Passing same value to more variables
I have written a script like this:
#!/bin/bash
set -e
dirs=()
dirs[0]=`export DIR_1=./dir1 && echo $DIR_1`
dirs[1]=`export DIR_2=./dir2 && echo $DIR_2`
for v in ${dirs[@]}; do
echo $v;
mkdir -p $v
done
# Here is my question: why it is null???
echo $DIR_1
How can I both do export the param and echo the value in same line?
bash
edited Nov 21 at 13:21
Biffen
4,16352129
asked Nov 20 at 6:54
Zhao Rocky
66431128
|
show 1 more comment
I have written a script like this:
#!/bin/bash
set -e
dirs=()
dirs[0]=`export DIR_1=./dir1 && echo $DIR_1`
dirs[1]=`export DIR_2=./dir2 && echo $DIR_2`
for v in ${dirs[@]}; do
echo $v;
mkdir -p $v
done
# Here is my question: why it is null???
echo $DIR_1
How can I both do export the param and echo the value in same line?
bash
edited Nov 21 at 13:21
Biffen
4,16352129
asked Nov 20 at 6:54
Zhao Rocky
66431128
...something like export DIR_1=./dir1; echo $DIR_1?... (instead of echo you can use any statement)
– linuxfan
Nov 20 at 7:00
what do you mean "instead of echo"?
– Zhao Rocky
Nov 20 at 7:08
1
Are you running the echo $DIR_1 after running your script? In which case it will be null. If you want to persist the value of environment variables you set it a script to be available after the script exits then run it as ". <script name>" That is a dot followed by a space followed by the script file name.
– Luv
Nov 20 at 7:23
Its hard to see what you are trying to achieve here.DIR_1=./dir1;dirs[0]=$DIR_1;echo $DIR_1saves you a whole subshell. What motivates the&&- assigning a constant to a variable will never fail surely.
– Niall Cosgrove
Nov 20 at 7:30
@NiallCosgrove yes, 3 lines. that's what I mean "I wanna do it in one line" ;) I donnot know if that's possible though ;) just like "a=b=1 // a=1 also b =1"
– Zhao Rocky
Nov 20 at 7:35
|
show 1 more comment
I have written a script like this:
#!/bin/bash
set -e
dirs=()
dirs[0]=`export DIR_1=./dir1 && echo $DIR_1`
dirs[1]=`export DIR_2=./dir2 && echo $DIR_2`
for v in ${dirs[@]}; do
echo $v;
mkdir -p $v
done
# Here is my question: why it is null???
echo $DIR_1
How can I both do export the param and echo the value in same line?
bash
edited Nov 21 at 13:21
Biffen
4,16352129
asked Nov 20 at 6:54
Zhao Rocky
66431128
I have written a script like this:
#!/bin/bash
set -e
dirs=()
dirs[0]=`export DIR_1=./dir1 && echo $DIR_1`
dirs[1]=`export DIR_2=./dir2 && echo $DIR_2`
for v in ${dirs[@]}; do
echo $v;
mkdir -p $v
done
# Here is my question: why it is null???
echo $DIR_1
How can I both do export the param and echo the value in same line?
bash
bash
edited Nov 21 at 13:21
Biffen
4,16352129
asked Nov 20 at 6:54
Zhao Rocky
66431128
edited Nov 21 at 13:21
Biffen
4,16352129
asked Nov 20 at 6:54
Zhao Rocky
66431128
edited Nov 21 at 13:21
Biffen
4,16352129
edited Nov 21 at 13:21
Biffen
4,16352129
edited Nov 21 at 13:21
Biffen
4,16352129
4,16352129
asked Nov 20 at 6:54
Zhao Rocky
66431128
asked Nov 20 at 6:54
Zhao Rocky
66431128
asked Nov 20 at 6:54
Zhao Rocky
66431128
66431128
...something like export DIR_1=./dir1; echo $DIR_1?... (instead of echo you can use any statement)
– linuxfan
Nov 20 at 7:00
what do you mean "instead of echo"?
