What does %1 in "kill %1" mean?

I know that $! stores the pid of the last (background) process, but what does %1 mean? I often see it together with kill and it has the same effect as kill $!. Can someone give me a hint? (Such small terms are so hard to google :-/)

Solution 1:

The % designator refers to the jobs in the current shell's job list, and returns the PID. Try help jobs.

Solution 2:

What you want to Google is man bash

There are a number of ways to refer to a job in the shell. The charac- ter % introduces a job name. Job number n may be referred to as %n. A job may also be referred to using a prefix of the name used to start it, or using a substring that appears in its command line. For exam- ple, %ce refers to a stopped ce job. If a prefix matches more than one job, bash reports an error. Using %?ce, on the other hand, refers to any job containing the string ce in its command line. If the substring matches more than one job, bash reports an error. The symbols %% and %+ refer to the shell's notion of the current job, which is the last job stopped while it was in the foreground or started in the back- ground. The previous job may be referenced using %-. When there is the current job only, %- refers to the shell's notion of the current job. In output pertaining to jobs (e.g., the output of the jobs com- mand), the current job is always flagged with a +, and the previous job with a -. A single % (with no accompanying job specification) also refers to the current job.

Simply naming a job can be used to bring it into the foreground: %1 is a synonym for ''fg %1'', bringing job 1 from the background into the foreground. Similarly, ''%1 &'' resumes job 1 in the background, equivalent to ''bg %1''.

TL;DR: %1 is job number 1.