我想绘制图像而不必将其保存在静态文件夹中。
我有一个用户 Messa 在这里发布的示例,运行得很好:
import io
import random
from flask import Response
from matplotlib.backends.backend_agg import FigureCanvasAgg as FigureCanvas
from matplotlib.figure import Figure
@app.route('/plot.png')
def plot_png():
fig = create_figure()
output = io.BytesIO()
FigureCanvas(fig).print_png(output)
return Response(output.getvalue(), mimetype='image/png')
def create_figure():
fig = Figure()
axis = fig.add_subplot(1, 1, 1)
xs = range(100)
ys = [random.randint(1, 50) for x in xs]
axis.plot(xs, ys)
return fig
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然后在 HTML 文件中:
<img src="/plot.png" alt="my plot">
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结果是完美的。我即时获取图像。
现在我想使用一些这样的参数:
@app.route('/<x_values_source>/<y_values_source>/plot.png')
def plot_png(x_values_source, y_values_source):
fig = create_figure(x_values_source, y_values_source)
output = io.BytesIO() …Run Code Online (Sandbox Code Playgroud)