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0

Here is my simple script

#!/bin/sh

thefile=/home/ark/arkserver/ShooterGame/Saved/SaveIsland/1288804998.arktribe



while inotifywait "${thefile}" ; do

    a="`strings ${thefile} | tail -n 5 | head -n 1 | sed 's/<[^>]*>//g'`"

echo $a



curl -H "Content-Type: application/json" -X POST -d '{"username": "Island", "content": $a}' https://discordapp.com/api/webhooks/5738674701/OjRQiAWHq5mX0Tn2MfBlF-mI41TWrVYVAbOfXpeZWqo8



done

Here is the output

Setting up watches.

Watches established.

/home/ark/arkserver/ShooterGame/Saved/SaveIsland/1288804998.arktribe ATTRIB

"Day 600, 21:31:17: Phorce1 demolished a 'Campfire'!"

{"code": 50006, "message": "Cannot send an empty message"}Setting up watches.

Watches established.

I cannot figure out how to get the curl command properly quoted with the variable inserted in the middle. This is what the command should look like. This works when typed out on the command line

curl -H "Content-Type: application/json" -X POST -d '{"username": "test", "content": "Day 600, 14:51:00: Phorce1 demolished a 'Campfire'!"}' https://discordapp.com/api/webhooks/594723628738674701/OjRQn2MfBlF-FJVl1cWwMlD6UQf0c

I have tried many different forms of quoting. The fact that it echos $a as properly quoted but still breaks something in the command that uses it confuses me.

(Yes, I mangled my webhook address)

bash variable-substitution

share|improve this question

edited 1 hour ago

Phorce1

asked 13 hours ago

Phorce1Phorce1

2377 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of Is there any way to print value inside variable inside single quote?
  
  – rush
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Bash doesn't expand variable in single quotes. You have either use " instead, either put additional ' around your variable.
  
  – rush
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @rush Please note the answer I selected here is quite different from the replies to that question. I tried escaping the quotes around the variable but the "many nested quotes" nature of he curl command line cause that to fail. So, no, this is not a duplicate of that question.
  
  – Phorce1
  5 hours ago
  

  
  
  
  

add a comment | 

0

Here is my simple script

#!/bin/sh

thefile=/home/ark/arkserver/ShooterGame/Saved/SaveIsland/1288804998.arktribe



while inotifywait "${thefile}" ; do

    a="`strings ${thefile} | tail -n 5 | head -n 1 | sed 's/<[^>]*>//g'`"

echo $a



curl -H "Content-Type: application/json" -X POST -d '{"username": "Island", "content": $a}' https://discordapp.com/api/webhooks/5738674701/OjRQiAWHq5mX0Tn2MfBlF-mI41TWrVYVAbOfXpeZWqo8



done

Here is the output

Setting up watches.

Watches established.

/home/ark/arkserver/ShooterGame/Saved/SaveIsland/1288804998.arktribe ATTRIB

"Day 600, 21:31:17: Phorce1 demolished a 'Campfire'!"

{"code": 50006, "message": "Cannot send an empty message"}Setting up watches.

Watches established.

I cannot figure out how to get the curl command properly quoted with the variable inserted in the middle. This is what the command should look like. This works when typed out on the command line

curl -H "Content-Type: application/json" -X POST -d '{"username": "test", "content": "Day 600, 14:51:00: Phorce1 demolished a 'Campfire'!"}' https://discordapp.com/api/webhooks/594723628738674701/OjRQn2MfBlF-FJVl1cWwMlD6UQf0c

I have tried many different forms of quoting. The fact that it echos $a as properly quoted but still breaks something in the command that uses it confuses me.

(Yes, I mangled my webhook address)

bash variable-substitution

share|improve this question

edited 1 hour ago

Phorce1

asked 13 hours ago

Phorce1Phorce1

2377 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Possible duplicate of Is there any way to print value inside variable inside single quote?
  
  – rush
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Bash doesn't expand variable in single quotes. You have either use " instead, either put additional ' around your variable.
  
  – rush
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @rush Please note the answer I selected here is quite different from the replies to that question. I tried escaping the quotes around the variable but the "many nested quotes" nature of he curl command line cause that to fail. So, no, this is not a duplicate of that question.
  
  – Phorce1
  5 hours ago
  

  
  
  
  

add a comment | 

Here is my simple script

#!/bin/sh

thefile=/home/ark/arkserver/ShooterGame/Saved/SaveIsland/1288804998.arktribe



while inotifywait "${thefile}" ; do

    a="`strings ${thefile} | tail -n 5 | head -n 1 | sed 's/<[^>]*>//g'`"

echo $a



curl -H "Content-Type: application/json" -X POST -d '{"username": "Island", "content": $a}' https://discordapp.com/api/webhooks/5738674701/OjRQiAWHq5mX0Tn2MfBlF-mI41TWrVYVAbOfXpeZWqo8



done

Here is the output

Setting up watches.

