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Tag: exec

Syntax error calling Python from C with fork-execv

I want to call a Python script as a child process from a C program using fork and execv. I have tested it calling bin/ls and it works: So I changed the C code to call the Create_Buffer() function in this Python file: The Python script has two functions; I want to call Create_Buffer(), so I changed my C file

Why is this twurl command (run on linux, by golang exec) not being authenticated?

I’m currently working on a Golang website (running on Ubuntu) that will update a twitter status. I used twurl customer key authentication on the system and I can successfully update the status if I type directly into the linux terminal. For example ssh/putty into target system type in terminal: twurl -d ‘status=is this thing on’ /1.1/statuses/update.json twitter status successfully updated

shell linux don’t execute commande

I have this command and it works fine on shell. I want to run it with php but it does not work what i missed : edited : $arg=”arg1″; $arg2= “arg2”; after this change no error on output but still commande not executed edited : the original commande is like this : and i want to around the probleme with

find: missing argument to -exec with ssh

Trying to remove directories from server: I tried with ; and ;; and ; and “{}” and -exec sh -c ‘rm -r “{}”‘ ; All get’s the error find: missing argument to “-exec'” Thanks Answer Directly running commands with ssh is often hard to impossible when you want to use special characters. This is because ssh remotely calls sh -c

C++ Command not recognized after fork() and execl() call (Linux)

I’m trying to create a program that can open a terminal browser and navigate through a website. It works fine up until the child process is called. The following error shows up “xdotool: Unknown command: search –onlyvisible –name Terminal windowactivate keydown Down” I’ve tested the command in a separate terminal and it indeed works but it doesn’t get recognized in

PHP exec sudo script not working despite having sudo rights

My php script: my visudo line (I tripled checked every path) If i remove the ‘sudo’ in the exec() it will execute (i put a touch aaa at the start, with sudo the file isn’t even created so the script doesn’t even begin to execute). The script has exec rights for everyone. I can’t manage to get any error output.

Remove folder structure (parents only) using ZIP command in PHP

I want to create a zip of all files in X folder with the name filename.zip using following command in php like: exec(zip -r “./Zips/filename.zip” “./Uploads/Data/X/”) but the created zip has the folder structure Uploads/Data/X. Please help me to get rid of these parent folders – Uploads/Data. Answer Ignoring folder structure when creating zip archive (Linux): By default, zip will

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