Posts made by jo

@frederik ah, that is super helpful!
How did I miss this great feature?
Thanks again

@frederik
Sad that the forum will be discontinued, so thanks a lot for keeping it up for so long and leaving it in read only mode.

I will still try to post one more question, since I am frequently wondering how to do that in a clever way.

Like in the example above I usually try to have all the settings (like the definition of the page dimensions) in one place and before doing any drawing.
But I can only get the width if there is already a page (or it will be the default 1000).
How could I get the dimenations of eg an A3 page without calling newPage('A3')?

Again thanks for drawbot and the forum. See you on discord!

aah, untested last minute changes. I changed the variable names to something more reasonable but forgot one instance. cell_h should be label_h.

PW, PH = 842, 1190  # A3 page width and height 

page_margin = 40 # the margin for the whole page 

amount = 50 # how many labels to draw in total
per_page = 10 # how many labels per page 

gap = 10 # the gap between the labels 

# the next line calculates the heigth for the label 
label_h = (PH - ((per_page-1)*gap) - 2 * page_margin ) / per_page


for i in range(amount):

    if i % per_page == 0: 
        newPage(PW, PH)
        y = PH - page_margin - label_h
        
    rect(page_margin, y, width() - 2*page_margin, label_h)
    
    y -= label_h + gap

you could new make an A3 page with
newPage('A3')
and then draw several of the textboxes on that page. basically take the newPage out of the loop and instead change the position of each new label.

PW, PH = 842, 1190  # A3 page width and height 

page_margin = 40 # the margin for the whole page 

amount = 50 # how many labels to draw in total
per_page = 10 # how many labels per page 

gap = 10 # the gap between the labels 

# the next line calculates the heigth for the label 
label_h = (PH - ((per_page-1)*gap) - 2 * page_margin ) / per_page


for i in range(amount):

    if i % per_page == 0: 
        newPage(PW, PH)
        y = PH - page_margin - cell_h
        
    rect(page_margin, y, width() - 2*page_margin, label_h)
    
    y -= label_h + gap

thanks for clarifying.
one thing that made me wonder when reading the unconventional UPM of 550: might this be related to this bug: https://github.com/typemytype/drawbot/issues/555#issuecomment-1948050444.

as @monomonnik suggests I also tried a simpler script with Mutator Sans and I think this does what you expect?

txt = 'lestextessociologiques'.upper()

fnt = 'MutatorSans.ttf'

fnt_s = 24
lh = 32

lh_shift = -3

rows = 11 


newPage('A4')
txt_l = len(txt)

fstr = FormattedString(fontSize = fnt_s, font = fnt)

for r_i in range(rows):
    for t_i in range(txt_l):
        fstr.append(txt[(t_i + r_i)%txt_l], font = fnt, fontSize = fnt_s, lineHeight = lh, fontVariations={'wght': 1000/rows * r_i, 'wdth': 1000/txt_l*t_i})
    fstr.append('\n')
    
    lh += lh_shift if r_i < rows/2 else -lh_shift
    
text(fstr, (20, height() - 100))

otherwise just using a text() for each line might me a way?

txt = 'lestextessociologiques'.upper()
fnt = '/Users/jo/Desktop/MutatorSans.ttf'

fnt_s = 24
lh = 50
lh_shift = -3

rows = 11 

 
newPage('A4')

txt_l = len(txt)

y = height() - 100

for r_i in range(rows):

    fstr = FormattedString(fontSize = fnt_s, font = fnt)

    for t_i in range(txt_l):
        fstr.append(txt[(t_i + r_i)%txt_l], font = fnt, fontSize = fnt_s, fontVariations={'wght': 1000/rows * r_i, 'wdth': 1000/txt_l*t_i})


    text(fstr, (10, y))
    lh += lh_shift if r_i < rows/2 else -lh_shift
    
    y -= lh

@MauriceMeilleur

not sure I understand your question correctly.
my assumption is you want to specify a position for the first baseline and have the textbox draw the text so the first line is at that given position no matter the lineheight or font size?
the following script should do that. I tested a few fonts, fontSizes and lineHeights.

