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passing argument from one function to another

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# 43  
Old 09-04-2012
Quote:
Originally Posted by nrjrasaxena

may i ask the significance of having "" ..???
"" will be useful if your variable having space...

My above given code works perfect for your file list..

Code:
echo "data1.list
data2.list
data3.list" | sed  's/data[a-zA-Z0-9]*/data1/g'
data1.list
data1.list
data1.list

This User Gave Thanks to pamu For This Post:
nrjrasaxena
# 44  
Old 09-04-2012
Quote:
Originally Posted by nrjrasaxena
..ummm, may i ask the significance of having "" ..???
Always keep in mind that the shell "expands" variables before evaluating a line. That means that an expression like "$var" will be replaced by the contents of this variable. The default content of a variable is the null string, now. Consider the following code:

Code:
x="foo"
if [ $x = foo ] ; then
...

The shell will do the following: it will first replace "$x" with "foo" and only then evaluate the line:

Code:
if [ foo = foo ] ; then

Of course this evaluates to "TRUE" and the if-branch will be executed (the else-branch in case it evaluates to "FALSE"). Now suppose that "$x" is not given a value:

Code:
# this commented so that it has no effect: # x="foo"
if [ $x = foo ] ; then
...

The shell would evaluate the line in question to:

Code:
if [  = foo ] ; then

and this is simply a syntactical error. The comparison lacks a part to which to compare to. Now consider the same with quotes:

Code:
x=foo
if [ "$x" = "foo" ] ; then
...

gives:
Code:
if [ "foo" = "foo" ] ; then

and

Code:
# this commented out # x=foo
if [ "$x" = "foo" ] ; then
...

gives

Code:
if [ "" = "foo" ] ; then

Which will now be "FALSE", but still be syntactically correct.

This is why it is good to ALWAYS quote your variables.

I hope this helps.

bakunin
This User Gave Thanks to bakunin For This Post:
nrjrasaxena
# 45  
Old 09-04-2012
Hi,
In the process of modification of my script, the step next problem is following, I Would like to incorporate the following changes..but does not seem to work !
I am not sure where to define the PATHNAME variable..??? I tried several options, did not work.

Sorry for bothering so much, its kind of first script that I am writing..:P

Code:
PATH525[1]="/eos/uscms/store/user/pooja04//analysis2012/525/data/doubleele/2012/datav1/"
PATH525[2]="/eos/uscms/store/user/pooja04//analysis2012/525/data/doubleele/2012/datav4/"
PATHZY[1]="/eos/uscms/store/user/pooja04//analysis2012/533/mc/zgammav1/"
                                                                                                                                                         
PATH533[1]="/eos/uscms/store/user/pooja04//analysis2012/533/data/datav14/"

CopyFiles() {
    if [ "$2" = "525" ]; then
        $PATHNAME = "$PATH525"
    elif [ "$2" = "533" ]; then
        $PATHNAME = "$PATH533"
    elif [ "$2" = "ZY" ]; then
        $PATHNAME = "$PATHZY"
    fi
    echo "pathname is $PATHNAME"
    for FileNameIndx in "${PATHNAME[@]}"
          do
          if [[ ! -e "dest_path/$FileNameIndx" ]]; then
              ls -ltr "$FileNameIndx" | grep root | awk '{print string path $9}' string="$CONSTANT" path="$FileNameIndx"  >> "$FileName"
              echo "$FileNameIndx is copied"
          else
              echo "Check the FileName in ${PATHNAME[@]}"
          fi
        done
}

thanks

---------- Post updated at 01:00 PM ---------- Previous update was at 12:54 PM ----------

Thanks for the detailed information..:-)

Quote:
Originally Posted by bakunin
Always keep in mind that the shell "expands" variables before evaluating a line. That means that an expression like "$var" will be replaced by the contents of this variable. The default content of a variable is the null string, now. Consider the following code:

Code:
x="foo"
if [ $x = foo ] ; then
...

The shell will do the following: it will first replace "$x" with "foo" and only then evaluate the line:

Code:
if [ foo = foo ] ; then

Of course this evaluates to "TRUE" and the if-branch will be executed (the else-branch in case it evaluates to "FALSE"). Now suppose that "$x" is not given a value:

Code:
# this commented so that it has no effect: # x="foo"
if [ $x = foo ] ; then
...

