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Top Forums Shell Programming and Scripting Update a field in a file based on condition Post 302478049 by kichu on Tuesday 7th of December 2010 04:14:20 AM
Old 12-07-2010
Update a field in a file based on condition

Hi

i am new to scripting. i have a file file.dat with content as :

Code:
 
CONTENT_STORAGE PERCENTAGE FLAG:
/storage_01 64% 0
/storage_02 17% 1

I need to update the value of FLAG for a particular CONTENT_STORAGE value

I have written the following code
Code:
 
 #!/bin/sh
threshold=20
alert="threshold"
#Get default file store 
defaultStorage1=DEFAULT_STORE 
#DEFAULT_STORE is obtained as storage_01 from database table
 
echo "file reading"
{
 
while read CONTENT_STORAGE PERCENTAGE FLAG
do
 
    #Code for reading through lines and ignoring all until CONTENT_STORAGE is found
 
    # Ignore all lines until content storage namematches the name given 
    if [[ ${CONTENT_STORAGE} != $defaultStorage1 ]]; then
 echo "here"
        continue
    fi   
echo "CONTENT_STORAGE = ${CONTENT_STORAGE} / PERCENTAGE = ${PERCENTAGE} / FLAG = ${FLAG}" 
done < file.dat 
}
echo "Getting the percentage og $defaultStorage1 *********************"
#check the percentage stored in defaultStorage
df -H | grep $defaultStorage1 | awk '{ print $4 " " $5}' | while read output;
do 
  #Get the used percentage
  usep=$(echo $output | awk '{ print $1}' | cut -d'%' -f1)
 echo "used percentage .................. $usep"
# Check if the used percentage is > threshold_value. Do nothing till its > threshold
   if [ $(($usep)) -le $(($threshold)) ]; then
 echo "Used percentage of $defaultStorage1 ................. $(($usep))"
       continue
 
else 
 
echo "Used percentage is greater than threshold"
 
    # Ignore all lines until content storage namematches the name given 
    if [[ ${CONTENT_STORAGE} != $defaultStorage1 ]]; then
        continue
    else
       $FLAG = 1  
#i am trying to update the FLAG field in here. But not happening. Also while creating the tmp file file.mp and rewriting to file.dat, the headings are ignored. i want to just modify the FLAG value for /storage1 to 1
    fi 
    echo $CONTENT_STORAGE $PERCENTAGE $FLAG >> file.tmp
  done < file.dat  
mv file.tmp file.dat
   # check the percentage of each other filestore with flag =0
   #get the least used filestore
   #update the default store
 
   fi
 
done

Thanks for all the help in advance
 

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ECHO(3) 								 1								   ECHO(3)

echo - Output one or more strings

SYNOPSIS
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)
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