– Zhao Rocky
Nov 20 at 7:08
1
Are you running the echo $DIR_1 after running your script? In which case it will be null. If you want to persist the value of environment variables you set it a script to be available after the script exits then run it as ". <script name>" That is a dot followed by a space followed by the script file name.
– Luv
Nov 20 at 7:23
Its hard to see what you are trying to achieve here.DIR_1=./dir1;dirs[0]=$DIR_1;echo $DIR_1saves you a whole subshell. What motivates the&&- assigning a constant to a variable will never fail surely.
– Niall Cosgrove
Nov 20 at 7:30
@NiallCosgrove yes, 3 lines. that's what I mean "I wanna do it in one line" ;) I donnot know if that's possible though ;) just like "a=b=1 // a=1 also b =1"
– Zhao Rocky
Nov 20 at 7:35
|
show 1 more comment
...something like export DIR_1=./dir1; echo $DIR_1?... (instead of echo you can use any statement)
– linuxfan
Nov 20 at 7:00
what do you mean "instead of echo"?
– Zhao Rocky
Nov 20 at 7:08
1
Are you running the echo $DIR_1 after running your script? In which case it will be null. If you want to persist the value of environment variables you set it a script to be available after the script exits then run it as ". <script name>" That is a dot followed by a space followed by the script file name.
– Luv
Nov 20 at 7:23
Its hard to see what you are trying to achieve here.DIR_1=./dir1;dirs[0]=$DIR_1;echo $DIR_1saves you a whole subshell. What motivates the&&- assigning a constant to a variable will never fail surely.
– Niall Cosgrove
Nov 20 at 7:30
@NiallCosgrove yes, 3 lines. that's what I mean "I wanna do it in one line" ;) I donnot know if that's possible though ;) just like "a=b=1 // a=1 also b =1"
– Zhao Rocky
Nov 20 at 7:35
...something like export DIR_1=./dir1; echo $DIR_1?... (instead of echo you can use any statement)
– linuxfan
Nov 20 at 7:00
...something like export DIR_1=./dir1; echo $DIR_1?... (instead of echo you can use any statement)
– linuxfan
Nov 20 at 7:00
what do you mean "instead of echo"?
– Zhao Rocky
Nov 20 at 7:08
what do you mean "instead of echo"?
– Zhao Rocky
Nov 20 at 7:08
1
1
Are you running the echo $DIR_1 after running your script? In which case it will be null. If you want to persist the value of environment variables you set it a script to be available after the script exits then run it as ". <script name>" That is a dot followed by a space followed by the script file name.
– Luv
Nov 20 at 7:23
Are you running the echo $DIR_1 after running your script? In which case it will be null. If you want to persist the value of environment variables you set it a script to be available after the script exits then run it as ". <script name>" That is a dot followed by a space followed by the script file name.
– Luv
Nov 20 at 7:23
Its hard to see what you are trying to achieve here.
DIR_1=./dir1;dirs[0]=$DIR_1;echo $DIR_1 saves you a whole subshell. What motivates the && - assigning a constant to a variable will never fail surely.– Niall Cosgrove
Nov 20 at 7:30
Its hard to see what you are trying to achieve here.
DIR_1=./dir1;dirs[0]=$DIR_1;echo $DIR_1 saves you a whole subshell. What motivates the && - assigning a constant to a variable will never fail surely.– Niall Cosgrove
Nov 20 at 7:30
@NiallCosgrove yes, 3 lines. that's what I mean "I wanna do it in one line" ;) I donnot know if that's possible though ;) just like "a=b=1 // a=1 also b =1"
– Zhao Rocky
Nov 20 at 7:35
@NiallCosgrove yes, 3 lines. that's what I mean "I wanna do it in one line" ;) I donnot know if that's possible though ;) just like "a=b=1 // a=1 also b =1"
– Zhao Rocky
Nov 20 at 7:35
|
show 1 more comment
2 Answers
2
active
oldest
votes
The "<<<" operator in bash is called a here string.
