问题
I'm using Wget to make http requests to a fresh web server. I am doing this to warm the MySQL cache. I do not want to save the files after they are served.
wget -nv -do-not-save-file $url
Can I do something like -do-not-save-file with wget?
回答1:
Use q flag for quiet mode, and tell wget to output to stdout with O- (uppercase o) and redirect to /dev/null to discard the output:
wget -qO- $url &> /dev/null
> redirects application output (to a file). if > is preceded by ampersand, shell redirects all outputs (error and normal) to the file right of >. If you don't specify ampersand, then only normal output is redirected.
./app &> file # redirect error and standard output to file
./app > file # redirect standard output to file
./app 2> file # redirect error output to file
if file is /dev/null then all is discarded.
This works as well, and simpler:
wget -O/dev/null -q $url
回答2:
Curl does that by default without any parameters or flags, I would use it for your purposes:
curl $url > /dev/null 2>&1
Curl is more about streams and wget is more about copying sites based on this comparison.
回答3:
You can use -O- (uppercase o) to redirect content to the stdout (standard output) or to a file (even special files like /dev/null /dev/stderr /dev/stdout )
wget -O- http://yourdomain.com
Or:
wget -O- http://yourdomain.com > /dev/null
Or: (same result as last command)
wget -O/dev/null http://yourdomain.com
来源：https://stackoverflow.com/questions/9691367/how-do-i-request-a-file-but-not-save-it-with-wget