# -----------------------------------
#  S E T T I N G S  

fnt = 'Helvetica'

fnt_s = 16  # the fontsize 
lh = fnt_s * 1.6 # the lineheight 

txt = 'Testing the position of the baseline and the setting of the line height in drawBot. '*30

tbx_x = 100  # x position of the textbox 

frst_bl_y = 657 # the exact position of the first baseline within the textbox 

tbx_w = 350 # textbox width
tbx_h = 400 # textbox heigth


# --------------
# F U N C T I O N S 

def calc_baseline_offset(fnt, fnt_s, lh):
    '''calculates the position of the first baseline of a placeholder textbox and returns the distance from the top edge of the textbox'''
    fstr = FormattedString('Hm', font = fnt, fontSize = fnt_s, lineHeight = lh) 
    temp_first_baseline = textBoxBaselines(fstr, (0, 0, fnt_s * 3, fnt_s * lh))[0][1]
    return temp_first_baseline - (fnt_s * lh)

# -----------------------------------
#  D R A W I N G S  


newPage('A4')

fstr = FormattedString(txt, font = fnt, fontSize = fnt_s, lineHeight = lh)

rect(tbx_x - 20, frst_bl_y, tbx_w + 40, -1) # drawing the first baseline 

y_shift  = calc_baseline_offset(fnt, fnt_s, lh)


tbx_y = frst_bl_y - tbx_h - y_shift # the y position for the textbox so that the first baseline stays at the given value

fill(1, 0, 0, .1)
rect(tbx_x, tbx_y, tbx_w, tbx_h)

textBox(fstr, (tbx_x, tbx_y, tbx_w, tbx_h))

# drawing all the baselines 
for x, y in textBoxBaselines(fstr, (tbx_x, tbx_y, tbx_w, tbx_h)):
    rect(tbx_x, y, tbx_w, -1)
    

I ran into the same issue a few weeks ago and @frederik actually posted another solution in the other QR code thread:
going through the pixels and checking the lightness/brightness:
https://forum.drawbot.com/topic/397/qr-code-with-type?_=1681732270808

Using a module has the nice advantage of checking the size of the QR code after generating it.

@monomonnik thanks, yes that is how i 'solved' it but when doing that with several files it gets a bit tedious. not a big deal but maybe it is can be easily stored during the export in drawbot.

hello,
Whenever I place a pdf in InDesign I get the error message:
'Cannot crop to trim box. Either the trim box is not defined, or is empty'.

Is there a way to include a trimbox in the pdfs saved from Drawbot or is this InDesign related?
thanks j

oké
I had some small hope but thanks for testing and clarifying!

not very elegant but kind of working:

txt = 'Some Text\nwith new lines\rIs a return\na new line\ror a new paragraph?'

paras = [[l + '\n' for l in p.split('\n')] for p in txt.split('\r')]

newPage(150, 150)

fstr = FormattedString()
for para in paras:
    for l in para:
        extra_space = 5 if l == para[-1] else 0
        fstr.append(l, paragraphBottomSpacing = extra_space)
text(fstr, (20, 110))

I tried to use paragraphBottomSpacing and was hoping that drawbot would interpret \n as still the same paragraph and only apply the extra white space once I use a \r.

So there should only be extra white space after the line 'with new lines' but not after the other newlines.

txt = 'Some Text\nwith new lines\rIs a return\na new line\ror a new paragraph?'
fstr = FormattedString(txt, paragraphBottomSpacing=3)
newPage(150, 150)
text(fstr, (20, 110))

hope that makes sense, thanks

you instantiate the BezierPath outside the loop and then draw the text in every round of the loop. just put the bp = BezierPath() inside the loop.

hi, I think the BezierPath text does not accept variable font settings. You can first draw the text into a FormattedString() and then add that to the BezierPath.

bp = BezierPath()
fstr = FormattedString("JUICY", font='Skia-Regular', fontSize=80, fontVariations = {'wght' : curr_value})
bp.text(fstr, (10, 10))

hope that helps!

