fix API key error

app.py

@@ -52,10 +52,21 @@ with gr.Blocks() as demo:
     """
     # 文章彙整工具
     輸入希望快速吸收資訊來源，即可使用問答的方式獲取資訊！
+
+    請輸入 Open AI API key 以使用本服務
     """
     )
+    openai_api_key_textbox = gr.Textbox(placeholder="Paste your OpenAI API key (sk-...) and hit Enter",
+                                    show_label=False, lines=1, type='password')
+    openai_api_key_textbox.change(set_openai_api_key,
+                                    inputs=[openai_api_key_textbox],
+                                    outputs=[])
+    openai_api_key_textbox.submit(set_openai_api_key,
+                                    inputs=[openai_api_key_textbox],
+                                    outputs=[])
+
     with gr.Tab("Articles"):
-        gr.Markdown("請在下列文字匡中輸入待彙整的資訊")
+        gr.Markdown("請在下列文字匡中輸入待彙整的文章")
         input1 = gr.Textbox(label="Articles 1")
         input2 = gr.Textbox(label="Articles 2")
         input3 = gr.Textbox(label="Articles 3")
@@ -71,15 +82,6 @@ with gr.Blocks() as demo:
         text_button.click(fn=create_embedding, inputs=[input1, input2, input3, input4, input5, input6, input7, input8, input9, input10], outputs=output_text, api_name="create_embedding")
 
     with gr.Tab("Chatbot"):
-        openai_api_key_textbox = gr.Textbox(placeholder="Paste your OpenAI API key (sk-...) and hit Enter",
-                                    show_label=False, lines=1, type='password')
-        openai_api_key_textbox.change(set_openai_api_key,
-                                      inputs=[openai_api_key_textbox],
-                                      outputs=[])
-        openai_api_key_textbox.submit(set_openai_api_key,
-                                      inputs=[openai_api_key_textbox],
-                                      outputs=[])
-
         def predict(message, history):
             bot_message = Bot(prompt = message)