what does these output mean?

Question

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]#

Answer 1

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.

Answer 2

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".