hi ı am triying to take the data of files in a folder with system function this is the code
char path[100],command[120];
scanf("%s",&path);
sprintf(command,"echo $(ls %s) > something.txt",path);
system(command);
but when I look to the something.txt there is no new line. This is the output, all on one line with many file names omitted:
acpi adjtime adobe apparmor.d arch-release asound.conf ati at-spi2 avahi bash.bash_logout ... wpa_supplicant X11 xdg xinetd.d xml yaourtrc
I did try -e -E -n options of echo but it didn't work. How to make a new line after each of these files?
You shouldn't use echo. Do just
sprintf(command,"ls %s > something.txt",path);
system(command);
When you use echo it outputs all command line arguments to the stdout, one by one, separated by the space character. Newline character (which is output of ls command) works as an argument separator, just as space.
There is one new line, which is to be expected. The echo command prints all its arguments on a single line separated by spaces, which is the output you see.
You need to execute the result of:
echo "$(ls %s)"
to preserve the newlines in the ls output. See Capturing multiple-line output to a Bash variable.
Use:
snprintf(command, sizeof(command), "echo \"$(ls %s)\" > something.txt", path);`
Of course, the echo is redundant; it would be better simply to run:
ls %s
and hence:
snprintf(command, sizeof(command), "ls %s > something.txt", path);
If you keep the echo, you should worry about the fact that you have more than 20 extra characters in the format string so the 120 should be more like 130 if not 140.
You should also use scanf("%99s", path) (no ampersand; add length limiting), and ideally check that it worked (if (scanf(...) == 1) { ... OK ... }).