reload is python 2
afaik in python 3.4 and newer you are supposed to use:

import importlib
importlib.reload(module)

good luck!

this should work already.

when you select the bool (True or False) and then press the Cmd + arrow keys the value should change.

you will need to use a loop to iterate over your list of characters and append them to a formattedString.

first initiate the formattedString:

f_str = FormattedString(font = myFont, fontSize = pointSize)

then iterate over your list:

for ch in wordText:

select a random color inside the loop and append the character to the formattedString

    ch_col = random.choice(colors)
    f_str.append(ch, fill = ch_col)

finally draw your textBox with the formattedString

textBox(f_str, box, align="left")

so the last four lines of your code should be:

f_str = FormattedString(font = myFont, fontSize = pointSize)

for ch in wordText:
    ch_col = random.choice(colors)
    f_str.append(ch, fill = ch_col)

textBox(f_str, box, align="left")

good luck!

@monomonnik
yes and no.
in the chessboard examples it does not really matter if you go by columns or by rows.
but yes the first returned value is the quotient i//n or int(i/n) and the second returned value is the remainder (modulo) i % n.

@monomonnik
perfect place to use divmod.
this:

x = i % 8
y = int(i/8)

could be written like:

x, y = divmod(i, 8)

it will return the quotient (x) and the remainder (y).

I am not quite sure what exactly you are asking but the get a random integer value inside a range you could use randint(min_value, max_value). This works for integers. To get a random value between 100 and 900 you would use randint(100, 900).
If you have a smaller range and want fractions as well you could use min_value + random() * (max_value - min_value) eg 100 + (900-100) * random()
I hope that helps, good luck!

here is some code to draw a 7 segment LCD and use the segments to count up to 127 (2**7-1).

class digit(object):
	"""a digit of the lcd display"""
	def __init__(self, pos, size, seg_l, seg_h):
		self.pos    = pos
		self.size   = size
		self.seg_l  = seg_l
		self.seg_h  = seg_h

	def draw_segment(self, indx, l, h):
		seg_poss = [
			(           0,  self.size),
			( self.size/2,  self.size/2),
			( self.size/2, -self.size/2),
			(           0, -self.size),
			(-self.size/2, -self.size/2),
			(-self.size/2,  self.size/2),
			(           0,  0), ]

		x = self.pos[0] + seg_poss[indx][0]
		y = self.pos[1] + seg_poss[indx][1]  

		if indx in [1, 2, 4, 5]:
			polygon(
				(x, y + l/2),
				(x - h/2, y + l/2 - h/2),
				(x - h/2, y - l/2 + h/2),
				(x, y - l/2),
				(x + h/2, y - l/2 + h/2),
				(x + h/2, y + l/2 - h/2),)
		else: 		
			polygon(
				(x + l/2, y),
				(x + l/2 - h/2, y + h/2),
				(x - l/2 + h/2, y + h/2),
				(x - l/2, y),
				(x - l/2 + h/2, y - h/2),
				(x + l/2 - h/2, y - h/2),)

	def draw(self, num):
		for s in range(7):
			if (num & (1<<s)):
				self.draw_segment(s, self.seg_l, self.seg_h)

# ---------------------------
 
int_to_bin = {
    0 :  63,
    1 :   6,
    2 :  91,
    3 :  79,
    4 : 102, 
    5 : 109, 
    6 : 125,
    7 :   7,
    8 : 127,
    9 : 111,
    }

# ---------------------------

PW, PH = 500, 500

d  = digit((0, 0), PW/3, 150, 30)
d_int = digit((PW*.5, -PH*.4), 16, 14, 4)


for n in range(128):
    assert (0 <= n < 128), "Values should be higher than -1 and smaller than 128!"    
    newPage(PW, PH)
    fill(.2)
    rect(0, 0, PW, PH)
    translate(PW/2, PH/2)
    skew(2)
    fill(0, 1, 0)
    d.seg_h = n+4
    d.draw(n)

    with savedState():
        for i in reversed(str(n)):
            translate(-30)
            d_int.draw(int_to_bin[int(i)])         

should make animations like this: