When I'm trying to run test.sh script I've always receive error from curl:
curl: (6) Couldn't resolve host 'application'
test.sh:
#!/bin/sh
CT="Content-Type:\ application/json"
TEST="curl http://127.0.0.1 -H $CT"
echo $TEST
RESPONSE=`$TEST`
echo $RESPONSE
But if I just run following command from console everything fine:
curl http://127.0.0.1 -H Content-Type:\ application/json
Could you please let me know what is wrong in script, as I understand something is wrong with 'space' escape, but have no idea how to fix it.
Also I've tried following combination, but result the same:
CT="Content-Type: application/json"
TEST="curl http://127.0.0.1 -H \"$CT\""
UPD:
bash / dash is only available on server. (/bin/sh --> bash)
GNU bash, version 4.2.10(1)-release (x86_64-pc-linux-gnu)
Run the following: (delete the space after Content-Type)
#!/bin/bash
CT="Content-Type:application/json"
TEST="curl http://127.0.0.1 -H $CT"
echo $TEST
RESPONSE=`$TEST`
echo $RESPONSE
You can try with: bash -c your_bash_file.sh It worked for me with the same problem
Related
I am trying to run a curl command within a bash, when running the curl outside separately its working fine, but same running with Bash throws bad request 400 error. Would appreciate if someone can point where I am going wrong
#!/bin/bash
TIMESTAMP=$(date +%Y-%m-%d_%H-%M-%S)
curl http://localhost/cron/abc -H 'content-type: application/json'
echo 'Cron Executed at:' "$TIMESTAMP" | tee /var/log/cron.txt
I want to pass multiple arguments through to curl. Some of these arguments are quoted and contain spaces.
I have tried like this:
ARGS="http://example.org -H 'My-Header: Foo'"
curl -vvv $ARGS
But my header is not set and I get an error at the end curl: (6) Could not resolve host: Foo'.
I have also tried quoting ARGS like this:
ARGS="http://example.org -H 'My-Header: Foo'"
curl -vvv "$ARGS"
But I get curl: (3) URL using bad/illegal format or missing URL.
If I just run curl with the arguments directly, then it works fine:
curl -vvv http://example.org -H 'My-Header: Foo'
How can I pass these arguments through to curl correctly?
There is a command called eval which evaluates all the arguments into one string and then runs it as a one big command.
Try eval curl $ARGS
I recommend you to checkout eval's man page ;)
I trying to do execute the following script in bash
#!/bin/bash
source chaves.sh
HEAD='"X-Cachet-Token:'$CACHET_KEY'"'
SEARCH="'{"'"status"'":1,"'"id"'":"'"7"'","'"enabled"'":true}'"
echo $SEARCH
if curl -s --head --request GET http://google.com.br | grep "200 OK" > /dev/null; then
echo 'rodou'
curl -X PUT -H '"Content-Type:application/json;"' -H '"'X-Cachet-Token:$CACHET_KEY'"' -d $SEARCH $CACHET_URL/7
else
echo 'não deu'
curl -X PUT -H '"Content-Type: application/json;"' -H $x -d '{"status":1,"id":"7","enabled":true}' $CACHET_URL/7
fi
But keep receiving a 400 bad request from the server.
When i try to run the same line (echo in the script, Ctrl+c and Ctrl+v) directly in terminal, the command run without problems.
The source file have the directions to path and a variable token i need to use, but as far as i have tested is reading ok.
edit 1 - hidding some sensitive content
edit 2 - posting the exit line (grabed trought Ctrl+c, Ctrl+v)
The command i neet to input in server is:
curl -X PUT -H "Content-Type:application/json;" -H
"X-Cachet-Token:4A7ixgkU4hcCWFReQ15G" -d
'{"status":1,"id":"7","enabled":true}'
http://XXX.XXX.XXX.XXX/api/v1/components/7
And the exit i grabed trought echo comand, give me the exact exit i want, but don't run inside script, only outside.
I'm a bit new to the curl, any help can be apreciate.
Sorry for the bad english and tks in advance.
I want to load a file from a clients webserver. This webserver is running local only. To get there I have to use ssh. I need the content as well as the return value (e.g. SSH connection broke, webserver down).
What do I have to change? My first try:
#!/bin/bash
RETURN=0
CONTENT=""
sshpass -p xxxxxx ssh root#172.17.1.33 "curl -X POST http://127.0.0.1:10000/status -H 'Content-Type: application/json' > $CONTENT | bash; RETURN=$?"
If you want to get the exit code of curl and the return value of curl:
#!/bin/bash
CONTENT=$(sshpass -p xxxxxx ssh root#172.17.1.33 "curl -X POST http://127.0.0.1:10000/status -H 'Content-Type: application/json'")
RETURN=$?
echo "$RETURN, $CONTENT"
In your script you set the variables on the server you ssh'ed into.
I know a similar question was posted, but I can't get it to work on my machine.
I tried the 1st answer from the mentioned question, i.e. response=$(curl --write-out %{http_code} --silent --output /dev/null servername) and when I echo $response I got 000 [Not sure if that is the desired output].
However, when trying to do so with my cURL command, I get no output.
This is my command:
curl -k --silent --ftp-pasv --ftp-ssl --user C:is_for_cookies --cert localcert_cert.pem --key certs/localcert_pkey.pem ftps://10.10.10.10:21/my_file.txt
and I use it with
x=$(curl -k --silent --ftp-pasv --ftp-ssl --user C:is_for_cookies --cert localcert_cert.pem --key certs/localcert_pkey.pem ftps://10.10.10.10:21/my_file.txt)
but when I try to echo $x all I get is a newline...
I know the cURL is failing, because when I run the same command, without --silent, I get curl: (7) Couldn't connect to server
This Q is tagged with both sh, bash because I've tried it on both with same results
I found this option which kind of helps (but I still don't know how to assign it to a variable, which should be easier than this...):
--stderr <file>
Redirect all writes to stderr to the specified file instead. If the file name is a plain '-', it is instead written to stdout.
If this option is used several times, the last one will be used.
When I use it like this:
curl -k --silent -S --stderr my_err_file --ftp-pasv --ftp-ssl --user C:is_for_cookies --cert localcert_cert.pem --key certs/localcert_pkey.pem ftps://10.10.10.10:21/my_file.txt
I can see the errors (i.e. curl: (7) Couldn't connect to server) inside that file.
I used --silent to suppress all output, and -S to un-suppress the errors, and the --stderr <file> to redirect them