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Top Forums UNIX for Dummies Questions & Answers help with shell scripting argument Post 302472265 by iluvsushi on Tuesday 16th of November 2010 04:50:45 PM
Old 11-16-2010
help with shell scripting argument

Code:
#!/bin/bash

echo "enter a file or directory name"
read name
if [ -f $name ]
        then
        echo " argument is file "
        ls -l $name | awk '{print $1,}'
        elif [ ???  $name ]
        echo " argument is a directory"
        ls -l $name | awk '{print $1}'
fi

what i am trying to do. get input file or directory name from user.
if it is a regular file, print permission or if it is a directory, print permissions.
i know " -f " means regular file but how to ask for directory. in other words i don't know
Code:
elif [ ???? $name]

any help will be appreciated.

Last edited by Scott; 11-16-2010 at 05:58 PM.. Reason: Please use code tags
 

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FTP_CHDIR(3)								 1							      FTP_CHDIR(3)

ftp_chdir - Changes the current directory on a FTP server

SYNOPSIS
bool ftp_chdir (resource $ftp_stream, string $directory) DESCRIPTION
Changes the current directory to the specified one. PARAMETERS
o $ftp_stream - The link identifier of the FTP connection. o $directory - The target directory. RETURN VALUES
Returns TRUE on success or FALSE on failure. If changing directory fails, PHP will also throw a warning. EXAMPLES
Example #1 ftp_chdir(3) example <?php // set up basic connection $conn_id = ftp_connect($ftp_server); // login with username and password $login_result = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass); // check connection if ((!$conn_id) || (!$login_result)) { die("FTP connection has failed !"); } echo "Current directory: " . ftp_pwd($conn_id) . " "; // try to change the directory to somedir if (ftp_chdir($conn_id, "somedir")) { echo "Current directory is now: " . ftp_pwd($conn_id) . " "; } else { echo "Couldn't change directory "; } // close the connection ftp_close($conn_id); ?> SEE ALSO
ftp_cdup(3), ftp_pwd(3). PHP Documentation Group FTP_CHDIR(3)
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