Login or Register to Ask a Question and Join Our Community


Shell Programming and Scripting

Bash variable not being passed

Login to Discuss or Reply to this Discussion in Our Community

 
Thread Tools Search this Thread
Top Forums Shell Programming and Scripting Bash variable not being passed
# 1  
Old 12-17-2015
Bash variable not being passed
In the bash below the variable date displays in the echo. However when I use it in the for loop it does not. Basically, the user inputs a date then that date is converted to the desired format of (month-day-year, no leading 0). That input is used in the for loop to return every file that matches the .bam extension. I can not seem to figure out why the $date variable is not being passed to the path to search in? Thank you Smilie.

Code:
printf "\n\n"
printf "What is the date to analyze : "; read id
    date -d "$id" +%-m-%-d-%Y ; read date #convert to month-date-year
    echo "$date"
    [ -z "$id" ] && printf "\n No date supplied. Leaving match function." && sleep 2 && return
        for f in /home/cmccabe/Desktop/NGS/bam/$date/*.bam ; do
        bname=`basename $f`
        pref=${bname%%.bam}
        echo "The bam files found in directory at $f are ${pref}"
done

# 2  
Old 12-17-2015
Did you enter a value to be read into the date variable?
This User Gave Thanks to RudiC For This Post:
cmccabe
# 3  
Old 12-17-2015
Yes, I entered 10/29/2015 and echo shows 10-29-2015, which is correct and is the variable $date. Also, if I hard code 10-29-2015 into the path it seems to work. Thank you Smilie.
Last edited by cmccabe; 12-17-2015 at 03:44 PM.. Reason: fixed format
# 4  
Old 12-17-2015
Instead of this:
Code:
date -d "$id" +%-m-%-d-%Y ; read date #convert to month-date-year

Wouldn't this be what you want?
Code:
date=$(date -d "$id" "+%m-%d-%Y")

This User Gave Thanks to Aia For This Post:
cmccabe
# 5  
Old 12-17-2015
Thank you @Aia that was it Smilie
# 6  
Old 12-17-2015
By the way, you can avoid calling an outside program to give you just the name of the file, which is more expensive.
Instead of bname=`basename $f` change to bname=${f##*/}
This User Gave Thanks to Aia For This Post:
cmccabe
Login or Register to Ask a Question

Previous Thread | Next Thread

10 More Discussions You Might Find Interesting

1. Shell Programming and Scripting

Variable not passed to the sed command

Hello, I am writing a script which is not giving the desired result. When I check the content of the 'InputFile_009_0.sh', it shows following with missing Index in this command sed -i "s/L1ITMBLT.root/L1ITMBLT_"".root/g" run_DttfFromCombinedPrimitives_cfg.py of . Any help? ... (13 Replies)
Discussion started by: emily
13 Replies

2. Shell Programming and Scripting

Multiple bash variables passed into nawk

I have a file that has 2 fields called b_file: 11977 DAR.V3.20150209.1.CSV 3295 DAR.V3.20150209.1.CSV 1721 DAR.V2.20150210.1.CSV I need to search a sftplog using the field 1, but want to maintain the relationship between field 1 and 2. I am passing field 1 as a parameter in a bash loop. ... (14 Replies)
Discussion started by: smenago
14 Replies

3. Shell Programming and Scripting

[BASH] getopts, OPTARG is passed but empty

EDIT: -- SOLVED -- Heyas, Getting used to optargs, but by far not understanding it. So i have that script that shall be 'changeable', trying to use the passed arguments to modify the script visuals. Passing: browser -t test -d sect $HOME Where -t should change the title, and -d... (0 Replies)
Discussion started by: sea
0 Replies

4. Shell Programming and Scripting

Variable passed as argument

I have a script. #!/bin/sh cur_$1_modify_time=Hello echo "cur_$1_modify_time" When I run like sh /root/script1 jj I expect value "Hello" being assigned to variable "cur_jj_modify_time" and output being "Hello" ie echoing $cur_jj_modify_time But the output comes as # sh... (3 Replies)
Discussion started by: anil510
3 Replies

5. Shell Programming and Scripting

In the sh file variable is not being passed on.

I am having difficulties with the fllowing script: !/bin/sh voicemaildir=/var/spool/asterisk/voicemail/$1/$2/INBOX/ echo `date` ':' $voicemaildir >> /var/log/voicemail-notify.log for audiofile in `ls $voicemaildir/*.wav`; do transcriptfile=${audiofile/wav/transcript} ... (4 Replies)
Discussion started by: ghurty
4 Replies

6. Shell Programming and Scripting

My variable cannot be passed through into my path

Hi, Can you please help. I am scripting in sh and I am trying to simply copy one directory to another but for some reason my variables are not recognised? echo "The latest version of the program is being found......." cd $SOFTWARE/src/$progname version=`ls $SOFTWARE/src/$progname | grep... (13 Replies)
Discussion started by: cyberfrog
13 Replies

7. Shell Programming and Scripting

bash array passed to oracle

Hi all.. Does anyone know have an example of passing the contents of a bash array to oracle? basically I am looking to loop through the contents of a file(not csv!) and store each line into a bash array. Once i have this I can then pass the array into an oracle procedure that accepts an array... (1 Reply)
Discussion started by: satnamx
1 Replies

8. UNIX for Dummies Questions & Answers

Assigning a Variable all the Parameters passed

Hi, I have a unix script which can accept n number of parameters . I can get the parameter count using the following command and assign it to a variable file_count=$# Is there a similar command through which i can assign a variable all the values that i have passed as a parameter ... (2 Replies)
Discussion started by: samit_9999
2 Replies

9. UNIX for Dummies Questions & Answers

variable passed to awk

Does anybody know how to print a variable passed to awk command? awk -F"|" 'BEGIN {print $job,"\n","Question \n"} {print $1,$2$4," ",$3}' "job=$job1" file1 I am trying to pass job the variable job1. the output is blank. ?? (3 Replies)
Discussion started by: whatisthis
3 Replies

10. UNIX for Dummies Questions & Answers

variable passed to awk

Anybody know what's wrong with this syntax? awk -v job="$job" 'BEGIN { FS="|"} {print $1,$2," ",$4," ",$3\n,$5,"\n"}' list It's keeping give me this message: awk: syntax error near line 1 awk: bailing out near line 1 It seems awk has problem with my BEGIN command. Any... (8 Replies)
Discussion started by: whatisthis
8 Replies
Login or Register to Ask a Question