Watches established.

/home/ark/arkserver/ShooterGame/Saved/SaveIsland/1288804998.arktribe ATTRIB

"Day 600, 21:31:17: Phorce1 demolished a 'Campfire'!"

{"code": 50006, "message": "Cannot send an empty message"}Setting up watches.

Watches established.

I cannot figure out how to get the curl command properly quoted with the variable inserted in the middle. This is what the command should look like. This works when typed out on the command line

curl -H "Content-Type: application/json" -X POST -d '{"username": "test", "content": "Day 600, 14:51:00: Phorce1 demolished a 'Campfire'!"}' https://discordapp.com/api/webhooks/594723628738674701/OjRQn2MfBlF-FJVl1cWwMlD6UQf0c

I have tried many different forms of quoting. The fact that it echos $a as properly quoted but still breaks something in the command that uses it confuses me.

(Yes, I mangled my webhook address)

bash variable-substitution

share|improve this question

edited 1 hour ago

Phorce1

asked 13 hours ago

Phorce1Phorce1

2377 bronze badges

#!/bin/sh

thefile=/home/ark/arkserver/ShooterGame/Saved/SaveIsland/1288804998.arktribe



while inotifywait "${thefile}" ; do

    a="`strings ${thefile} | tail -n 5 | head -n 1 | sed 's/<[^>]*>//g'`"

echo $a



curl -H "Content-Type: application/json" -X POST -d '{"username": "Island", "content": $a}' https://discordapp.com/api/webhooks/5738674701/OjRQiAWHq5mX0Tn2MfBlF-mI41TWrVYVAbOfXpeZWqo8



done

Setting up watches.

Watches established.

/home/ark/arkserver/ShooterGame/Saved/SaveIsland/1288804998.arktribe ATTRIB

"Day 600, 21:31:17: Phorce1 demolished a 'Campfire'!"

{"code": 50006, "message": "Cannot send an empty message"}Setting up watches.

Watches established.

curl -H "Content-Type: application/json" -X POST -d '{"username": "test", "content": "Day 600, 14:51:00: Phorce1 demolished a 'Campfire'!"}' https://discordapp.com/api/webhooks/594723628738674701/OjRQn2MfBlF-FJVl1cWwMlD6UQf0c

share|improve this question

edited 1 hour ago

Phorce1

asked 13 hours ago

Phorce1Phorce1

2377 bronze badges

share|improve this question

edited 1 hour ago

Phorce1

asked 13 hours ago

Phorce1Phorce1

2377 bronze badges

asked 13 hours ago

Phorce1Phorce1

2377 bronze badges

asked 13 hours ago

Phorce1Phorce1

2377 bronze badges

1 Answer
1

active

oldest

votes

2

Single quotes prevent variable expansion.

I would do this:

url=https://discordapp.com/api/webhooks/5738674701/OjRQiAWHq5mX0Tn2MfBlF-mI41TWrVYVAbOfXpeZWqo8

while inotifywait "$thefile" ; do

    a=$(strings "$thefile" | tail -n 5 | head -n 1 | sed 's/<[^>]*>//g')

    echo "$a"

    payload=$(printf '{"username": "Island", "content": "%s"}' "$a")

    curl -H "Content-Type: application/json" -X POST -d "$payload" "$url" 

done

You don't need to escape double quotes inside of single quotes. In fact, if you do, the escape will be preserved as a literal character.

Always quote your variables, unless you know exactly what might happen if you don't.

(This is a matter of personal style) You don't need to brace a variable inside of quotes, unless you need to disambiguate the variable name from surrounding text:

echo "$aVariable"

echo "${aVariable}_abc"

share|improve this answer

answered 13 hours ago

glenn jackmanglenn jackman

54.7k66 gold badges7676 silver badges115115 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you. I write little scripts so seldom I had to go look up a refresher on printf to figure out what the %s was doing. I was also looking at other similar questions that suggested creating a function for the curl command but this, fixing my quoting using variables, feels more right to me.
  
  – Phorce1
  13 hours ago
  
  
  

  
  
  
  

add a comment | 

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2

Single quotes prevent variable expansion.

