leading zero in \foreach of a TikZ calendar

Question

The calendar TikZ library creates for each day a node called (<name>-<year>-<month>-<day>). But there is a leading zero for <day> from 1 to 9, that is, to use the node of the first day we have to refer to it by (<name>-<year>-<month>-01).

Since I'm trying to connect some nodes using a \foreach loop, I'm having trouble with the leading zero. For example,

! Package pgf Error: No shape named cal-2015-01-2 is known.
! Package pgf Error: No shape named cal-2015-09-4 is known.

But if I insert the leading zero (cal-2015-09-0\x.north east) the error changes to

! Package pgf Error: No shape named cal-2015-01-016 is known.

So, I think that working with integer numbers with exactly two digits could solve the problem.

I'd like to have the same two dashed lines as in the 1st calendar below, but translated to separate the other weekends.

MWE

\documentclass[11pt]{report}
\usepackage[margin=1cm,landscape,a4paper]{geometry}
\usepackage{tikz}
\usetikzlibrary{calendar,calc,math}

\begin{document}

\begin{tikzpicture}%
  [every day/.style={anchor=mid},
   every node/.style={inner sep=2pt,rectangle,thick}
  ]
\calendar (cal)%
 [dates=\the\year-1-1 to \the\year-12-last,
  month list,
  month label left,
  month text=\textcolor{black}{\%mt},
  month yshift=1.7em,
  black
 ]
if (Saturday) [blue]
if (Sunday)   [red]
;
\draw[dashed]
  ( $(cal-\the\year-01-02.north east)!.5!(cal-\the\year-01-03.north west)$ ) --
  ( $(cal-\the\year-12-04.south east)!.5!(cal-\the\year-12-05.south west)$ )
%%
  ( $(cal-\the\year-01-04.north east)!.5!(cal-\the\year-01-05.north west)$ ) --
  ( $(cal-\the\year-12-06.south east)!.5!(cal-\the\year-12-07.south west)$ )
;
\end{tikzpicture}

\vfill\noindent\hrulefill\vfill

\begin{tikzpicture}%
  [every day/.style={anchor=mid},
   every node/.style={inner sep=2pt,rectangle,thick}
  ]
\calendar (cal)%
 [dates=\the\year-1-1 to \the\year-12-last,
  month list,
  month label left,
  month text=\textcolor{black}{\%mt},
  month yshift=1.7em,
  black
 ]
if (Saturday) [blue]
if (Sunday)   [red]
;
%%  use only one foreach below
%% the first has problems with leading zeros on day numbers
%%
\foreach \i in {0,7,...,31}{% should be tested if is >31
%\foreach \i in {16,23,...,31}{% should be tested if is >31
%%
%% January 02 and 04 are starting days
  \tikzmath{integer \xaa; \xaa = 2+\i;}
  \tikzmath{integer \xab; \xab = \xaa+1;}
  \tikzmath{integer \xba; \xba = 4+\i;}
  \tikzmath{integer \xbb; \xbb = \xba+1;}
%% December 04 and 06 are ending days
  \tikzmath{integer \yaa; \yaa = 4+\i;}
  \tikzmath{integer \yab; \yab = \yaa+1;}
  \tikzmath{integer \yba; \yba = 6+\i;}
  \tikzmath{integer \ybb; \ybb = \yba+1;}
\draw[dashed]
  ( $(cal-\the\year-01-\xaa.north east)!.5!(cal-\the\year-01-\xab.north west)$ ) --
  ( $(cal-\the\year-12-\xba.south east)!.5!(cal-\the\year-12-\xbb.south west)$ )
%%
  ( $(cal-\the\year-01-\yaa.north east)!.5!(cal-\the\year-01-\yab.north west)$ ) --
  ( $(cal-\the\year-12-\yba.south east)!.5!(cal-\the\year-12-\ybb.south west)$ )
;
}
\end{tikzpicture}

\end{document}

Answer 1

Update

Elegant solution to the whole problem:

Here is another simplified solution which avoids the two-digits problem and adds the last two columns finally requested by the OP. The idea is that calendar columns are separated by some default distance. Here, for an 11pt-document, it is 3.5ex. We make sure it is 3.5ex by setting explicitly the day xshift=3.5ex option and passing it to the calendar command.

