Bash printf control characters

[Johnny Rosenberg]

2014-05-26 12:34 GMT+02:00 Patrick Asselman <iceblink at seti.nl>:

>
> You could tell printf to use a certain width for your %A variable. The
> variable is printed aligned to the right of its assigned space, and the
> space in front is filled with space characters.
>
> example:
> printf "%20s" söndag
>

Doesn't work. The spaces end up at the beginning of the string, not at the
end of it.



Johnny Rosenberg


>
> (haven't tested it, but should work)
>
>
> Patrick
>
>
>
> On 2014-05-24 19:14, Johnny Rosenberg wrote:
>
>> I wrote this bash function, just for no particular reason, learning I
>> guess:
>>
>> function lvdati () {
>>     while [[ 1 ]]; do
>>         printf 'r%(%F %T (%A))T' -1
>>         sleep 0.1
>>      done
>> }
>>
>> It displays the current date and time in the terminal (use Ctrl+c to
>> terminate it) in a certain format. On my machine, it looks like this,
>> for instance:
>> 2014-05-24 18:58:50 (lördag)
>>
>> What I don't want to happen is this:
>> 2014-05-22 23:59:59 (torsdag)
>> Then a second later:
>> 2014-05-23 00:00:00 (fredag))
>>
>> Since ”fredag” (friday) is one character shorter than ”torsdag”
>> (thursday), the last character of the torsdag line will still remain,
>> won't it?
>> So I need something at the end of the string that works like Ctrl+k
>> in the terminal (erase everything on the right side of the cursor),
>> but how do I implement it?
>>
>> Let's say that Ctrl+k='k' (it isn't, but just imagine it…), then I would
>> try:
>> function lvdati () {
>>     while [[ 1 ]]; do
>>         printf 'r%(%F %T (%A))Tk' -1
>>
>>         sleep 0.1
>>     done
>> }
>>
>> Suggestions (except adding one or more spaces to the string)?
>>
>> Johnny Rosenberg
>>
>
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