I would do this:

url=https://discordapp.com/api/webhooks/5738674701/OjRQiAWHq5mX0Tn2MfBlF-mI41TWrVYVAbOfXpeZWqo8

while inotifywait "$thefile" ; do

    a=$(strings "$thefile" | tail -n 5 | head -n 1 | sed 's/<[^>]*>//g')

    echo "$a"

    payload=$(printf '{"username": "Island", "content": "%s"}' "$a")

    curl -H "Content-Type: application/json" -X POST -d "$payload" "$url" 

done

You don't need to escape double quotes inside of single quotes. In fact, if you do, the escape will be preserved as a literal character.

Always quote your variables, unless you know exactly what might happen if you don't.

(This is a matter of personal style) You don't need to brace a variable inside of quotes, unless you need to disambiguate the variable name from surrounding text:

echo "$aVariable"

echo "${aVariable}_abc"

share|improve this answer

answered 13 hours ago

glenn jackmanglenn jackman

54.7k66 gold badges7676 silver badges115115 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you. I write little scripts so seldom I had to go look up a refresher on printf to figure out what the %s was doing. I was also looking at other similar questions that suggested creating a function for the curl command but this, fixing my quoting using variables, feels more right to me.
  
  – Phorce1
  13 hours ago
  
  
  

  
  
  
  

add a comment | 

2

Single quotes prevent variable expansion.

I would do this:

url=https://discordapp.com/api/webhooks/5738674701/OjRQiAWHq5mX0Tn2MfBlF-mI41TWrVYVAbOfXpeZWqo8

while inotifywait "$thefile" ; do

    a=$(strings "$thefile" | tail -n 5 | head -n 1 | sed 's/<[^>]*>//g')

    echo "$a"

    payload=$(printf '{"username": "Island", "content": "%s"}' "$a")

    curl -H "Content-Type: application/json" -X POST -d "$payload" "$url" 

done

You don't need to escape double quotes inside of single quotes. In fact, if you do, the escape will be preserved as a literal character.

Always quote your variables, unless you know exactly what might happen if you don't.

(This is a matter of personal style) You don't need to brace a variable inside of quotes, unless you need to disambiguate the variable name from surrounding text:

echo "$aVariable"

echo "${aVariable}_abc"

share|improve this answer

answered 13 hours ago

glenn jackmanglenn jackman

54.7k66 gold badges7676 silver badges115115 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thank you. I write little scripts so seldom I had to go look up a refresher on printf to figure out what the %s was doing. I was also looking at other similar questions that suggested creating a function for the curl command but this, fixing my quoting using variables, feels more right to me.
  
  – Phorce1
  13 hours ago
  
  
  

  
  
  
  

add a comment | 

Single quotes prevent variable expansion.

I would do this:

url=https://discordapp.com/api/webhooks/5738674701/OjRQiAWHq5mX0Tn2MfBlF-mI41TWrVYVAbOfXpeZWqo8

while inotifywait "$thefile" ; do

    a=$(strings "$thefile" | tail -n 5 | head -n 1 | sed 's/<[^>]*>//g')

    echo "$a"

    payload=$(printf '{"username": "Island", "content": "%s"}' "$a")

    curl -H "Content-Type: application/json" -X POST -d "$payload" "$url" 

done

You don't need to escape double quotes inside of single quotes. In fact, if you do, the escape will be preserved as a literal character.

Always quote your variables, unless you know exactly what might happen if you don't.

(This is a matter of personal style) You don't need to brace a variable inside of quotes, unless you need to disambiguate the variable name from surrounding text:

echo "$aVariable"

echo "${aVariable}_abc"

share|improve this answer

answered 13 hours ago

glenn jackmanglenn jackman

54.7k66 gold badges7676 silver badges115115 bronze badges

url=https://discordapp.com/api/webhooks/5738674701/OjRQiAWHq5mX0Tn2MfBlF-mI41TWrVYVAbOfXpeZWqo8

while inotifywait "$thefile" ; do

    a=$(strings "$thefile" | tail -n 5 | head -n 1 | sed 's/<[^>]*>//g')

    echo "$a"

    payload=$(printf '{"username": "Island", "content": "%s"}' "$a")

    curl -H "Content-Type: application/json" -X POST -d "$payload" "$url" 

done

echo "$aVariable"

echo "${aVariable}_abc"

share|improve this answer

answered 13 hours ago

glenn jackmanglenn jackman

54.7k66 gold badges7676 silver badges115115 bronze badges

answered 13 hours ago

glenn jackmanglenn jackman

54.7k66 gold badges7676 silver badges115115 bronze badges

answered 13 hours ago

glenn jackmanglenn jackman

54.7k66 gold badges7676 silver badges115115 bronze badges