Programming :: Assigning Variable Is Messing Up?
Nov 5, 2010I am trying to run this and was into issues
function() {
node=echo "10.11.12.13" | awk '{split ($0, a, "."); print a[1]}'
}
[code]....
I am trying to run this and was into issues
function() {
node=echo "10.11.12.13" | awk '{split ($0, a, "."); print a[1]}'
}
[code]....
Using things like awk/sed, but have managed to cobble together what I needed so far without a problem. The only thing I'm struggling with is to assign the content of a particular line as a variable, and then to use it again throughout the file.
For example, if what I want is the first line of the file to become the variable "from1", and then to replace the word "Subject" in the file with the string "Message from [from1]". What I thought would work
I tried a few diff combinations but nothing seems to work. All I get is the terminal hanging indefinitely.
I am having all sorts of trouble trying to assign a variable within an awk script with the system command. I know there is a lot of ways around this problem, but for efficiency reasons, I would like to, within my awk script, do something like
system(x=3)
or
system(x=NR)
and, latter on the shell script which calls the awk script, use the variable $x. But nothing is passed to x. I have already tried things like
command = "x=3"
system(command)
and also used a pipeline within the system to pipe it to /bin/sh In fact tried a lot of stuff like that, using $(( )) etc etc etc I can create directories e write to files (yes, i could write to a file and read from there, but I dont think it is efficient, plus I am puzzled).
Kindly take a look at the code below :
Code:
#include <stdio.h>
struct test
{
int i;
int j;
[Code]....
why i am getting this error. I know the error is occuring because i have assigned values to obj.i and obj.j outside main(). But i want to know why do that result in an error. From my part i have created an object 'obj' of stucture 'test' and assigned values to its variables.
Kernel 2.6.21.5, GNU/Linux (Slackware 12.0).In this script,
Code:
semoi@darkstar:~$ cat rename4.sh
#!/bin/bash
INPUT="k3b_audio_0_04"
[code]...
Why doesn't "var1=`echo $var2 | grep pattern`" work ?
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e.g.
a=`find . -name "hello.txt" -type f`
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I'm trying to read content of file to variable and use this variable in for loop. The problem is, when I have c++ comment style in file - /*. Spaces in line are also interpreted as separated lines.
For example:
Code:
Changing $files to "$files" eliminate these problems but causes that whole content of variable is treated as one string (one execution of loop).
my script has a variable which comes in the form +00.00 +0.00 -00.00 or -0.00 (the numbers can be any in that form) for any that have a + symbol I need to remove the +, but if it has a - symbol it needs to stay.
i need to make a new variable with the string from the old variable btut without any plus sign. I have tried a lot of different ways with no success, each thing I tried either left the + or removed the entire string. I think this should work but doesn't
foo=+12.40
bar=${foo#+}
how I can search within a variable and assign the results to a new variable. I'll use the following as an example -
cars="Audi BMW Cadillac Chevy Dodge Ferrari Ford Mercedes"
list=`echo ${cars} | egrep -o '<A?+|<C+'`
with the echo command I get the following output assigned to list -
A
C
C
What I'd like to get for output is -
Audi
Cadillac
Chevy
how I could do this regardless of upper/lower case letters?
I am assigning a file path to a variable like
$1 means it should accept some file name from cmd prompt but in my case its not working.
included shell script inside c program, and i wanted to assign the value of c variable to shell variable..Can any one please suggest me how to do it?
View 8 Replies View RelatedI am trying to execute a Unix Command in perl and assigning its output to an array:
Code:
@File_List=exec("ls -1 /tmp");
but it is not working. I have tried the perl function system() also but its return code is
[code]...
How can I pass by reference, a parameter, without assigning a new object? In my first example, var = "changed" creates a new local object. In the second, mylist.append will affect the reference target. How to i achieve the same effect with example 1?
Code:
#!/bin/python
var = "unchanged"
def print_string( var ):
[Code].....
I am working on some minor programming projects. and wouldt like to keep track of them by version numbers. E.g. 1.0.0
Is there a tool or other method for automatically assigning version numbers to software ?
I am currently using Mercurial. as far as i can see it only assigns revision numbers. E.g. 5
This loop is part of a bash script which takes multiple arguments.
Code:
for ((i=1;i<=$number;++i)) ; do
offset=$(($i+5))
[code]...
i want to pass variable in mysql qyery in c programming
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[code]....
I have beat this enough and don't get what should have been a very simple thing to do. I build a variable;
Code:
CLIST=java,lua,python,php,perl,ruby,tcl
CLIST will be used by another bash script but I need to replace the commas with a space. I
[code]...
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View 2 Replies View RelatedI have been searching most of today and am stuck on getting a variable into an awk portion of my bash script. I have this working:
Code:
#!/bin/sh
SRC=/var/log/mail.log
DEST=/var/www/output/myFile.txt
VAR=userName@myDomain.tld
[code]....
Can awk take a shell variable? Or do I have to do something completely different?
In C++ what does the suffix, "*" mean appended to a variable type, e.g., "char* variable1;"?
View 3 Replies View RelatedFollowing is the way I saw a variable initialized in C
Code:
static const unsigned int rtl8139_rx_config =
RxCfgRcv64K |
(RX_FIFO_THRESH << RxCfgFIFOShift) |
(RX_DMA_BURST << RxCfgDMAShift);
on following link
[URL]
I have initialized variables in past but above initialization I could not understand what is it?
problem statement:
pattern_search="Exam Name"
sed -n "/$pattern_search/,/hello/"p tmp5 | awk '{if ( $4 != 0 && $4 ~ /[0-9]+.*[0-9]*/ ) print "$pattern_search" " " $0 }'
"tmp5" is a file. this is printing output as
$pattern_search value1
i.e value of $pattern_search is not getting substituted. i am expecting output as
Exam Name value1
how shall I print each variable separately using a generalized form. I tried writing the following within a for loop...Code:echo $(echo a$(echo $i)$(echo $j))which did yield no result. So what shall I write??
View 3 Replies View RelatedI have the following input:
Code:
Event 1............................................................
full_name: JENNY_JENNINGS genre: f
age: 32
[code]....
But as you can see in the input, in the 2nd "Event", the line containing "age" is not present, but in the output my code is printing the 1rst age value twice. The correct output should be blank in the age field for 2nd line in the output like this:
Code:
full_name|genre|age|code
JENNY_JENNINGS|f|32|15a
JOHN_JOHNSON|m||23c
MARY_JEAN|f|25|11d
What is wrong in my code? how can I fix it? * I�m using ubuntu 10.10
How are environment variable set in tcl? I tried "set $env(MYVAR) xxxx" but it didn't work.
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sed -n '/Sales ID: ${array[$i]}/,/Totals:/p'
that command creates empty files. so my guess is that its not recognizing the array as an array but as text?
how would i be able to utilize the array in the command? i got it, didnt think that if i doubled up the single quotes that it would work, but this worked:
sed -n '/Sales ID: '${array[$i]'}/,/Totals:/p'