Set a variable equal to the output of a command containing non-command words
I'm writing a small script in which I want to set the value of a variable equal to the output of a command. However, the command in question is a call to another script with command-line arguments. I'm using backticks as you normally should in this scenario, but the problem is that the the computer gives an error, in which it tries to interpret the command-line arguments as commands.
#!/bin/bash
filename="$1"
while read p; do
echo "This is the gene we are looking at: ""$p"
lookIn= `./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri`
echo "$lookIn"
#grep "$p" "$lookIn""/""prokka_""$lookIn""/*.tsv" | awk '{print $1}'
done < $filename
I'm trying to set variable lookIn equal to the output of ./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri, where ./findGeneIn is a script, and the words burgdorferi,...,parkeri are command line arguments for ./findGeneIn.
The issue, is that I get an error saying "burgdorferi: command not found". So it's trying to interpret those arguments as commands. How do I get it to not do that?
bash arguments
edited Nov 25 '18 at 1:39
John Kugelman
245k54406459
asked Nov 25 '18 at 1:31
The_QuestionerThe_Questioner
142314
add a comment |
I'm writing a small script in which I want to set the value of a variable equal to the output of a command. However, the command in question is a call to another script with command-line arguments. I'm using backticks as you normally should in this scenario, but the problem is that the the computer gives an error, in which it tries to interpret the command-line arguments as commands.
#!/bin/bash
filename="$1"
while read p; do
echo "This is the gene we are looking at: ""$p"
lookIn= `./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri`
echo "$lookIn"
#grep "$p" "$lookIn""/""prokka_""$lookIn""/*.tsv" | awk '{print $1}'
done < $filename
I'm trying to set variable lookIn equal to the output of ./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri, where ./findGeneIn is a script, and the words burgdorferi,...,parkeri are command line arguments for ./findGeneIn.
The issue, is that I get an error saying "burgdorferi: command not found". So it's trying to interpret those arguments as commands. How do I get it to not do that?
bash arguments
edited Nov 25 '18 at 1:39
John Kugelman
245k54406459
asked Nov 25 '18 at 1:31
The_QuestionerThe_Questioner
142314
2
Remove the space afterlookIn=. Also, use$( )instead of backticks, and quote the filename (done < "$filename"). shellcheck.net is good at spotting common mistakes like these.
– Gordon Davisson
Nov 25 '18 at 1:42
add a comment |
I'm writing a small script in which I want to set the value of a variable equal to the output of a command. However, the command in question is a call to another script with command-line arguments. I'm using backticks as you normally should in this scenario, but the problem is that the the computer gives an error, in which it tries to interpret the command-line arguments as commands.
#!/bin/bash
filename="$1"
while read p; do
echo "This is the gene we are looking at: ""$p"
lookIn= `./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri`
echo "$lookIn"
#grep "$p" "$lookIn""/""prokka_""$lookIn""/*.tsv" | awk '{print $1}'
done < $filename
I'm trying to set variable lookIn equal to the output of ./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri, where ./findGeneIn is a script, and the words burgdorferi,...,parkeri are command line arguments for ./findGeneIn.
The issue, is that I get an error saying "burgdorferi: command not found". So it's trying to interpret those arguments as commands. How do I get it to not do that?
bash arguments
edited Nov 25 '18 at 1:39
John Kugelman
245k54406459
asked Nov 25 '18 at 1:31
The_QuestionerThe_Questioner
142314
I'm writing a small script in which I want to set the value of a variable equal to the output of a command. However, the command in question is a call to another script with command-line arguments. I'm using backticks as you normally should in this scenario, but the problem is that the the computer gives an error, in which it tries to interpret the command-line arguments as commands.
#!/bin/bash
filename="$1"
while read p; do
echo "This is the gene we are looking at: ""$p"
lookIn= `./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri`
echo "$lookIn"
#grep "$p" "$lookIn""/""prokka_""$lookIn""/*.tsv" | awk '{print $1}'
done < $filename
I'm trying to set variable lookIn equal to the output of ./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri, where ./findGeneIn is a script, and the words burgdorferi,...,parkeri are command line arguments for ./findGeneIn.
