prevent unhandled error when getting URL content

scrape.py

@@ -9,20 +9,24 @@ def get_url_content(url: str) -> str:
     :param url: The URL to retrieve content from.
     :return: The content of the URL as a string.
     """
+    try:
+        response = requests.get(
+            url,
+            headers={
+                "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 Safari/537.3"
+            }
+        )
 
-    response = requests.get(
-        url,
-        headers={
-            "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 Safari/537.3"
-        }
-    )
+        if response.status_code != 200:
+            print(f"Failed to retrieve content from {url}. Status code: {response.status_code} - {response.reason}")
+            return ""
 
-    if response.status_code != 200:
-        print(f"Failed to retrieve content from {url}. Status code: {response.status_code} - {response.reason}")
-        return ""
+        # parse the html content using BeautifulSoup
+        parser = BeautifulSoup(response.text, 'html.parser')
 
-    # parse the html content using BeautifulSoup
-    parser = BeautifulSoup(response.text, 'html.parser')
+        # extract text from the parsed HTML
+        return parser.text.strip() if parser.text else ""
 
-    # extract text from the parsed HTML
-    return parser.text.strip() if parser.text else ""
+    except Exception as e:
+        print(f"An error occurred while retrieving content from {url}: {e}")
+        return ""