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Using spd-say in a bash script function
The Next CEO of Stack OverflowRedirecting output when I execute “bash -c …” from a C programfor loop in bash functionWhat's the best way to make a bash function in a script as a parameter when running via command line?Execute only if it is a bash functionFunction for archiving arbitrary files with encryptionHow to make a function available to the command `parallel` (GNU)?How to catch and handle nonzero exit status within a Bash function?Bash function assign value to passed parameterPassing multiple arguments to sudo within functionHow to calculate say output duration?
I'm sure this is fairly elementary, but I can't figure it out.
My script:
#!/bin/bash
sez ()
echo $1
spd-say "$1"
sez "does this work"
sez "this does work"
What I'm trying to make happen is use spd-say in a function to make the computer talk to me.
The echo portion of my function works. It outputs both lines of text that I feed to it in the expected order. However, the spd-say part doesn't. It only ever says the last line. I'm assuming it's because the second command is "overwriting" the output of the first because it's trying to run them in parallel to the same output. I've tried adding ;wait, &&, and various other things to the end of the sez command, on the next line after, within the function on the spd-say command, etc, but everything I'm trying isn't helping.
What am I doing wrong?
bash shell-script audio function text-to-speech
edited Aug 20 '18 at 22:35
Rui F Ribeiro
41.8k1483142
asked Oct 9 '16 at 15:06
lightwinglightwing
498
add a comment |
I'm sure this is fairly elementary, but I can't figure it out.
My script:
#!/bin/bash
sez ()
echo $1
spd-say "$1"
sez "does this work"
sez "this does work"
What I'm trying to make happen is use spd-say in a function to make the computer talk to me.
The echo portion of my function works. It outputs both lines of text that I feed to it in the expected order. However, the spd-say part doesn't. It only ever says the last line. I'm assuming it's because the second command is "overwriting" the output of the first because it's trying to run them in parallel to the same output. I've tried adding ;wait, &&, and various other things to the end of the sez command, on the next line after, within the function on the spd-say command, etc, but everything I'm trying isn't helping.
What am I doing wrong?
bash shell-script audio function text-to-speech
edited Aug 20 '18 at 22:35
Rui F Ribeiro
41.8k1483142
asked Oct 9 '16 at 15:06
lightwinglightwing
498
add a comment |
I'm sure this is fairly elementary, but I can't figure it out.
My script:
#!/bin/bash
sez ()
echo $1
spd-say "$1"
sez "does this work"
sez "this does work"
What I'm trying to make happen is use spd-say in a function to make the computer talk to me.
The echo portion of my function works. It outputs both lines of text that I feed to it in the expected order. However, the spd-say part doesn't. It only ever says the last line. I'm assuming it's because the second command is "overwriting" the output of the first because it's trying to run them in parallel to the same output. I've tried adding ;wait, &&, and various other things to the end of the sez command, on the next line after, within the function on the spd-say command, etc, but everything I'm trying isn't helping.
What am I doing wrong?
bash shell-script audio function text-to-speech
edited Aug 20 '18 at 22:35
Rui F Ribeiro
41.8k1483142
asked Oct 9 '16 at 15:06
lightwinglightwing
498
I'm sure this is fairly elementary, but I can't figure it out.
My script:
#!/bin/bash
sez ()
echo $1
spd-say "$1"
sez "does this work"
sez "this does work"
What I'm trying to make happen is use spd-say in a function to make the computer talk to me.
The echo portion of my function works. It outputs both lines of text that I feed to it in the expected order. However, the spd-say part doesn't. It only ever says the last line. I'm assuming it's because the second command is "overwriting" the output of the first because it's trying to run them in parallel to the same output. I've tried adding ;wait, &&, and various other things to the end of the sez command, on the next line after, within the function on the spd-say command, etc, but everything I'm trying isn't helping.
What am I doing wrong?
bash shell-script audio function text-to-speech
bash shell-script audio function text-to-speech
edited Aug 20 '18 at 22:35
Rui F Ribeiro
41.8k1483142
asked Oct 9 '16 at 15:06
lightwinglightwing
498
edited Aug 20 '18 at 22:35
Rui F Ribeiro
41.8k1483142
asked Oct 9 '16 at 15:06
lightwinglightwing
498
edited Aug 20 '18 at 22:35
Rui F Ribeiro
41.8k1483142
edited Aug 20 '18 at 22:35
Rui F Ribeiro
41.8k1483142
edited Aug 20 '18 at 22:35
Rui F Ribeiro
41.8k1483142
41.8k1483142
asked Oct 9 '16 at 15:06
lightwinglightwing
498
asked Oct 9 '16 at 15:06
lightwinglightwing
498
asked Oct 9 '16 at 15:06
lightwinglightwing
498
498
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
I had a similar problem and I found a workaround. Instead of using spd-say I used espeak directly.
edited Dec 4 '16 at 14:05
Jeff Schaller♦
44.3k1162143
answered Dec 4 '16 at 10:00
alagrisalagris
465
add a comment |
spd-say requires a suitable shell environment. cron doesn't execute tasks in your regular shell environment.