The issue, is that I get an error saying "burgdorferi: command not found". So it's trying to interpret those arguments as commands. How do I get it to not do that?
bash arguments
bash arguments
edited Nov 25 '18 at 1:39
John Kugelman
245k54406459
asked Nov 25 '18 at 1:31
The_QuestionerThe_Questioner
142314
edited Nov 25 '18 at 1:39
John Kugelman
245k54406459
asked Nov 25 '18 at 1:31
The_QuestionerThe_Questioner
142314
edited Nov 25 '18 at 1:39
John Kugelman
245k54406459
edited Nov 25 '18 at 1:39
John Kugelman
245k54406459
edited Nov 25 '18 at 1:39
John Kugelman
245k54406459
245k54406459
asked Nov 25 '18 at 1:31
The_QuestionerThe_Questioner
142314
asked Nov 25 '18 at 1:31
The_QuestionerThe_Questioner
142314
asked Nov 25 '18 at 1:31
The_QuestionerThe_Questioner
142314
142314
2
Remove the space afterlookIn=. Also, use$( )instead of backticks, and quote the filename (done < "$filename"). shellcheck.net is good at spotting common mistakes like these.
– Gordon Davisson
Nov 25 '18 at 1:42
add a comment |
2
Remove the space afterlookIn=. Also, use$( )instead of backticks, and quote the filename (done < "$filename"). shellcheck.net is good at spotting common mistakes like these.
– Gordon Davisson
Nov 25 '18 at 1:42
2
2
Remove the space after
lookIn=. Also, use $( ) instead of backticks, and quote the filename (done < "$filename"). shellcheck.net is good at spotting common mistakes like these.– Gordon Davisson
Nov 25 '18 at 1:42
Remove the space after
lookIn=. Also, use $( ) instead of backticks, and quote the filename (done < "$filename"). shellcheck.net is good at spotting common mistakes like these.– Gordon Davisson
Nov 25 '18 at 1:42
add a comment |
1 Answer
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lookIn= `./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri`
^
Delete the extra space. Assignments must not have spaces around the equal sign.
With the space there, Bash parses the line as var=value command, which runs a command with $var temporarily set to "value". Or in this case, it interprets result of the backticks as a command name and lookIn= as an empty variable assignment.
answered Nov 25 '18 at 1:38
John KugelmanJohn Kugelman
245k54406459
add a comment |
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lookIn= `./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri`
^
Delete the extra space. Assignments must not have spaces around the equal sign.
With the space there, Bash parses the line as var=value command, which runs a command with $var temporarily set to "value". Or in this case, it interprets result of the backticks as a command name and lookIn= as an empty variable assignment.
answered Nov 25 '18 at 1:38
John KugelmanJohn Kugelman
245k54406459
add a comment |
lookIn= `./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri`
^
Delete the extra space. Assignments must not have spaces around the equal sign.
With the space there, Bash parses the line as var=value command, which runs a command with $var temporarily set to "value". Or in this case, it interprets result of the backticks as a command name and lookIn= as an empty variable assignment.
answered Nov 25 '18 at 1:38
John KugelmanJohn Kugelman
245k54406459
add a comment |
lookIn= `./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri`
^
Delete the extra space. Assignments must not have spaces around the equal sign.
With the space there, Bash parses the line as var=value command, which runs a command with $var temporarily set to "value". Or in this case, it interprets result of the backticks as a command name and lookIn= as an empty variable assignment.
answered Nov 25 '18 at 1:38
John KugelmanJohn Kugelman
245k54406459
lookIn= `./findGeneIn "$p" burgdorferi afzelii garinii hermsii miyamotoi parkeri`
^
Delete the extra space. Assignments must not have spaces around the equal sign.
With the space there, Bash parses the line as var=value command, which runs a command with $var temporarily set to "value". Or in this case, it interprets result of the backticks as a command name and lookIn= as an empty variable assignment.
answered Nov 25 '18 at 1:38
John KugelmanJohn Kugelman
245k54406459
answered Nov 25 '18 at 1:38
John KugelmanJohn Kugelman
245k54406459
answered Nov 25 '18 at 1:38
John KugelmanJohn Kugelman
245k54406459
answered Nov 25 '18 at 1:38
John KugelmanJohn Kugelman
245k54406459
245k54406459
add a comment |
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2
Remove the space after
lookIn=. Also, use$( )instead of backticks, and quote the filename (done < "$filename"). shellcheck.net is good at spotting common mistakes like these.– Gordon Davisson
Nov 25 '18 at 1:42