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2

I'm trying to calculate Euler's numb. but I'm having problems trying to display the result. This is what I have:

#Using a switch an case

"Euler's Number")
szAnswer=$(zenity --info --text "Enter a number")
result = "(1+1/$szAnswer)^$szAnswer" | bc -l
zenity --info --text "Euler's Numb: $result"

I'm able to input a number and all, but when it comes to giving me the output result it just stays blank. Any help is welcomed.

bash sh zenity

share|improve this question

asked 4 hours ago

escobarverasescobarveras

153

add a comment | 

2

I'm trying to calculate Euler's numb. but I'm having problems trying to display the result. This is what I have:

#Using a switch an case

"Euler's Number")
szAnswer=$(zenity --info --text "Enter a number")
result = "(1+1/$szAnswer)^$szAnswer" | bc -l
zenity --info --text "Euler's Numb: $result"

I'm able to input a number and all, but when it comes to giving me the output result it just stays blank. Any help is welcomed.

bash sh zenity

share|improve this question

asked 4 hours ago

escobarverasescobarveras

153

add a comment | 

I'm trying to calculate Euler's numb. but I'm having problems trying to display the result. This is what I have:

#Using a switch an case

"Euler's Number")
szAnswer=$(zenity --info --text "Enter a number")
result = "(1+1/$szAnswer)^$szAnswer" | bc -l
zenity --info --text "Euler's Numb: $result"

I'm able to input a number and all, but when it comes to giving me the output result it just stays blank. Any help is welcomed.

bash sh zenity

share|improve this question

asked 4 hours ago

escobarverasescobarveras

153

#Using a switch an case

"Euler's Number")
szAnswer=$(zenity --info --text "Enter a number")
result = "(1+1/$szAnswer)^$szAnswer" | bc -l
zenity --info --text "Euler's Numb: $result"

share|improve this question

asked 4 hours ago

escobarverasescobarveras

153

share|improve this question

asked 4 hours ago

escobarverasescobarveras

153

asked 4 hours ago

escobarverasescobarveras

153

asked 4 hours ago

escobarverasescobarveras

153

1 Answer
1

active

oldest

votes

2

The problem is result = "(1+1/$szAnswer)^$szAnswer" | bc -l line. It reads:

• execute command result with parameters = and "(1+1/$szAnswer)^$szAnswer"
  


• connect the stdout stream of the result command to bc command's stdin stream

Probably you're wondering why result is a command in this case. That's because variable assignments in shell scripting are made without spaces separating variable name and assigned value. You also want to send "(1+1/$szAnswer)^$szAnswer" to stdin of bc -l command,so you need something capable of writing to stdout

What should be done is

result=$( echo "(1+1/$szAnswer)^$szAnswer" | bc -l )`

Now you have result variable being assigned output of echo "(1+1/$szAnswer)^$szAnswer" | bc -l pipeline. The $(...) structure is called command substitution, and is generally used when command's output has to be reused in place of the command itself.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Sergiy KolodyazhnyySergiy Kolodyazhnyy

76.1k9159334

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  For bash, a here string would be another option: result=$(bc -l <<< "(1+1/$szAnswer)^$szAnswer")
  
  – steeldriver
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, here-string would probably be even more preferable for ksh and bash
  
  – Sergiy Kolodyazhnyy
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  TBH I hadn't even noticed the sh tag ...
  
  – steeldriver
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @steeldriver It's negligible since there's a high chance OP is using bash and since they haven't shown the full script.
  
  – Sergiy Kolodyazhnyy
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you for the help and also for explaining how are things being executed.
  
