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docs/src/howto/tables-grids.md

@@ -7,14 +7,16 @@ end
 
 You often want to position graphics at regular locations on the drawing. The positions can be provided by:
 
-- `Tiler`: a rectangular grid which you specify by enclosing area, and the number of rows and columns
-- `Partition`: a rectangular grid which you specify by enclosing area, and the width and height of each cell
-- `Grid` a rectangular grid, points supplied on demand
-- `Table`: a rectangular grid which you specify by providing row and column numbers, row heights and column widths
+- `Tiler`: a rectangular grid iterator which you specify by enclosing area, and the number of rows and columns
+- `Partition`: a rectangular grid iterator which you specify by enclosing area, and the width and height of each cell
+- `Grid` a rectangular grid iterator, where points are supplied on demand
+- `Table`: a rectangular grid iterator which you specify by providing row and column numbers, row heights and column widths
 
 These are types which act as iterators. Their job is to provide you with centerpoints; you'll probably want to use these in combination with the cell's widths and heights.
 
-There are also functions to make hexagonal grids. See [Hexagonal grids](@ref).
+There are functions to make hexagonal grids. See [Hexagonal grids](@ref).
+
+There is [EquilateralTriangleGrid](@ref), a grid iterator which generates the vertices for a rectangular grid of equilateral triangles.
 
 ## Tiles and partitions
 
@@ -490,3 +492,32 @@ for (n, h) in enumerate(hexspiral(hexagon, 10))
 end
 end # hide
 ```
+
+## EquilateralTriangleGrid
+
+To create a grid of equilateral triangles, you can use
+[`EquilateralTriangleGrid`](@ref). Provide the starting point, the side length,
+and the number of rows and columns. The iterator provides a tuple of the
+vertices and number for each triangle.
+
+```@example
+using Luxor, Colors
+@drawsvg begin # hide
+    background("black")
+    nrows = 12
+    ncols = 20
+    side = 45
+    startpoint = boxtopleft() + (50, 50)
+    eqtg = EquilateralTriangleGrid(startpoint, side, nrows, ncols)
+
+    for e in eqtg
+        vertices, trianglenumber = e
+        sethue(HSB(rescale(trianglenumber, 0, nrows * ncols, 0, 360), 
+            0.7, 0.8))
+        poly(vertices, :fillpreserve)
+        sethue("black")
+        strokepath()
+        text(string(trianglenumber), halign=:center, polycentroid(vertices))
+    end
+end # hide
+```

src/basics.jl

@@ -39,7 +39,7 @@ rescale(x, (from_min, from_max), (to_min, to_max))
 ```
 
 Examples
-```jldoctest
+```julia
 julia> rescale(15, 0, 100, 0, 1)
 0.15
 

src/colors_styles.jl

@@ -10,7 +10,7 @@ Use `sethue()` for changing colors without changing current opacity level.
 
 `sethue()` and `setcolor()` return the three or four values that were used:
 
-```jldoctest
+```julia
 julia> setcolor(sethue("red")..., .8)
 (1.0N0f8, 0.0N0f8, 0.0N0f8, 0.8)
 

src/drawings.jl

@@ -1507,7 +1507,7 @@ The default drawing is 256 by 256 points.
 
 You don't need `finish()` (the macro calls it), and it's not previewed by `preview()`.
 
-```jldoctest
+```julia
 julia> m = @imagematrix begin
                sethue("red")
                box(O, 20, 20, :fill)
@@ -1561,7 +1561,7 @@ Transparency
 The default value for the cells in an image matrix is
 transparent black. (Luxor's default color is opaque black.)
 
-```jldoctest
+```julia
 julia> @imagematrix begin
        end 2 2
 2×2 reinterpret(ColorTypes.ARGB32, ::Matrix{UInt32}):
@@ -1572,7 +1572,7 @@ julia> @imagematrix begin
 Setting the background to a partially or completely
 transparent value may give unexpected results:
 
-```jldoctest
+```julia
 julia> @imagematrix begin
            background(1, 0.5, 0.0, 0.5) # semi-transparent orange
        end 2 2
@@ -1584,7 +1584,7 @@ julia> @imagematrix begin
 here the semi-transparent orange color has been partially
 applied to the transparent background.
 
-```jldoctest
+```julia
 julia> @imagematrix begin
            sethue(1., 0.5, 0.0)
            paint()

src/hexagons.jl

@@ -215,7 +215,7 @@ For more information about making hexagons and hexagonal grids, see [Luxor.Hexag
 
 ## Example
 
-```jldoctest
+```julia
 julia> h = HexagonOffsetEvenR(0, 0, 70.0);
 
 julia> hexneighbors(h)

src/point.jl

@@ -54,7 +54,7 @@ Base.length(::Point) = 2
 
 Transform a point `pt` by the 3×3 matrix `m`.
 
-```jldoctest
+```julia
 julia> M = [2 0 0; 0 2 0; 0 0 1]
 3×3 Matrix{Int64}:
  2  0  0
@@ -362,7 +362,7 @@ Find angle of a line starting at `pointA` and ending at `pointB`.
 
 Return a value between 0 and 2pi. Value will be relative to the current axes.
 
-```jldoctest
+```julia
 julia> slope(O, Point(0, 100)) |> rad2deg # y is positive down the page
 90.0
 
@@ -429,7 +429,7 @@ intersectionlinecircle(p1::Point, p2::Point, cpoint::Point, r) =
 
 Convert a tuple of two numbers to a Point of x, y Cartesian coordinates.
 
-```jldoctest
+```julia
 julia> @polar 10, pi/4
 Point(7.0710678118654755, 7.071067811865475)
 ```
@@ -452,7 +452,7 @@ end
 
 Convert a point specified in polar form (radius and angle) to a Point.
 
-```jldoctest
+```julia
 julia> polar(10, pi/4)
 Point(7.0710678118654755, 7.071067811865475)
 ```

src/shapes.jl

@@ -341,7 +341,7 @@ drawn below the positive x-axis.
 If you just want the raw points, use keyword argument `vertices=true`, which
 returns the array of points. Compare:
 
-```jldoctest
+```julia
 julia> ngon(0, 0, 4, 4, 0, vertices=true) # returns the polygon's points:
 4-element Vector{Point}:
  Point(2.4492935982947064e-16, 4.0)

src/tiles-grids.jl

@@ -135,7 +135,7 @@ For a column, set the `xspacing` to 0:
     grid = GridRect(O, 0, 40)
 
 To get points from the grid, use `nextgridpoint(g::Grid)`.
-```jldoctest
+```julia
 julia> grid = GridRect(O, 0, 40);
 
 julia> nextgridpoint(grid)