Next, we perform a \foreach loop to draw two vertical lines and repeat after a week by adding an xshift of 7*3.5ex as this:

\foreach \i in {0,...,4}{%
\pgfmathparse{3.5*7*\i} \edef\w{\pgfmathresult}
\pgfmathparse{\w+2*3.5} \edef\x{\pgfmathresult}
\draw[dashed]
([xshift=\w ex]$(cal-\the\year-01-02.north east)!.5!(cal-\the\year-01-03.north west)$) --
([xshift=\w ex]$(cal-\the\year-12-04.south east)!.5!(cal-\the\year-12-05.south west)$)
([xshift=\x ex]$(cal-\the\year-01-02.north east)!.5!(cal-\the\year-01-03.north west)$) --
([xshift=\x ex]$(cal-\the\year-12-04.south east)!.5!(cal-\the\year-12-05.south west)$);
} 

That loop draws all vertical lines as this:

And the total simplified code now is this:

\documentclass[11pt]{report}
\usepackage[margin=1cm,landscape,a4paper]{geometry}
\usepackage{tikz}
\usetikzlibrary{calendar,calc,math}
\begin{document}

\begin{tikzpicture}%
  [every day/.style={anchor=mid},
   every node/.style={inner sep=2pt,rectangle,thick}
  ]
\calendar (cal)%
 [dates=\the\year-1-1 to \the\year-12-last,
  month list,
  month label left,
  month text=\textcolor{black}{\%mt},
  month yshift=1.7em,
  black
 ]
if (Saturday) [blue]
if (Sunday)   [red];

\draw[dashed]
  ( $(cal-\the\year-01-02.north east)!.5!(cal-\the\year-01-03.north west)$ ) --
  ( $(cal-\the\year-12-04.south east)!.5!(cal-\the\year-12-05.south west)$ )
%%
  ( $(cal-\the\year-01-04.north east)!.5!(cal-\the\year-01-05.north west)$ ) --
  ( $(cal-\the\year-12-06.south east)!.5!(cal-\the\year-12-07.south west)$ );
\end{tikzpicture}

\vfill\noindent\hrulefill\vfill

\begin{tikzpicture}%
  [every day/.style={anchor=mid},
   every node/.style={inner sep=2pt,rectangle,thick}
  ]
\calendar[day xshift=3.5ex](cal)%
 [dates=\the\year-1-1 to \the\year-12-last,
  month list,
  month label left,
  month text=\textcolor{black}{\%mt},
  month yshift=1.7em,
  black
 ]
if (Saturday) [blue]
if (Sunday)   [red];

% use only one foreach below
% the first has problems with leading zeros on day numbers

\foreach \i in {0,...,4}{%
\pgfmathparse{3.5*7*\i} \edef\w{\pgfmathresult}
\pgfmathparse{\w+2*3.5} \edef\x{\pgfmathresult}
\draw[dashed]
([xshift=\w ex]$(cal-\the\year-01-02.north east)!.5!(cal-\the\year-01-03.north west)$) --
([xshift=\w ex]$(cal-\the\year-12-04.south east)!.5!(cal-\the\year-12-05.south west)$)
%
([xshift=\x ex]$(cal-\the\year-01-02.north east)!.5!(cal-\the\year-01-03.north west)$) --
([xshift=\x ex]$(cal-\the\year-12-04.south east)!.5!(cal-\the\year-12-05.south west)$);
}
\end{tikzpicture}

\end{document}

Solution 1

Inspired by the error message mentioned by @clemens, a straightforward solution could be to avoid days greater than 31. To achieve this, observe the largest node number \ybb = \yba+1 which is equal to \i+7 because \yba=\i+6. This latter value should maximize to 31. So, max{ \i }= 24, which simply implies the condition:

\ifnum\i<25\draw ... \else\fi 

Here is the full code with the second solution (which gives the same results as above):

\documentclass[11pt]{report}
\usepackage[margin=1cm,landscape,a4paper]{geometry}
\usepackage{tikz}
\usetikzlibrary{calendar,calc,math}
\makeatletter
\let\twodigits\two@digits
\makeatother

\begin{document}

\begin{tikzpicture}%
  [every day/.style={anchor=mid},
   every node/.style={inner sep=2pt,rectangle,thick}
  ]
\calendar (cal)%
 [dates=\the\year-1-1 to \the\year-12-last,
  month list,
  month label left,
  month text=\textcolor{black}{\%mt},
  month yshift=1.7em,
  black
 ]
if (Saturday) [blue]
if (Sunday)   [red];

