Are we then yet?
JavaScript, 282 bytes
(x,v=x.split(/\D/g),l=v.length-2,[a,b,c,d]=("10e5,01,-,12,01,-,"+new Date(v[0],v[1],0).getDate()+",00,T,24,00,:,60,00,:,60,000,.,1000").split`,`.slice(l*3,l*3+4),t=(v[l+1]-b)/d*20+.5|0,n=v[l],o=((n|0)+1)%a,r=l?('0'+o).slice(-2):o)=>n+c+b+' '+'#'.repeat(t)+'-'.repeat(20-t)+' '+r+c+b
Passes all the tests
(
x,
v=x.split(/\D/g),
l=v.length-2,
[a,b,c,d]=("10e5,01,-,12,01,-,"+new Date(v[0],v[1],0).getDate()+",00,T,24,00,:,60,00,:,60,000,.,1000").split`,`.slice(l*3,l*3+4),
t=(v[l+1]-b)/d*20+.5|0,
n=v[l],
o=((n|0)+1)%a,
r=l?('0'+o).slice(-2):o
) =>n+c+b+' '+'#'.repeat(t)+'-'.repeat(20-t)+' '+r+c+b
Test function prints nothing for pass, values for fail.
function test(value,expected){
if (f(value)!=expected)
{
console.log(value);
console.log(f(value));
console.log(expected);
}
}
The test cases:
test('2016-12-12T12:17','12:00 ######-------------- 13:00') ;
test('2016-12-12','12-01 #######------------- 01-01') ;
test('0000-01-01T00:00:00.000','00.000 -------------------- 01.000') ;
test('0000-01-01T00:00','00:00 -------------------- 01:00') ;
test('1899-12-31T23','31T00 ###################- 01T00') ;
test('1899-12-31','12-01 ###################- 01-01') ;
test('1899-12','1899-01 ##################-- 1900-01') ;
test('1982-05-15T17:15','17:00 #####--------------- 18:00') ;
test('1982-05-15T17','15T00 ##############------ 16T00') ;
test('1982-05','1982-01 #######------------- 1983-01') ;
test('9999-12-31T23:59:59.999','59.000 #################### 00.000') ;
test('9999-12','9999-01 ##################-- 10000-01') ;
test('2000-01-06','01-01 ###----------------- 02-01') ;
test('2000-02-06','02-01 ###----------------- 03-01') ;
test('2001-02-06','02-01 ####---------------- 03-01') ;
test('1742-09-10','09-01 ######-------------- 10-01') ;
Pyth, 213 bytes
My first code in pyth! Behold:
+%h=N:[d"%.4d"\-=Z"%.2d"\-Z\TZ\:Z\:Z\."%.3d")-*2lKr:w"[-T:.]"d7 3*2lKJ@K_2+@N1+%eN=b<lK4+d+*\#=Gs+*20c-eKb@=H[0^9T12?q2=k@K1+28+q0%=YhK4-q0%Y400q0%Y100+30%+k/k8 2 24 60 60 999)lK.5+*\--20G+d+%hN%+J1@H-lK1+@N1%eNb
My pyth code is closely based off of my previous python answer. Here is the ungolfed version with comments:
"K is the input, as a list of numbers"
Kr:w"[-T:.]"d7
"Y=year"
=YhK
"k=month"
=k@K1
"H = a list of denominators"
=H[0 ^9T 12 ?q2k+28+q0%Y4-q0%Y400q0%Y100 +30%+k/k8 2 24 60 60 999)
"J is the second-to-last number of the input"
J@K_2
"b is the +1 starting point for months and days"
=b<lK4
"G is the number of hashtags in the statusbar"
=Gs+*[email protected]
"N is the formatted string"
=N:[d"%.4d"\-=Z"%.2d"\-Z\TZ\:Z\:Z\."%.3d")-*2lK3 *2lK
+%hNJ+@N1+%eNb+d+*\#G+*\--20G+d+%hN%+J1@H-lK1+@N1%eNb
Testing multiple values is easily accomplished by making the code loop, and adding a newline print to the end:
Wp+%h=N:[d"%.4d"\-=Z"%.2d"\-Z\TZ\:Z\:Z\."%.3d")-*2lKr:w"[-T:.]"d7 3*2lKJ@K_2+@N1+%eN=b<lK4+d+*\#=Gs+*20c-eKb@=H[0^9T12?q2=k@K1+28+q0%=YhK4-q0%Y400q0%Y100+30%+k/k8 2 24 60 60 999)lK.5+*\--20G+d+%hN%+J1@H-lK1+@N1+%eNb"\n"
Then I ran cat testinput | pyth code.pyth > output
and diff output testoutput
Or try it online.
Python 2, 371 bytes
This challenge was surprisingly difficult! It seemed like I was going to be just under 300 until I worked out the output string formatting.
The kind of cool part is that my answer does not use any date package:
import re
s=raw_input()
S=[int(i)for i in re.sub('[-T:.]',' ',s).split()]
l=len(S)
y,m=S[:2]
d=[0,20<<9,12,28+(y%4==0!=y%100)+(y%400==0)if m==2else 30+(m+m/8)%2,24,60,60,999]
a,n=S[-2:]
b=1-(1if l>3else 0)
h=int(20.*(n-b)/d[l]+.5)
x,y,z='- %.4d - %.2d - %.2d T %.2d : %.2d : %.2d . %.3d'.split()[l*2-3:l*2]
print x%a+y+z%b+' '+'#'*h+'-'*(20-h)+' '+x%((a+1)%d[l-1])+y+z%b