Reputation: 139872
especially the last two with '-' and '+' respectively, what does it mean on earth?
[1] Done php start.php bots/admin32.bot.php
[2] Done php start.php bots/admin36.bot.php
[3] Done php start.php bots/admin10.bot.php
[4] Done php start.php bots/admin11.bot.php
[5] Done php start.php bots/admin13.bot.php
[6] Done php start.php bots/admin3.bot.php
[7]- Done php start.php bots/admin4.bot.php
[8]+ Done php start.php bots/admin7.bot.php
[root@www2 robot]#
Upvotes: 1
Views: 536
Reputation: 204778
From the Bash Reference Manual,
Job number
n
may be referred to as ‘%n
’. 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 background. A single ‘%
’ (with no accompanying job specification) also refers to the current job. The previous job may be referenced using ‘%-
’. If there is only a single job, ‘%+
’ and ‘%-
’ can both be used to refer to that job. In output pertaining to jobs (e.g., the output of the jobs command), the current job is always flagged with a ‘+
’, and the previous job with a ‘-
’.
In Bash's output when starting and stopping jobs,
[n]+
status
means "job %n
(aka job %+
) is now status".
Upvotes: 8
Reputation: 137176
I think your command contains the character &
, which is causing the command to run in the background. When the command finishes, it prints this output.
Example:
$ echo 1 &
[1] 16021
$
[1]+ Done echo 1
To prevent this from happening, quote the &
character. It's probably part of a URL in your case, so you can use:
$ wget "http://www.example.com/index.php?a=1&b=2"
As for the +
and -
signs, they refer to the current and previous jobs respectively. See this page on job control in bash for more info.
Upvotes: 1