From 573c9d616da3f32dc0d20c18712d9a7626092dc3 Mon Sep 17 00:00:00 2001
From: cancelself <332509+cancelself@users.noreply.github.com>
Date: Fri, 5 Sep 2025 19:52:06 -0700
Subject: [PATCH] Handle invalid Office Open XML file error

Adds a check for the specific error string indicating an invalid Office Open XML file. If detected, a FailedConversionAttempt is recorded and the next converter is tried, improving error handling for unsupported file formats.

See https://github.com/microsoft/markitdown/issues/1408
---
 .../markitdown/src/markitdown/_markitdown.py  | 19 +++++++++++++++++++
 1 file changed, 19 insertions(+)

diff --git a/packages/markitdown/src/markitdown/_markitdown.py b/packages/markitdown/src/markitdown/_markitdown.py
index 702b10c6..d8509b44 100644
--- a/packages/markitdown/src/markitdown/_markitdown.py
+++ b/packages/markitdown/src/markitdown/_markitdown.py
@@ -607,6 +607,25 @@ def _convert(
                         file_stream.seek(cur_pos)
 
                 if res is not None:
+                    # Check for Office Open XML error string and raise if found
+                    if (
+                        res.text_content.strip()
+                        == "This is not a valid Office Open XML file."
+                    ):
+                        failed_attempts.append(
+                            FailedConversionAttempt(
+                                converter=converter,
+                                exc_info=(
+                                    FileConversionException,
+                                    FileConversionException(
+                                        "Invalid Office Open XML file detected."
+                                    ),
+                                    None,
+                                ),
+                            )
+                        )
+                        continue  # Try next converter
+
                     # Normalize the content
                     res.text_content = "\n".join(
                         [line.rstrip() for line in re.split(r"\r?\n", res.text_content)]