I need to login to an application using cURL and find out the time taken to login to an application using cURL command.
What I have tried so far is write a shell script
curl -k -X -s "http://username:passowrd#servername:portnumber" -w 'Lookup time:\t%{time_namelookup} s
Total time:\t%{time_total} s
' -o /dev/null
IS this correct way to do it?
Related
I'm trying to explore the curl connect or retrieve the data with query API link but when I run my GET command in curl it runs without any error shown in -vv but the process stop. below the actual statement at the end of the posted process, may I know if this is normal or I need to add an additional arguments or parameters on my command for me to retrieve the data? If yes may I know how to do it?
Connection #1 to host myserver.com left intact
My command
curl -vv -k -X GET -H --user user123:password1 -H "Accept: application/json" https://myapplink/statement/EXE=sample
Hi I am writing a auto script in test.sh , attempting to download a file. It works fine when I use all hard code string. But it does not work with variables. Belong are my code example:
#!/bin/bash
USER="admin"
PWD="adminpass"
curl -v -k -u ${USER}:${PWD} ${NEXUS_URL}/${SP1}/60/${SP1}-60.zip --output ${SP1}-60.zip
Above code not working not able to download my file, but if I put it as :
curl -v -k -u "admin":"adminpass" ${NEXUS_URL}/${SP1}/60/${SP1}-60.zip
--output ${SP1}-60.zip
Then it works. So how do I get the variable credential working with this curl command?
Thanks
Option 1
The parameter expansion will not include the double quotes. You can use:
#!/bin/bash
USER='"'admin'"' #single quote, double quote, single quote
PASS='"'adminpass'"'
curl -v -k -u ${USER}:${PASS} ${NEXUS_URL}/${SP1}/60/${SP1}-60.zip
Option 2
Alternatively, you can create a .netrc file and use curl -n as follows:
Documentation from https://ec.haxx.se/usingcurl-netrc.html
Create .netrc containing the following and place it in your home directory.
machine http://something.com
login admin
password adminpass
Run the command
curl -n -k --output ${SP1}-60.zip
curl will automatically look for the .netrc file. You can also specify the file path with curl --netrc-file <netrc_file_path>
I need to migrate from a windows based proxy to a linux one.
In the old server Squid uses mswin_auth.exe to authenticate user against an Active Directory domain. For this reason my users now enter ntdomain\username in the browser popup for proxy authentication.
In the linux Centos server Squid will use basic_ldap_auth, in this case the ntdomain must not be entered by the user. It will be very annoying for my users to change the old habit. Is there a way to automatically remove the ntdomain from the entered username?
In squid.conf I have
auth_param basic program /usr/lib64/squid/basic_ldap_auth -R -b "dc=ntdomain,dc=parentd,dc=it" -D "CN=squid,OU=Squid,OU=Sede,DC=ntdomain,DC=parentd,DC=it" -W /etc/squid/squid.adpwd -f sAMAccountName=%s -h vfdc1.ntdomain.parentd.it
...
external_acl_type ldap_group %LOGIN /usr/lib64/squid/ext_ldap_group_acl -R -b "dc=ntdomain,dc=parentd,dc=it" -D "CN=squid,OU=Squid,OU=Sede,DC=ntdomain,DC=parentd,DC=it" -W /etc/squid/squid.adpwd -f "(&(objectclass=person) (sAMAccountname=%u)(memberof:1.2.840.113556.1.4.1941:=cn=%g,OU=Squid,OU=Sede,DC=ntdomain,DC=parentd,DC=it))" -h vfdc1.ntdomain.parentd.it -S
thanks in advance
Use -K option in your request (it Strip Kerberos realm from usernames):
external_acl_type ldap_group %LOGIN /usr/lib64/squid/ext_ldap_group_acl -R -K -b "dc=ntdomain,dc=parentd,dc=it" -D "CN=squid,OU=Squid,OU=Sede,DC=ntdomain,DC=parentd,DC=it" -W /etc/squid/squid.adpwd -f "(&(objectclass=person) (sAMAccountname=%u)(memberof:1.2.840.113556.1.4.1941:=cn=%g,OU=Squid,OU=Sede,DC=ntdomain,DC=parentd,DC=it))" -h vfdc1.ntdomain.parentd.it and all be fine.
