CODEWITHSUNDEEP
shell
p zankat
p zankat

Reputation: 135

how to include variable value in command argument in the shell script?

This script statements in certmgmt.sh-

 update_keystore(){   
     keystore_file_path="/data/cert/keystore.jks"
     kestore_password="pass@word"
     temp_cert_path="/data/cert/temp_cert_file.cer"
     cert_alias="root";

     command="keytool -importcert -file $temp_cert_path -keystore $keystore_file_path -alias $cert_alias  -keypass $kestore_password -storepass $kestore_password -noprompt"
     $("$command")
 }

This is the output of function execution

pz-active-user# /usr/certmgmt.sh update_keystore
usr/certmgmt.sh: line 1381: keytool -importcert -file /data/cert/temp_cert_file.cer -keystore /data/cert/keystore.jks -alias root  -keypass pass@word -storepass pass@word -noprompt: No such file or directory

But if I run this directly from the terminal, it works.

pz-active-user# keytool -importcert -file /data/cert/temp_cert_file.cer -keystore /data/cert/keystore.jks -alias root  -keypass pass@word -storepass pass@word -noprompt

Upvotes: 0

Views: 137

Answers (1)

John Kugelman
John Kugelman

Reputation: 361675

Get rid of the command variable and the $(...) command substitution. You can call keytool directly like you did at the command line:

update_keystore() {   
     keystore_file_path="/data/cert/keystore.jks"
     keystore_password="pass@word"
     temp_cert_path="/data/cert/temp_cert_file.cer"
     cert_alias="root"

     keytool -importcert -file "$temp_cert_path" -keystore "$keystore_file_path" -alias "$cert_alias" -keypass "$keystore_password" -storepass "$keystore_password" -noprompt
}

Upvotes: 2

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