find is not shell, it does not expand * inside commandlines. It does inside -name and -path arguments however.
\ is taken literally inside single quotes anyway, so probably won't do what you want it to do.
I don't think find can do everything you want all by itself, but it can at least find the names for you, and print them one-by-one. Shell can do the rest.
Code:
find ./ -path '*/05_scripts/*.aep Logs' | while read LINE
do
echo mv "$LINE" $(basedir $LINE)/.Logfiles
done
Remove the 'echo' once you've tested and are sure it does what you want.
Last edited by Corona688; 01-11-2013 at 04:44 PM..
cat .servers | while read LINE; do
ssh jason@$LINE $1
done
exit 1
./command.ksh "ls -l ~jason"
Why does this ONLY iterate on the first server in the list? It's not doing the command on all the servers in the list, what am I missing?
Thanks!
JP (2 Replies)
Howdie everyone...
I have a shell script RemoveFiles.sh
Inside this file, it only has two commands as below:
rm -f ../../reportToday/temp/*
rm -f ../../report/*
My problem is that when i execute this script, nothing happened. Files remained unremoved. I don't see any error message as it... (2 Replies)
Hello, the ls -d command to only list directories in a directory doesn't seem to work on Solaris and the man command says to use that combination: ls -d
Anyone have the same problem and find a resolve?
Thanks
BobK (9 Replies)
Hi. I've been playing around a bit. This isn't for any practical purpose-- it's really just a theoretical exercise. I wrote this little thing:
foreach num ( 6 5 4 )
awk -v "number=$num" 'BEGIN{for(x=0;x<$number;x++) printf "-"; printf "\n"}'
end
I would expect the following output:
... (3 Replies)
Hi
I have put alias ll='ls -la' in .profile file but it doesn't work.
On hand it works it looks like the .profile file is not beeing read.
How to check whitch file is loaded? ,profile? .bash_profile?
My system: SunOS mion 5.10 Generic
Shell: /bin/pfksh
Thanks (2 Replies)
i want to get the value for column 4rth when i =4. please guide what i am doing wrong. thanks
var=`cat file.csv`
for i in $var; do {
if ; then
var4=$var4+$i
fi
echo $i
}
done
I am geting this error message "0403-009 The specified number is not valid for this command." (8 Replies)
I edited sudoers like this:vi /etc/sudoers
subex ALL =(root) NOPASSWD: /usr/ccs/bin/pstack
But the respective user still is prompted for password, and even when the right password is used, the command is still not launched.$sudo usr/ccs/bin/pstack 26557
We trust you have received the usual... (5 Replies)
Hi,
I am using korn shell.
until ]
do
echo "\$# = " $#
echo "$1"
shift
done
To the above script, I passed 2 parameters and the program control doesn't enter inside "until" loop. If I change it to until ] then it does work.
Why numeric comparison is not working with -ne and works... (3 Replies)
Discussion started by: ab_2010
3 Replies
LEARN ABOUT PHP
echo
ECHO(3) 1 ECHO(3)echo - Output one or more stringsSYNOPSIS
void echo (string $arg1, [string $...])
DESCRIPTION
Outputs all parameters.
echo is not actually a function (it is a language construct), so you are not required to use parentheses with it. echo (unlike some other
language constructs) does not behave like a function, so it cannot always be used in the context of a function. Additionally, if you want
to pass more than one parameter to echo, the parameters must not be enclosed within parentheses.
echo also has a shortcut syntax, where you can immediately follow the opening tag with an equals sign. Prior to PHP 5.4.0, this short syn-
tax only works with the short_open_tag configuration setting enabled.
I have <?=$foo?> foo.
PARAMETERS
o $arg1
- The parameter to output.
o $...
-
RETURN VALUES
No value is returned.
EXAMPLES
Example #1
echo examples
<?php
echo "Hello World";
echo "This spans
multiple lines. The newlines will be
output as well";
echo "This spans
multiple lines. The newlines will be
output as well.";
echo "Escaping characters is done "Like this".";
// You can use variables inside of an echo statement
$foo = "foobar";
$bar = "barbaz";
echo "foo is $foo"; // foo is foobar
// You can also use arrays
$baz = array("value" => "foo");
echo "this is {$baz['value']} !"; // this is foo !
// Using single quotes will print the variable name, not the value
echo 'foo is $foo'; // foo is $foo
// If you are not using any other characters, you can just echo variables
echo $foo; // foobar
echo $foo,$bar; // foobarbarbaz
// Some people prefer passing multiple parameters to echo over concatenation.
echo 'This ', 'string ', 'was ', 'made ', 'with multiple parameters.', chr(10);
echo 'This ' . 'string ' . 'was ' . 'made ' . 'with concatenation.' . "
";
echo <<<END
This uses the "here document" syntax to output
multiple lines with $variable interpolation. Note
that the here document terminator must appear on a
line with just a semicolon. no extra whitespace!
END;
// Because echo does not behave like a function, the following code is invalid.
($some_var) ? echo 'true' : echo 'false';
// However, the following examples will work:
($some_var) ? print 'true' : print 'false'; // print is also a construct, but
// it behaves like a function, so
// it may be used in this context.
echo $some_var ? 'true': 'false'; // changing the statement around
?>
NOTES
Note
Because this is a language construct and not a function, it cannot be called using variable functions.
SEE ALSO print(3), printf(3), flush(3), Heredoc syntax.
PHP Documentation Group ECHO(3)