This is my solution: I created to small bash script based on sed to strip the domain and used it squid.conf as in
auth_param basic program /usr/local/bin/squid_auth
where the file /usr/local/bin/squid_auth is
#!/bin/bash
/usr/bin/sed -u "s/^ve[\\]//i"|/usr/bin/sed -u "s/^ve%5c//i"|/usr/lib64/squid/basic_ldap_auth -R -b "dc=ve,dc=dipvvf,dc=it" -D "CN=squid,OU=Squid,OU=Sede,DC=ve,DC=dipvvf,DC=it" -W /etc/squid/squid.adpwd -f sAMAccountName=%s -h vfdc1.ve.dipvvf.it
The domain name is ve
I'm learning shell scripting! for the same I've tried downloading the facebook page using curl on ubuntu terminal.
t.sh content
vi#vi-Dell-7537(Desktop) $ cat t.sh
curlCmd="curl \"https://www.facebook.com/vivekkumar27june88\""
echo $curlCmd
($curlCmd) > ~/Desktop/fb.html
Getting error when running the script as
vi#vi-Dell-7537(Desktop) $ ./t.sh
curl "https://www.facebook.com/vivekkumar27june88"
curl: (1) Protocol "https not supported or disabled in libcurl
But if the run the command directly then it is working fine.
vi#vi-Dell-7537(Desktop) $ curl "https://www.facebook.com/vivekkumar27june88"
<!DOCTYPE html>
<html lang="hi" id="facebook" class="no_js">
<head><meta chars.....
I will appreciate if anyone let me know the mistake I am doing in the script.
I've verified that curl library have ssl enabled.
A command embedded within a parenthesis runs as a sub-shell so your environment variables will be missing.
Try eval:
curlCmd="curl 'https://www.facebook.com/vivekkumar27june88' > ~/Desktop/fb.html"
eval $curlCmd
Create your script t.sh as this single line only:
curl -k "https://www.facebook.com/vivekkumar27june88" -o ~/Desktop/fb.html
As per man curl:
-k, --insecure
(SSL) This option explicitly allows curl to perform "insecure" SSL connections transfers.
All SSL connections are attempted to be made secure by using the CA certificate bundle
installed by default. This makes all connections considered "insecure" fail unless -k,
--insecure is used.
-o file
Store output in the given filename.
As #Chepner said, go read BashFAQ #50: I'm trying to put a command in a variable, but the complex cases always fail!. To summarize, how you should do things like this depends on what your goal is.
If you don't need to store the command, don't! Storing commands is difficult to get right, so if you don't need to, just skip that mess and execute it directly:
curl "https://www.facebook.com/vivekkumar27june88" > ~/Desktop/fb.html
If you want to hide the details of the command, or are going to use it a lot and don't want to write it out each time, use a function:
curlCmd() {
curl "https://www.facebook.com/vivekkumar27june88"
}
curlCmd > ~/Desktop/fb.html
If you need to build the command piece-by-piece, use an array instead of a plain string variable:
curlCmd=(curl "https://www.facebook.com/vivekkumar27june88")
for header in "${extraHeaders[#]}"; do
curlCmd+=(-H "$header") # Add header options to the command
done
if [[ "$useSilentMode" = true ]]; then
curlCmd+=(-s)
fi
"${curlCmd[#]}" > ~/Desktop/fb.html # This is the standard idiom to expand an array
If you want to print the command, the best way to do it is usually with set -x:
set -x
curl "https://www.facebook.com/vivekkumar27june88" > ~/Desktop/fb.html
set +x
...but you can also do something similar with the array approach if you need to:
printf "%q " "${curlCmd[#]}" # Print the array, quoting as needed
printf "\n"
"${curlCmd[#]}" > ~/Desktop/fb.html
Install following softwares in ubuntu 14.04
sudo apt-get install php5-curl
sudo apt-get install curl
then run sudo service apache2 restart
check your phpinfo() is enable with curl "cURL support: enabled"
Then check your command in shell script
RESULT=curl -L "http://sitename.com/dashboard/?show=api&action=queue_proc&key=$JOBID" 2>/dev/null
echo $RESULT
You will get response;
Thanks you.
I am working on a bash script to configure openldap and add ldif script with users and groups.
How can I write the password from the bash script ?
This is the script I run when it asks for password:
ldapadd -h localhost -D "cn=admin,dc=sysadmin1,dc=hioa,dc=no" -W -x -f /etc/ldap/base.ldif
EDIT:
I tried this and created a passwd.txt file with the password:
ldapadd -h localhost -D "cn=admin,dc=sysadmin1,dc=hioa,dc=no" -W -x -y'passwd.txt' -f /etc/ldap/base.ldif
But gets this error:
Warning: Password file passwd.txt is publicly readable/writeable
ldap_bind: Invalid credentials (49)
man ldapadd.
-W
Prompt for simple authentication. This is used instead of specifying the password on the command line.
-w passwd
Use passwd as the password for simple authentication.
-y passwdfile
Use complete contents of passwdfile as the password for simple authentication.
So seems you are looking for option of -w or -y, not -W
There're two possibilities:
ldapadd reads the password from the standard input.
ldapadd reads the password directly from the current TTY.
In the first case it's enough to use something like this echo 'MySecretPassword' | ldapadd -h localhost -D "cn=admin,dc=sysadmin1,dc=hioa,dc=no" -W -x -f /etc/ldap/base.ldif. The second one is more complicated because you need a tool like expect. Check if the simple redirection works first.