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How can I use the input_url parameter? #326

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opened 2026-02-16 00:17:29 -05:00 by yindo · 1 comment
yindo commented 2026-02-16 00:17:29 -05:00
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Originally created by @motoneurona on GitHub (Nov 12, 2024).

I'm trying to use an URL as input. The website indicates that it is possible.

image

But I am not able to use that parameter. Am I missing something? Thank you.

Originally created by @motoneurona on GitHub (Nov 12, 2024). I'm trying to use an URL as input. The website indicates that it is possible. ![image](https://github.com/user-attachments/assets/60ab96e6-e8c8-44e9-a2b6-c644b2d6a52e) But I am not able to use that parameter. Am I missing something? Thank you.
yindo closed this issue 2026-02-16 00:17:29 -05:00
yindo commented 2026-02-16 00:17:29 -05:00
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@logan-markewich commented on GitHub (Nov 12, 2024):

Added in latest release. pip install -U llama-parse

parser.load_data(input_url) or parser.load_data([url1, url2]) or even mixing parser.load_data([file_path, input_url])

The URLs work best when pointing to direct files. Improvements to actual websites will come over time

@logan-markewich commented on GitHub (Nov 12, 2024): Added in latest release. `pip install -U llama-parse` `parser.load_data(input_url)` or `parser.load_data([url1, url2])` or even mixing `parser.load_data([file_path, input_url])` The URLs work best when pointing to direct files. Improvements to actual websites will come over time
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Reference: run-llama/llama_cloud_services#326
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