WebMar 28, 2024 · How to use this command on a windows command prompt i tried the command curl -I -s -w %{http_code} server:8080/page/ and it throws curl: no URL specified! – mo-ta-to. Dec 19, 2024 at 15:02. ... Curl to return http status code along with the response. 578. Converting a POSTMAN request to Curl. WebAug 1, 2024 · 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
stdout - Can I make cURL fail with an exitCode different than 0 if …
WebMay 17, 2024 · 14. In case you want to: Return Exit code 0: If command completed successfully (code 0). OR if command did not complete yet (code 124), but that's OK too. Return Exit code 1: If command had a failure before timeout reached. Then try this: WebMar 13, 2024 · Letting curl echo the contents of the file and piping it to bash accounts for the text output of the curl command and allows bash to execute it. I'll bet that, if you try this, you will get the same results: $ ( cat / [path]/simple.sh ); echo $? Share Improve this answer Follow edited Mar 13, 2024 at 22:14 ilkkachu 129k 15 231 386 infected movie 2009
process - 127 Return code from $? - Stack Overflow
WebMar 29, 2012 · CURL error code 7 (CURLE_COULDNT_CONNECT) is very explicit ... it means Failed to connect () to host or proxy. The following code would work on any system: $ch = curl_init ("http://google.com"); // initialize curl handle curl_setopt ($ch, CURLOPT_FOLLOWLOCATION, 1); $data = curl_exec ($ch); print ($data); WebMay 21, 2015 · Return code for curl used in a command substitution Ask Question Asked 7 years, 10 months ago Modified 4 years, 3 months ago Viewed 57k times 9 I've script as below :- (edited) httpUrl="http://www.nnin.com" rep=$ (curl -v -X POST -d "UID=username&PWD=pass" $httpUrl) status=$? if [ "$?" WebJul 17, 2014 · It works and return me the HTTP code 201 ("created"). Now I try to use this command in a bash script replacing a part of the url with variable: Now I try to use this command in a bash script replacing a part of the url with variable: infected mouth ulcer antibiotics