You can try changing your script such as in this example:
#!/bin/bash
source $HOME/.profile
sez ()
echo $1
spd-say "$1"
sez "does this work"
sez "this does work"
See also this question for more information: http://unix.stackexchange.com/questions/67940/cron-ignores-variables-defined-in-bashrc-and-bash-profile
answered 2 days ago
user344053user344053
1
New contributor
user344053 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
I had a similar problem and I found a workaround. Instead of using spd-say I used espeak directly.
edited Dec 4 '16 at 14:05
Jeff Schaller♦
44.3k1162143
answered Dec 4 '16 at 10:00
alagrisalagris
465
add a comment |
I had a similar problem and I found a workaround. Instead of using spd-say I used espeak directly.
edited Dec 4 '16 at 14:05
Jeff Schaller♦
44.3k1162143
answered Dec 4 '16 at 10:00
alagrisalagris
465
add a comment |
I had a similar problem and I found a workaround. Instead of using spd-say I used espeak directly.
edited Dec 4 '16 at 14:05
Jeff Schaller♦
44.3k1162143
answered Dec 4 '16 at 10:00
alagrisalagris
465
I had a similar problem and I found a workaround. Instead of using spd-say I used espeak directly.
edited Dec 4 '16 at 14:05
Jeff Schaller♦
44.3k1162143
answered Dec 4 '16 at 10:00
alagrisalagris
465
edited Dec 4 '16 at 14:05
Jeff Schaller♦
44.3k1162143
edited Dec 4 '16 at 14:05
Jeff Schaller♦
44.3k1162143
edited Dec 4 '16 at 14:05
Jeff Schaller♦
44.3k1162143
44.3k1162143
answered Dec 4 '16 at 10:00
alagrisalagris
465
answered Dec 4 '16 at 10:00
alagrisalagris
465
answered Dec 4 '16 at 10:00
alagrisalagris
465
465
add a comment |
add a comment |
spd-say requires a suitable shell environment. cron doesn't execute tasks in your regular shell environment.
You can try changing your script such as in this example:
#!/bin/bash
source $HOME/.profile
sez ()
echo $1
spd-say "$1"
sez "does this work"
sez "this does work"
See also this question for more information: http://unix.stackexchange.com/questions/67940/cron-ignores-variables-defined-in-bashrc-and-bash-profile
answered 2 days ago
user344053user344053
1
New contributor
user344053 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
spd-say requires a suitable shell environment. cron doesn't execute tasks in your regular shell environment.
You can try changing your script such as in this example:
#!/bin/bash
source $HOME/.profile
sez ()
echo $1
spd-say "$1"
sez "does this work"
sez "this does work"
See also this question for more information: http://unix.stackexchange.com/questions/67940/cron-ignores-variables-defined-in-bashrc-and-bash-profile
answered 2 days ago
user344053user344053
1
New contributor
user344053 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
spd-say requires a suitable shell environment. cron doesn't execute tasks in your regular shell environment.
You can try changing your script such as in this example:
#!/bin/bash
source $HOME/.profile
sez ()
echo $1
spd-say "$1"
sez "does this work"
sez "this does work"
See also this question for more information: http://unix.stackexchange.com/questions/67940/cron-ignores-variables-defined-in-bashrc-and-bash-profile
answered 2 days ago
user344053user344053
1
New contributor
user344053 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
spd-say requires a suitable shell environment. cron doesn't execute tasks in your regular shell environment.
You can try changing your script such as in this example:
#!/bin/bash
source $HOME/.profile
sez ()
echo $1
spd-say "$1"
sez "does this work"
sez "this does work"
See also this question for more information: http://unix.stackexchange.com/questions/67940/cron-ignores-variables-defined-in-bashrc-and-bash-profile
answered 2 days ago
user344053user344053
1
New contributor
user344053 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
user344053user344053
1
New contributor
user344053 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
user344053user344053
1
answered 2 days ago
user344053user344053
1
1
New contributor
user344053 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user344053 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user344053 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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