How can I get directory for specified DN by one ldapsearch request?
I mean - I have few databases. OpenLDAP configured with cn=config. For each DN - it hve own ldif-file, where it's olcDbDirectory specified.
Can I obtain olcDbDirectory value for each DN?
For backup script - I need to set varibale which contains directory, and this variable changes every time for every DN, wich backuped/restored at this moment.
So - in bash I just found solution to create function like:
#!/bin/bash
getDir () {
file=`grep -R "$1" /etc/openldap/slapd.d/ | cut -d":" -f 1 | tail -n 1`
echo $file
dir=`cat $file | grep "olcDbDirectory" | awk '{print $2}'`
echo $dir
}
getDir testdb;
$ ./dn.sh
/etc/openldap/slapd.d/cn=config/olcDatabase={9}bdb.ldif
/var/lib/ldap/testdb
But this solution seems not tidy... And I'd preffered to use something like:
getDir () {
dir=`ldapsearch -x -D "cn-root,cn=config" "*somefilter*"
}
Here it is:
$ ldapsearch -x -LLL -D 'cn=root,cn=config' -w PassWord -b 'cn=config' '(&(olcDbDirectory=*)(olcSuffix='testdb'))' olcDbDirectory | grep "olcDbDirectory" | cut -d":" -f 2
/var/lib/ldap/testdb
Of in bash function:
#!/bin/bash
getDir () {
dirtodel=`ldapsearch -x -LLL -D 'cn=root,cn=config' -w PassWord -b 'cn=config' '(&(olcDbDirectory=*)(olcSuffix='${1}'))' olcDbDirectory | grep "olcDbDirectory" | cut -d":" -f 2`
echo $dirtodel
}
getDir 'dc=testdb'
Result:
$ ./dn.sh
/var/lib/ldap/testdb
Related
Objective:
I'm trying to write a script that will fetch two URLs from a GitHub release page and do something different with each one.
So far:
Here's what I've got so far.
λ curl -s https://api.github.com/repos/mozilla-iot/gateway/releases/latest | grep "browser_download_url.*tar.gz" | cut -d : -f 2,3 | tr -d \"
This will return the following:
"https://github.com/mozilla-iot/gateway/releases/download/0.8.1/gateway-8c29257704ddb021344bdaaa790909a0eacf3293bab94e02859828a6fd9b900a.tar.gz"
"https://github.com/mozilla-iot/gateway/releases/download/0.8.1/node_modules-921bd0d58022aac43f442647324b8b58ec5fdb4df57a760e1fc81a71627f526e.tar.gz"
I want to be able to create some directories, pull in the first one, navigate in the directories from the newly pulled zip after extracting it, and then pull in the second.
fetching the first line is easy by piping the output to head -n1. for solving your problem, you need more than just fetching the first URL of the cURL output. give this a try:
#!/bin/bash
# fetch your URLs
answer=`curl -s https://api.github.com/repos/mozilla-iot/gateway/releases/latest | grep "browser_download_url.*tar.gz" | cut -d : -f 2,3 | tr -d \"`
# get URLs and file names
first_file=`echo "$answer" | grep -Eo '.+?\.tar\.gz' | head -n1 | tr -d " "`
second_file=`echo "$answer" | grep -Eo '.+?\.tar\.gz' | head -n2 | tail -1 | tr -d " "`
first_file_name=`echo "$answer" | grep -Eo '[^/]+?\.tar\.gz' | head -n1 `
second_file_name=`echo "$answer" | grep -Eo '[^/]+?\.tar\.gz' | head -n2 | tail -1`
#echo $first_file
#echo $first_file_name
#echo $second_file_name
#echo $second_file
# download first file
wget "$first_file"
# extracting first one that must be in the current directory.
# else, change the directory first and put the path before $first_file!
tar -xzf "$first_file_name"
# do your stuff with the second file
You can simply pipe the URLs to xargs curl;
curl -s https://api.github.com/repos/mozilla-iot/gateway/releases/latest |
grep "browser_download_url.*tar.gz" |
cut -d : -f 2,3 | tr -d \" |
xargs curl -O
Or if you want to do some more manipulation on each URL, perhaps loop over the results:
curl ... | grep ... | cut ... | tr ... |
while IFS= read -r url; do
curl -O "$url"
: maybe do things with "$url" here
done
The latter could easily be extended to someting like
... | while IFS= read -r url; do
d=${url##*/}
mkdir -p "$d"
( cd "$d"
curl -O "$url"
tar zxf *.tar.gz
# end of subshell means effects of "cd" end
)
done
I am helping debug some code that exec the following script is there any reason why its not writing a file to the server? - if that what it does. All the $ data and permissions are ok:
Script:
#!/bin/bash
RANGE=$1
ALLOCATION=`echo $RANGE | cut -f1,2,3 -d'.'`
/sbin/ip rule add from $1 lookup $2
echo $ALLOCATION
rm /path/too/file/location/$ALLOCATION
for i in `seq 3 254`
do
echo $ALLOCATION.$i >> /path/too/file/location/$ALLOCATION
done
ETH=`/sbin/ifconfig | grep eth0 | tail -n1 | cut -f2 -d':' | cut -f1 -d' '`
I am trying to build a Bash CGI Script that takes in coordinates as parameters from the url and uses osmosis to extract the map and the splitter and mkgmap to make the map so that it can be opened with Qlandkarte. My problem being is that when i type wget localhost/cgi-bin/script.pl?top=42&left=10&bottom=39&right=9&file=map.osm the linux terminal reads the file with the coordinates. How can I make wget just activate the script so it takes the coordinates and executes the commands. And also when the map is created at the end how can a return the file that was created by the script.