The shell would evaluate the line in question to:

Code:
if [  = foo ] ; then

and this is simply a syntactical error. The comparison lacks a part to which to compare to. Now consider the same with quotes:

Code:
x=foo
if [ "$x" = "foo" ] ; then
...

gives:
Code:
if [ "foo" = "foo" ] ; then

and

Code:
# this commented out # x=foo
if [ "$x" = "foo" ] ; then
...

gives

Code:
if [ "" = "foo" ] ; then

Which will now be "FALSE", but still be syntactically correct.

This is why it is good to ALWAYS quote your variables.

I hope this helps.

bakunin
# 46  
Old 09-04-2012
hi

variable assignment should be like this

Code:
Pathname = "$paths"

---------- Post updated at 11:32 PM ---------- Previous update was at 11:31 PM ----------

Quote:
Originally Posted by nrjrasaxena
Hi,
In the process of modification of my script, the step next problem is following, I Would like to incorporate the following changes..but does not seem to work !
I am not sure where to define the PATHNAME variable..??? I tried several options, did not work.

Sorry for bothering so much, its kind of first script that I am writing..:P

Code:
PATH525[1]="/eos/uscms/store/user/pooja04//analysis2012/525/data/doubleele/2012/datav1/"
PATH525[2]="/eos/uscms/store/user/pooja04//analysis2012/525/data/doubleele/2012/datav4/"
PATHZY[1]="/eos/uscms/store/user/pooja04//analysis2012/533/mc/zgammav1/"
                                                                                                                                                         
PATH533[1]="/eos/uscms/store/user/pooja04//analysis2012/533/data/datav14/"

CopyFiles() {
    if [ "$2" = "525" ]; then
        $PATHNAME = "$PATH525"
    elif [ "$2" = "533" ]; then
        $PATHNAME = "$PATH533"
    elif [ "$2" = "ZY" ]; then
        $PATHNAME = "$PATHZY"
    fi
    echo "pathname is $PATHNAME"
    for FileNameIndx in "${PATHNAME[@]}"
          do
          if [[ ! -e "dest_path/$FileNameIndx" ]]; then
              ls -ltr "$FileNameIndx" | grep root | awk '{print string path $9}' string="$CONSTANT" path="$FileNameIndx"  >> "$FileName"
              echo "$FileNameIndx is copied"
          else
              echo "Check the FileName in ${PATHNAME[@]}"
          fi
        done
}

thanks

---------- Post updated at 01:00 PM ---------- Previous update was at 12:54 PM ----------

Thanks for the detailed information..:-)
# 47  
Old 09-04-2012
Quote:
Originally Posted by pamu
hi

variable assignment should be like this

Code:
Pathname = "$paths"

Hi,
thanks for such quick response.
ummm, I did declare it within the function and as a global variable..did not work !!
# 48  
Old 09-04-2012
Quote:
Originally Posted by nrjrasaxena
Hi,
thanks for such quick response.
ummm, I did declare it within the function and as a global variable..did not work !!

Declare variable in function thats fine just make sure before using thats variable you should assign value to it

and in above code you are doing

Code:
$A=$b # this will not assign value to A.

#just do like this.
A=$b

hope this helps you Smilie
# 49  
Old 09-04-2012
Quote:
Originally Posted by pamu
Declare variable in function thats fine just make sure before using thats variable you should assign value to it

and in above code you are doing

Code:
$A=$b # this will not assign value to A.

#just do like this.
A=$b

hope this helps you Smilie
How about this..??
Code:
CopyFiles() {
    PATHNAME="$paths"

    if [ "$2" = "525" ]; then
        PATHNAME="$PATH525"
    elif [ "$2" = "533" ]; then
        PATHNAME="$PATH533"
    elif [ "$2" = "ZY" ]; then
        PATHNAME="$PATHZY"
    fi
    echo "pathname is $PATHNAME"
    for FileNameIndx in "${PATHNAME[@]}"
          do
          if [[ ! -e "dest_path/$FileNameIndx" ]]; then
              ls -ltr "$FileNameIndx" | grep root | awk '{print string path $9}' string="$CONSTANT" path="$FileNameIndx"  >> "$FileName"
              echo "$FileNameIndx is copied"
          else
              echo "Check the FileName in ${PATHNAME[@]}"
 fi
        done
}

Not Working...!! The echo command not showing the value of the PATHNAME variable.
Last edited by nrjrasaxena; 09-04-2012 at 03:25 PM..
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