python - Streamlit 按钮更改禁用状态

Question

如何在 Streamlit 中获得正确启用/禁用的按钮？下面的当前代码适用于 1 单击延迟(不需要)。

所需的操作 - 单击“button_a”时启用“button_c”，单击“button_b”时禁用“button_c”。

当前操作--在单击button_a然后单击任何其他按钮后启用button_c，或者在单击button_b然后单击任何其他按钮后禁用。

import streamlit as st

if 'but_a' not in st.session_state:
    st.session_state.disabled = True

    
button_a = st.button('a', key='but_a')

button_b = st.button('b', key='but_b')

button_c = st.button('c', key='but_c', disabled=st.session_state.disabled)

st.write(button_a, button_b, button_c)

if button_a:
    st.write("clicked A")
    st.session_state.disabled = False

if button_b:
    st.write('clicked B')
    st.session_state.disabled = True
    

Answer 1

我认为所有问题都是当您按下任何按钮时它会再次运行所有代码，并再次创建所有按钮，并且您必须在创建 button_c 之前更改状态。

if button_a:
    st.session_state.disabled = False

if button_b:
    st.session_state.disabled = True

button_c = st.button('c', key='but_c', disabled=st.session_state.disabled)

---

完整的工作代码:

import streamlit as st

if 'but_a' not in st.session_state:
    st.session_state.disabled = True

print('before:', st.session_state)

button_a = st.button('a', key='but_a')

button_b = st.button('b', key='but_b')

# change state before creating button `c` (but after creating button `a` and `b`)

if button_a:
    st.session_state.disabled = False

if button_b:
    st.session_state.disabled = True

button_c = st.button('c', key='but_c', disabled=st.session_state.disabled)

st.write(button_a, button_b, button_c)

# display text after displaying all buttons

if button_a:
    st.write("clicked A - Activate C")

if button_b:
    st.write("clicked B - Deactivate C")

print('after:', st.session_state)