  – escobarveras
  4 hours ago
  

  
  
  
  

 | 
show 1 more comment

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2

The problem is result = "(1+1/$szAnswer)^$szAnswer" | bc -l line. It reads:

• execute command result with parameters = and "(1+1/$szAnswer)^$szAnswer"
  


• connect the stdout stream of the result command to bc command's stdin stream

Probably you're wondering why result is a command in this case. That's because variable assignments in shell scripting are made without spaces separating variable name and assigned value. You also want to send "(1+1/$szAnswer)^$szAnswer" to stdin of bc -l command,so you need something capable of writing to stdout

What should be done is

result=$( echo "(1+1/$szAnswer)^$szAnswer" | bc -l )`

Now you have result variable being assigned output of echo "(1+1/$szAnswer)^$szAnswer" | bc -l pipeline. The $(...) structure is called command substitution, and is generally used when command's output has to be reused in place of the command itself.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Sergiy KolodyazhnyySergiy Kolodyazhnyy

76.1k9159334

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  For bash, a here string would be another option: result=$(bc -l <<< "(1+1/$szAnswer)^$szAnswer")
  
  – steeldriver
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, here-string would probably be even more preferable for ksh and bash
  
  – Sergiy Kolodyazhnyy
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  TBH I hadn't even noticed the sh tag ...
  
  – steeldriver
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @steeldriver It's negligible since there's a high chance OP is using bash and since they haven't shown the full script.
  
  – Sergiy Kolodyazhnyy
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you for the help and also for explaining how are things being executed.
  
  – escobarveras
  4 hours ago
  

  
  
  
  

 | 
show 1 more comment

2

The problem is result = "(1+1/$szAnswer)^$szAnswer" | bc -l line. It reads:

• execute command result with parameters = and "(1+1/$szAnswer)^$szAnswer"
  


• connect the stdout stream of the result command to bc command's stdin stream

Probably you're wondering why result is a command in this case. That's because variable assignments in shell scripting are made without spaces separating variable name and assigned value. You also want to send "(1+1/$szAnswer)^$szAnswer" to stdin of bc -l command,so you need something capable of writing to stdout

What should be done is

result=$( echo "(1+1/$szAnswer)^$szAnswer" | bc -l )`

Now you have result variable being assigned output of echo "(1+1/$szAnswer)^$szAnswer" | bc -l pipeline. The $(...) structure is called command substitution, and is generally used when command's output has to be reused in place of the command itself.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Sergiy KolodyazhnyySergiy Kolodyazhnyy

76.1k9159334

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  For bash, a here string would be another option: result=$(bc -l <<< "(1+1/$szAnswer)^$szAnswer")
  
  – steeldriver
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, here-string would probably be even more preferable for ksh and bash
  
  – Sergiy Kolodyazhnyy
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  TBH I hadn't even noticed the sh tag ...
  
  – steeldriver
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @steeldriver It's negligible since there's a high chance OP is using bash and since they haven't shown the full script.
  
  – Sergiy Kolodyazhnyy
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you for the help and also for explaining how are things being executed.
  
  – escobarveras
  4 hours ago
  

  
  
  
  

 | 
show 1 more comment

The problem is result = "(1+1/$szAnswer)^$szAnswer" | bc -l line. It reads:

• execute command result with parameters = and "(1+1/$szAnswer)^$szAnswer"
  


• connect the stdout stream of the result command to bc command's stdin stream

Probably you're wondering why result is a command in this case. That's because variable assignments in shell scripting are made without spaces separating variable name and assigned value. You also want to send "(1+1/$szAnswer)^$szAnswer" to stdin of bc -l command,so you need something capable of writing to stdout

What should be done is

result=$( echo "(1+1/$szAnswer)^$szAnswer" | bc -l )`

Now you have result variable being assigned output of echo "(1+1/$szAnswer)^$szAnswer" | bc -l pipeline. The $(...) structure is called command substitution, and is generally used when command's output has to be reused in place of the command itself.

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Sergiy KolodyazhnyySergiy Kolodyazhnyy

76.1k9159334

result=$( echo "(1+1/$szAnswer)^$szAnswer" | bc -l )`

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Sergiy KolodyazhnyySergiy Kolodyazhnyy

76.1k9159334

answered 4 hours ago

Sergiy KolodyazhnyySergiy Kolodyazhnyy

76.1k9159334

answered 4 hours ago

Sergiy KolodyazhnyySergiy Kolodyazhnyy

76.1k9159334