\S+)(\s+\S+\s+\S+)(\s*#.*)?\s*$#\1/dev/shm\2nodev\3\4##" /etc/fstab
[root#ip-172-35-24-37 ec2-user]#
And it worked.
You can use something else besides # for a delimiter like ! or ? or %. Just don't use / if you have dir paths.

How do I insert a newline/linebreak after a line using sed

It took me a while to figure out how to do this, so posting in case anyone else is looking for the same.
For adding a newline after a pattern, you can also say:
sed '/pattern/{G;}' filename
Quoting GNU sed manual:
G
Append a newline to the contents of the pattern space, and then append the contents of the hold space to that of the pattern space.
EDIT:
Incidentally, this happens to be covered in sed one liners:
# insert a blank line below every line which matches "regex"
sed '/regex/G'
This sed command:
sed -i '' '/pid = run/ a\
\
' file.txt
Finds the line with: pid = run
file.txt before
; Note: the default prefix is /usr/local/var
; Default Value: none
;pid = run/php-fpm.pid
; Error log file
and adds a linebreak after that line inside file.txt
file.txt after
; Note: the default prefix is /usr/local/var
; Default Value: none
;pid = run/php-fpm.pid
; Error log file
Or if you want to add text and a linebreak:
sed -i '/pid = run/ a\
new line of text\
' file.txt
file.txt after
; Note: the default prefix is /usr/local/var
; Default Value: none
;pid = run/php-fpm.pid
new line of text
; Error log file
A simple substitution works well:
sed 's/pattern.*$/&\n/'
Example :
$ printf "Hi\nBye\n" | sed 's/H.*$/&\nJohn/'
Hi
John
Bye
To be standard compliant, replace \n by backslash newline :
$ printf "Hi\nBye\n" | sed 's/H.*$/&\
> John/'
Hi
John
Bye
sed '/pattern/a\\r' file name
It will add a return after the pattern while g will replace the pattern with a blank line.
If a new line (blank) has to be added at end of the file use this:
sed '$a\\r' file name
Another possibility, e.g. if You don't have an empty hold register, could be:
sed '/pattern/{p;s/.*//}' file
Explanation:
/pattern/{...} = apply sequence of commands, if line with pattern found,
p = print the current line,
; = separator between commands,
s/.*// = replace anything with nothing in the pattern register,
then automatically print the empty pattern register as additional line)
The easiest option -->
sed 'i\
' filename

replace and delete characters in a shell script

I have a shell script to automate builds of my programm. I need to transform versioning numbers like V4_5_1-RC1 to 4.5.1-RC1. The V should be removed and the _ should be replaced with .. I tried several things, for example with sed:
$NAMEEXT = "V4_5_1-RC1"
$lffNameRSC = ${sed -e "s/V//g" <<< $NAMEEXT}
$lffNameRSC = ${sed -e "s/_/./g" <<< $lffNameRSC}
echo $lffNameRSC
but I'm getting errors.
./makerelease.sh: line 113: ${sed -e "s/V//g" <<< $NAMEEXT}: bad substitution
./makerelease.sh: line 114: ${sed -e "s/_/./g" <<< $lffNameRSC}: bad substitution
there should be no spaces around =
there should be $(..) instead of ${..} to evaluate the command
there should not be $ in variable assignment statement
With
#!/bin/bash
NAMEEXT="V4_5_1-RC1"
lffNameRSC=$(sed -e "s/V//g" <<< $NAMEEXT)
lffNameRSC=$(sed -e "s/_/./g" <<< $lffNameRSC)
echo $lffNameRSC
you will get
4.5.1-RC1
And, by the way, it could be done easier, like
$> echo "V4_5_1-RC1" | sed "s/V//g; s/_/./g"
4.5.1-RC1
Bash parameter expansion can do what you want without any external tools:
NAMEEXT="V4_5_1-RC1"
version=${NAMEEXT#V} # remove the leading V
version=${version//_/.} # replace all _ with .
echo $version # ==> 4.5.1-RC1

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