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command:
/usr/bin/ruby “${SRCROOT}/generate-games.rb” “${SRCROOT}/sample-data.csv”
error: /usr/bin/ruby: No such file or directory --
“/Users/rays/Desktop/GenerateTestData/generate-games.rb” (LoadError)
Command PhaseScriptExecution failed with a nonzero exit codet I
Yet I confirmed the script is where it should be.
You are using UTF-8 quotes: “...” instead of standard ASCII quotes "...". Your shell will will only recognize ASCII quotes as delimiters, and therefore will interpret the UTF-8 quotes as part of the file name (which they obviously are not).
Conclusion:
Fix your quotes and that should fix your problem.
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I have a filename.txt with the following contents:
HEL50-C2
DEN50-C1
SFO5-C3
BRU50-C1
SCL50-C1
I have a program which runs this:
FOR /F %%G IN (filename.txt) DO (ECHO %%G)
But when I run it, it only returns a modified version of the last line of the file:
SCL50-C1C1
What am I missing? I think it has something to do with the hyphens, but I'm not sure how to account for them.
Turns out my file had only carriage return (\r) at the end of each line. When I added carriage return and line feed (\r\n) the script ran fine.
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I copied and running the command on my ubuntu 18.04 from here
https://kite.com/linux/ but got an error like:
$ bash -c “$(wget -q -O – https://linux.kite.com/dls/linux/current)”
bash: “”: command not found
$ type quote
quote is a function
quote ()
{
local quoted=${1//\'/\'\\\'\'};
printf "'%s'" "$quoted"
}
Any suggestions of the error?
You probably copied and pasted that from some word processor or website that turned the straight, regular, ASCII quotes " into pretty, curly Unicode quotes “ ”. Bash doesn't understand those. Just type them in by hand to fix the problem.
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I'm sure this is quite simple. However, it's just not working for me. What stupid thing am I doing wrong please? I am running the following shell script like this:
bash test1.sh
Here's the code:
#!/bin/bash
bluesman_a="Magic Slim"
bluesman_b=($echo "$bluesman_a" | sed "/s/Slim/Sam/")
echo $bluesman_b
I get:
syntax error near unexpected token `|'
Thanks for your time
You need to use "$(...)" to wrap a command to assign the output to a variable and you need to remove the first / in the sed replacement command. Also, you do not need to use echo to pass a variable to sed.
bluesman_b="$(sed 's/Slim/Sam/' <<< "$bluesman_a")"
Or, to replace Slim with Sam just once, use
bluesman_b="${bluesman_a/Slim/Sam}"
See 10.1. Manipulating Strings.
See the online Bash demo
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So I have a bash script
!#/bin/bash
while [ true ];do
ls -lah /sth/ | grep sth*
sleep 0.001
done
exit 0
I thoought thai it was ok but when I run it I get
line 7: syntax error: unexpected end of file
But the code has only 6 lines?
What may be a problem? I edited the file in linux, deleted unnecessary spaces but still my scropt doesn't work.
The shebang line is wrong. You are not running it under Bash at all.
#!/bin/bash
Notice the order of the sharp (#) and the bang (!).
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As per this answer: Unix Bash Shell Programming if directory exists, I'm trying to check if a directory exists. However, when I run this, I get line 1: [-d: command not found. What am I doing wrong here?
if [-d "~/.ssl"]; then
echo '~/.ssl directory already exists'
else
sudo mkdir ~/.ssl/
fi
[-d
is not a command.
[ -d
is the test command with the -d option.
Space matters.
(Also, the [ command needs to end with a ] parameter, which likewise has to be separated from other arguments by whitespace.)
That's the crux of the matter. There is another issue, though: If you quote the tilde, it doesn't expand. (This is one of the rare place where you may want to avoid quotes.) Quotes are great, though, so why not write "$HOME/.ssl"? (There's a subtle difference between ~ and "$HOME", but it doesn't matter for most uses.)
Honestly, all you really need is probably:
if mkdir -p ~/.ssl; then
# Do stuff with new directory
else
# Handle failure (but keep in mind `mkdir` will have its own error output)
fi