我将如何计算在 PIL 中保持两个绘制句子之间的空间所需的 X 坐标？

Question

I'm trying to make a Python script to generate a fake Discord message, and this is my code so far:

from PIL import Image, ImageFont, ImageDraw

text = "Hello"
name = "A username...."
date = "Today at 10:30 AM"

comment = Image.open("message.png")

text_font = ImageFont.truetype("whitneybook.otf", 31)
name_font = ImageFont.truetype("whitneymedium.otf", 32)
date_font = ImageFont.truetype("whitneymedium.otf", 25)
    
draw = ImageDraw.Draw(comment)

draw.text((129, 70), text, (215, 215, 215), font=text_font)
draw.text((129, 25), name, (250, 250, 250), font=name_font)
draw.text((356, 32), date, (120, 120, 120), font=date_font)

comment.show()

And this is the result:

This code works, but I want the time and name to be able to be dynamic. If I change the name variable to "A very long username", I get this result:

How can I calculate the X-coordinate required to stop the username and date from overlapping?

Answer 1

Font objects have a getsize method, which should serve your purpose. One way or another, this solves all your problems.

In the simplest case, you just want to find the x-coordinate for the date:

date_x = name_font.getsize(name)[0] + 129 + 32
draw.text((date_x, 32), date, (120, 120, 120), font=date_font)

The offset in date_x is 129 for the left offset of the text, and 32 for the name-date padding.

Now you probably don't want to run your date off the edge of the image. In fact, let's say you want to always ensure a 10px right margin for it. In that case, you may need to adjust the x-coordinate a bit by inserting the following between the other two lines:

max_date_x = comment.width - 10 -  date_font.getsize(date)[0]
date_x = min(date_x, max_date_x)

Finally, you may want to truncate the username if it exceeds a certain length. This is a bit less trivial, since each letter is a different size. The amount of space you are targetting is date_x - 129 , but also accounting for an ellipsis after the truncated name and some padding. Let's do a linear search:

name_width = name_font.getsize(name)[0]
name_space = date_x - 129 - 10
n = 0
while name_width > name_space:
    n += 1
    name_width = name_font.getsize(name[:-n] + '...')
display_name = name[:-n] + '...' if n else name

You could do the same thing with a binary search for the correct value of n . Don't forget to draw display_name instead of name in this version.