If you want to do multiple assignments on a single line use read as follows
read -r var1 var2 var3 <<< $(echo val1 val2 val3)
Each word in the string you echo will get assigned to each variable.
answered Nov 20 at 7:57
Niall Cosgrove
1,1681922
I accept it. Though I have to write the same value twice
– Zhao Rocky
Nov 20 at 7:59
@ZhaoRocky Yes that's true, but I cant think of a language where you can saya,b=1- though I'm willing to be contradicted if someone knows of one.
– Niall Cosgrove
Nov 20 at 8:02
lol.................
– Zhao Rocky
Nov 20 at 8:05
add a comment |
After my comment, I post this reply which explains what I meant with "instead of echo..."
The original line of code is:
dirs[0]=`export DIR_1=./dir1 && echo $DIR_1`
Here, the export does not work because it is executed under another instance of the shell, as you used the back ticks. In the child shell the export is executed, but the modified environment can not be "ported back".
Instead, to obtain the same, but working, result, you can write:
export DIR_1=./dir1; dirs[0]=$(echo $DIR_1)
The above line is equivalent to your original one, it makes exactly the same thing, included the strange way to assign to dir[0], but the export is executed in the local shell and exported to all the future childs. What I meant was you can write any command after the "export DIR_1...", even dirs[0]=...
edited Nov 21 at 13:22
Biffen
4,16352129
answered Nov 21 at 12:06
linuxfan
2,0722719
@Biffen thank you for editing, these backticks drive me crazy with my keyboard!
– linuxfan
Nov 21 at 13:33
I probably should have talked about this in my own answer. Good work :)
– Niall Cosgrove
Nov 21 at 13:53
@linuxfan thanks a lot for making that clear. I agree with you. Just for fun to know if any possible in one expression ;)
– Zhao Rocky
Nov 22 at 7:56
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
The "<<<" operator in bash is called a here string.
If you want to do multiple assignments on a single line use read as follows
read -r var1 var2 var3 <<< $(echo val1 val2 val3)
Each word in the string you echo will get assigned to each variable.
answered Nov 20 at 7:57
Niall Cosgrove
1,1681922
I accept it. Though I have to write the same value twice
– Zhao Rocky
Nov 20 at 7:59
@ZhaoRocky Yes that's true, but I cant think of a language where you can saya,b=1- though I'm willing to be contradicted if someone knows of one.
– Niall Cosgrove
Nov 20 at 8:02
lol.................
– Zhao Rocky
Nov 20 at 8:05
add a comment |
The "<<<" operator in bash is called a here string.
If you want to do multiple assignments on a single line use read as follows
read -r var1 var2 var3 <<< $(echo val1 val2 val3)
Each word in the string you echo will get assigned to each variable.
answered Nov 20 at 7:57
Niall Cosgrove
1,1681922
I accept it. Though I have to write the same value twice
– Zhao Rocky
Nov 20 at 7:59
@ZhaoRocky Yes that's true, but I cant think of a language where you can saya,b=1- though I'm willing to be contradicted if someone knows of one.
– Niall Cosgrove
Nov 20 at 8:02
lol.................
– Zhao Rocky
Nov 20 at 8:05
add a comment |
The "<<<" operator in bash is called a here string.
If you want to do multiple assignments on a single line use read as follows
read -r var1 var2 var3 <<< $(echo val1 val2 val3)
Each word in the string you echo will get assigned to each variable.
answered Nov 20 at 7:57
Niall Cosgrove
1,1681922
The "<<<" operator in bash is called a here string.
If you want to do multiple assignments on a single line use read as follows
read -r var1 var2 var3 <<< $(echo val1 val2 val3)
Each word in the string you echo will get assigned to each variable.
answered Nov 20 at 7:57
Niall Cosgrove
1,1681922
answered Nov 20 at 7:57
Niall Cosgrove
1,1681922
answered Nov 20 at 7:57
Niall Cosgrove
1,1681922
answered Nov 20 at 7:57
Niall Cosgrove
1,1681922
1,1681922
I accept it. Though I have to write the same value twice
– Zhao Rocky
Nov 20 at 7:59
@ZhaoRocky Yes that's true, but I cant think of a language where you can saya,b=1- though I'm willing to be contradicted if someone knows of one.
– Niall Cosgrove
Nov 20 at 8:02
lol.................
– Zhao Rocky
Nov 20 at 8:05
add a comment |
I accept it. Though I have to write the same value twice
– Zhao Rocky
Nov 20 at 7:59
@ZhaoRocky Yes that's true, but I cant think of a language where you can saya,b=1- though I'm willing to be contradicted if someone knows of one.
– Niall Cosgrove
Nov 20 at 8:02
lol.................
– Zhao Rocky
Nov 20 at 8:05
I accept it. Though I have to write the same value twice
– Zhao Rocky
Nov 20 at 7:59
I accept it. Though I have to write the same value twice
– Zhao Rocky
Nov 20 at 7:59
@ZhaoRocky Yes that's true, but I cant think of a language where you can say
a,b=1 - though I'm willing to be contradicted if someone knows of one.– Niall Cosgrove
Nov 20 at 8:02
@ZhaoRocky Yes that's true, but I cant think of a language where you can say
a,b=1 - though I'm willing to be contradicted if someone knows of one.– Niall Cosgrove
Nov 20 at 8:02
lol.................
– Zhao Rocky
Nov 20 at 8:05
lol.................
– Zhao Rocky
Nov 20 at 8:05
add a comment |
After my comment, I post this reply which explains what I meant with "instead of echo..."
The original line of code is:
dirs[0]=`export DIR_1=./dir1 && echo $DIR_1`
Here, the export does not work because it is executed under another instance of the shell, as you used the back ticks. In the child shell the export is executed, but the modified environment can not be "ported back".
Instead, to obtain the same, but working, result, you can write:
export DIR_1=./dir1; dirs[0]=$(echo $DIR_1)
The above line is equivalent to your original one, it makes exactly the same thing, included the strange way to assign to dir[0], but the export is executed in the local shell and exported to all the future childs. What I meant was you can write any command after the "export DIR_1...", even dirs[0]=...
edited Nov 21 at 13:22
Biffen
4,16352129
answered Nov 21 at 12:06
linuxfan
2,0722719
@Biffen thank you for editing, these backticks drive me crazy with my keyboard!
– linuxfan
Nov 21 at 13:33
I probably should have talked about this in my own answer. Good work :)
– Niall Cosgrove
Nov 21 at 13:53
@linuxfan thanks a lot for making that clear. I agree with you. Just for fun to know if any possible in one expression ;)
– Zhao Rocky
Nov 22 at 7:56
add a comment |
After my comment, I post this reply which explains what I meant with "instead of echo..."
The original line of code is:
dirs[0]=`export DIR_1=./dir1 && echo $DIR_1`
Here, the export does not work because it is executed under another instance of the shell, as you used the back ticks. In the child shell the export is executed, but the modified environment can not be "ported back".
Instead, to obtain the same, but working, result, you can write:
export DIR_1=./dir1; dirs[0]=$(echo $DIR_1)
The above line is equivalent to your original one, it makes exactly the same thing, included the strange way to assign to dir[0], but the export is executed in the local shell and exported to all the future childs. What I meant was you can write any command after the "export DIR_1...", even dirs[0]=...
edited Nov 21 at 13:22
Biffen
4,16352129
answered Nov 21 at 12:06
linuxfan
2,0722719
@Biffen thank you for editing, these backticks drive me crazy with my keyboard!
– linuxfan
Nov 21 at 13:33
I probably should have talked about this in my own answer. Good work :)
– Niall Cosgrove
Nov 21 at 13:53
@linuxfan thanks a lot for making that clear. I agree with you. Just for fun to know if any possible in one expression ;)
– Zhao Rocky
Nov 22 at 7:56
add a comment |
After my comment, I post this reply which explains what I meant with "instead of echo..."
The original line of code is:
dirs[0]=`export DIR_1=./dir1 && echo $DIR_1`
Here, the export does not work because it is executed under another instance of the shell, as you used the back ticks. In the child shell the export is executed, but the modified environment can not be "ported back".
Instead, to obtain the same, but working, result, you can write:
export DIR_1=./dir1; dirs[0]=$(echo $DIR_1)
The above line is equivalent to your original one, it makes exactly the same thing, included the strange way to assign to dir[0], but the export is executed in the local shell and exported to all the future childs. What I meant was you can write any command after the "export DIR_1...", even dirs[0]=...
edited Nov 21 at 13:22
Biffen
4,16352129
answered Nov 21 at 12:06
linuxfan
2,0722719
After my comment, I post this reply which explains what I meant with "instead of echo..."
The original line of code is:
dirs[0]=`export DIR_1=./dir1 && echo $DIR_1`
Here, the export does not work because it is executed under another instance of the shell, as you used the back ticks. In the child shell the export is executed, but the modified environment can not be "ported back".
Instead, to obtain the same, but working, result, you can write:
export DIR_1=./dir1; dirs[0]=$(echo $DIR_1)
The above line is equivalent to your original one, it makes exactly the same thing, included the strange way to assign to dir[0], but the export is executed in the local shell and exported to all the future childs. What I meant was you can write any command after the "export DIR_1...", even dirs[0]=...
edited Nov 21 at 13:22
Biffen
4,16352129
answered Nov 21 at 12:06
linuxfan
2,0722719
edited Nov 21 at 13:22
Biffen
4,16352129
edited Nov 21 at 13:22
Biffen
4,16352129
edited Nov 21 at 13:22
Biffen
4,16352129
4,16352129
answered Nov 21 at 12:06
linuxfan
2,0722719
answered Nov 21 at 12:06
linuxfan
2,0722719
answered Nov 21 at 12:06
linuxfan
2,0722719
2,0722719
@Biffen thank you for editing, these backticks drive me crazy with my keyboard!
– linuxfan
Nov 21 at 13:33
I probably should have talked about this in my own answer. Good work :)
– Niall Cosgrove
Nov 21 at 13:53
@linuxfan thanks a lot for making that clear. I agree with you. Just for fun to know if any possible in one expression ;)
– Zhao Rocky
Nov 22 at 7:56
add a comment |
@Biffen thank you for editing, these backticks drive me crazy with my keyboard!
– linuxfan
Nov 21 at 13:33
I probably should have talked about this in my own answer. Good work :)
– Niall Cosgrove
Nov 21 at 13:53
@linuxfan thanks a lot for making that clear. I agree with you. Just for fun to know if any possible in one expression ;)
– Zhao Rocky
Nov 22 at 7:56
@Biffen thank you for editing, these backticks drive me crazy with my keyboard!
– linuxfan
Nov 21 at 13:33
@Biffen thank you for editing, these backticks drive me crazy with my keyboard!
– linuxfan
Nov 21 at 13:33
I probably should have talked about this in my own answer. Good work :)
– Niall Cosgrove
Nov 21 at 13:53
I probably should have talked about this in my own answer. Good work :)
– Niall Cosgrove
Nov 21 at 13:53
@linuxfan thanks a lot for making that clear. I agree with you. Just for fun to know if any possible in one expression ;)
– Zhao Rocky
Nov 22 at 7:56
@linuxfan thanks a lot for making that clear. I agree with you. Just for fun to know if any possible in one expression ;)
– Zhao Rocky
Nov 22 at 7:56
add a comment |
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...something like export DIR_1=./dir1; echo $DIR_1?... (instead of echo you can use any statement)
– linuxfan
Nov 20 at 7:00
what do you mean "instead of echo"?
– Zhao Rocky
Nov 20 at 7:08
1
Are you running the echo $DIR_1 after running your script? In which case it will be null. If you want to persist the value of environment variables you set it a script to be available after the script exits then run it as ". <script name>" That is a dot followed by a space followed by the script file name.
– Luv
Nov 20 at 7:23
Its hard to see what you are trying to achieve here.
DIR_1=./dir1;dirs[0]=$DIR_1;echo $DIR_1saves you a whole subshell. What motivates the&&- assigning a constant to a variable will never fail surely.– Niall Cosgrove
Nov 20 at 7:30
@NiallCosgrove yes, 3 lines. that's what I mean "I wanna do it in one line" ;) I donnot know if that's possible though ;) just like "a=b=1 // a=1 also b =1"
– Zhao Rocky
Nov 20 at 7:35