Thanks
#!/bin/bash
TOP=`echo "$QUERY_STRING" | grep -oE "(^|[?&])top=[0.0-9.0]+" | cut -f 2 -d "=" | head -n1`
LEFT=`echo "$QUERY_STRING" | grep -oE "(^|[?&])left=[0.0-9.0]+" | cut -f 2 -d "=" | head -n1`
BOTTOM=`echo "$QUERY_STRING" | grep -oE "(^|[?&])bottom=[0.0-9.0]+" | cut -f 2 -d "=" | head -n1`
RIGHT=`echo "$QUERY_STRING" | grep -oE "(^|[?&])right=[0.0-9.0]+" | cut -f 2 -d "=" | head -n1`
FILE=`echo "$QUERY_STRING" | grep -oE "(^|[?&])file=[^&]+" | sed "s/%20/ /g" | cut -f 2 -d "="`
$(sudo osmosis --read-xml file=bulgaria.osm --bounding-box top=$TOP left=$LEFT bottom=$BOTTOM right=$RIGHT --write-xml file=$FILE)
$(sudo java -Xmx900m -jar splitter.jar --max-nodes=110000 $FILE)
$(sudo java -ea -Xmx900m -jar mkgmap.jar --tdbfile --route -c template.args)
echo "Content-type: text/html"
echo ""
I am trying a simple shell script like the following:
#!/bin/bash
up_cap=$( cat result.txt | cut -d ":" -f 6,7 | sort -n | cut -d " " -f 2 | sort -n)
down_cap=$( cat result.txt | cut -d : -f 6,7 | sort -n | cut -d " " -f 6| sort -n)
for value in "${down_cap[#]}";do
if [ $value > 80000 ]; then
cat result.txt | grep -B 1 "$value"
fi
done
echo " All done, exiting"
when I execute the above script as ./script.sh, I get the error:
./script.sh: line 5: [: too many arguments
All done, exiting
I have googled enough, and still not able to rectify this.
You want
if [ "$value" -gt 80000 ]; then
You use -gt for checking if A is bigger than B, not >. The quotation marks I merely added to prevent the script from failing in case $value is empty.
Try to declare variable $value explicitly:
declare -i value
So, with the dominikh's and mine additions the code should look like this:
#!/bin/bash
up_cap=$( cat result.txt | cut -d ":" -f 6,7 | sort -n | cut -d " " -f 2 | sort -n)
down_cap=$( cat result.txt | cut -d : -f 6,7 | sort -n | cut -d " " -f 6| sort -n)
for value in "${down_cap[#]}";do
declare -i value
if [ $value -gt 80000 ]; then
cat result.txt | grep -B 1 "$value"
fi
done
echo " All done, exiting"
I'm busy writing a basic migration scripts for some WP sites, and i'm trying to automate creating mysql database and user credentials from reading the wp-config file
how can i read the following variables from the wp-config file so that i effectively end up with 3 variables within a bash script:
/** The name of the database for WordPress */
define('DB_NAME', 'somedbname');
/** MySQL database username */
define('DB_USER', 'someusername');
/** MySQL database password */
define('DB_PASSWORD', 'somerandompassword');
eg my output should effectively give me:
WPDBNAME=somedbname
WPDBUSER=somedbuser
WPDBPASS=somerandompassword
Try this:
WPDBNAME=`cat wp-config.php | grep DB_NAME | cut -d \' -f 4`
WPDBUSER=`cat wp-config.php | grep DB_USER | cut -d \' -f 4`
WPDBPASS=`cat wp-config.php | grep DB_PASSWORD | cut -d \' -f 4`
If you want to use wp-config details to connect to mysql you can use;
mysql -u `cat wp-config.php | grep DB_USER | cut -d \' -f 4` -p`cat wp-config.php | grep DB_PASSWORD | cut -d \' -f 4` -h `cat wp-config.php | grep DB_HOST | cut -d \' -f 4` `cat wp-config.php | grep DB_NAME | cut -d \' -f 4`
you can use awk:
awk -F"[()']" '/^define/{printf "%s=\"%s\"\n", $3, $5;}' < foo.php
This will give you:
DB_NAME="somedbname"
DB_USER="someusername"
DB_PASSWORD="somerandompassword"
Note, that this solution will work with variables containing spaces.
find . -name "wp-config.php" -print0 | xargs -0 -r grep -e "DB_NAME" -e "DB_USER" -e "DB_PASSWORD"
If you are trying to dump the MySQL database, you can also use wp-cli db export function:
wp db export --path=PATHTOYOURWP
Here is an example of the universal way to pass php variables to bash without the need to parse:
#!/bin/bash
source <(php -r 'require("wp-config.php"); echo("DB_NAME=".DB_NAME."; DB_USER=".DB_USER."; DB_PASSWORD=".DB_PASSWORD); ')
mysqldump --user $DB_USER --password="$DB_PASSWORD" $DB_NAME | gzip > $DB_NAME-$(date +%Y%m%d%H%M%S).sql.gz
Algorithm explanation in a few words:
run your php
print variables as var=value
source output
use variables in bash