Always display WebAgg URL regardless of browser open status #30786
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I am not familiar with webagg, but if a url successfully opens in a browser, can the user not see the url in the browser? |
ianhi
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Nov 26, 2025
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ianhi
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I'm in general in favor of more aggressive printing out of the url for all the reasons that you note. However, I don't think this is the right approach as leaves anyone using webagg as part of an application with a print statement that they cannot suppress, something that that they may not want.
Ideally there would be an escape hatch for that case.
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Summary
Ensures the WebAgg server URL is always displayed to users, even when the browser successfully opens automatically.
Problem
Currently, the URL is only shown when:
webagg.open_in_browserisFalse, ORThis means users never see the URL when the browser opens successfully, making it difficult to:
Solution
Moved the
print(f"To view figure, visit {url}")statement outside the conditional block, ensuring it always executes regardless of the browser open attempt.Changes
lib/matplotlib/backends/backend_webagg.pyFigureManagerWebAgg.pyplot_show()Benefits
Testing
Tested with:
webagg.open_in_browser = True→ URL displays, browser openswebagg.open_in_browser = False→ URL displays, browser doesn't openFixes #15650