\draw[dashed]
  ( $(cal-\the\year-01-02.north east)!.5!(cal-\the\year-01-03.north west)$ ) --
  ( $(cal-\the\year-12-04.south east)!.5!(cal-\the\year-12-05.south west)$ )
%%
  ( $(cal-\the\year-01-04.north east)!.5!(cal-\the\year-01-05.north west)$ ) --
  ( $(cal-\the\year-12-06.south east)!.5!(cal-\the\year-12-07.south west)$ );
\end{tikzpicture}

\vfill\noindent\hrulefill\vfill

\begin{tikzpicture}%
  [every day/.style={anchor=mid},
   every node/.style={inner sep=2pt,rectangle,thick}
  ]
\calendar(cal)%
 [dates=\the\year-1-1 to \the\year-12-last,
  month list,
  month label left,
  month text=\textcolor{black}{\%mt},
  month yshift=1.7em,
  black
 ]
if (Saturday) [blue]
if (Sunday)   [red];

%%  use only one foreach below
%% the first has problems with leading zeros on day numbers
%%
\foreach \i in {0,7,...,31}{% should be tested if is >31
%\foreach \i in {16,23,...,31}{% should be tested if is >31
%%
%% January 02 and 04 are starting days
  \tikzmath{integer \xaa; \xaa = 2+\i;}
  \tikzmath{integer \xab; \xab = \xaa+1;}
  \tikzmath{integer \xba; \xba = 4+\i;}
  \tikzmath{integer \xbb; \xbb = \xba+1;}
%% December 04 and 06 are ending days
  \tikzmath{integer \yaa; \yaa = 4+\i;}
  \tikzmath{integer \yab; \yab = \yaa+1;}
  \tikzmath{integer \yba; \yba = 6+\i;}
  \tikzmath{integer \ybb; \ybb = \yba+1;}
\ifnum\i<25
\draw[dashed]
  ($(cal-\the\year-01-\twodigits\xaa.north east)!.5!(cal-\the\year-01-\twodigits\xab.north west)$) --
  ($(cal-\the\year-12-\twodigits\xba.south east)!.5!(cal-\the\year-12-\twodigits\xbb.south west)$)
%%
  ($(cal-\the\year-01-\twodigits\yaa.north east)!.5!(cal-\the\year-01-\twodigits\yab.north west)$) --
  ($(cal-\the\year-12-\twodigits\yba.south east)!.5!(cal-\the\year-12-\twodigits\ybb.south west)$);
\else\fi
}
\end{tikzpicture}

\end{document}

Answer 2

LaTeX has (the expandable) \two@digits which takes an integer as input an adds a leading 0 if the number is less than 10. So you could place

\makeatletter
\let\twodigits\two@digits
\makeatother

somewhere in your preamble and then use

\draw[red] 
  ($(cal-2015-09-\twodigits\x.north east)!.5!(cal-2015-09-05.north west)$)--
  ($(cal-2016-02-\twodigits\y.south east)!.5!(cal-2016-02-06.south west)$) ;

in your first example or

$\frac{\twodigits\i}{\twodigits\x}$

in your second example.

---

Edit in light of the updated question – here the code part is now:

\draw[dashed]
  ( $(cal-\the\year-01-\twodigits\xaa.north east)!.5!(cal-\the\year-01-\twodigits\xab.north west)$ ) --
  ( $(cal-\the\year-12-\twodigits\xba.south east)!.5!(cal-\the\year-12-\twodigits\xbb.south west)$ )
%%
  ( $(cal-\the\year-01-\twodigits\yaa.north east)!.5!(cal-\the\year-01-\twodigits\yab.north west)$ ) --
  ( $(cal-\the\year-12-\twodigits\yba.south east)!.5!(cal-\the\year-12-\twodigits\ybb.south west)$ )
;

BTW: this will still give errors

! Package pgf Error: No shape named cal-2015-12-32 is known.

but now it's not due to missing zeros but because 4+\i eventually is bigger than 31.