问当selectInput()被checkboxInput()禁用时，始终从它中选择特定的项
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Stack Overflow用户

提问于 2020-03-21 01:37:24

回答 1查看 217关注 0票数 1

假设我在Shiny应用程序下面-

library(shiny)
library(shinyjs)

shinyApp(
  ui = fluidPage(
    useShinyjs(),
    checkboxInput("checkbox", label = "Choice a value", value = TRUE),
    selectInput("variable", "Variable:",
                c("Cylinders" = "cyl",
                  "Transmission" = "am",
                  "Gears" = "gear"))
  ),
  server = function(input, output) {
    observeEvent(input$checkbox, {
            toggleState("variable")
        })
  }
)

现在，如果单击了checkbox，那么总是从variable中选择Cylinders。

按照上述方法，这是不可能发生的。当前当用户选择其他值i.g时。Gears在启用variable之后，然后用户继续使用checkbox禁用variable，Gears的选择仍然保留在UI中，我不想这样做，而且每当variable被禁用时，都想立即返回Cylinders。

任何指针如何完成这一点将受到高度赞赏。

bretauv's 回复后的更新-

他的回答对shiny's本机selectInput()很有用，但是如果我使用shinyWidgets包的pickerInput()，如下所示，似乎不起作用-

library(shiny)
library(shinyjs)
library(shinyWidgets)

shinyApp(
  ui = fluidPage(
    useShinyjs(),
    checkboxInput("checkbox", label = "Choice a value", value = FALSE),
    pickerInput("variable", "Variable:",
                c("Cylinders" = "cyl",
                  "Transmission" = "am",
                  "Gears" = "gear"),
                 width = "50%",
                selected = "gear")
  ),
  server = function(input, output, session) {
    observe({
            if (input$checkbox) {
                    toggleState(id = "variable")
                } else {
                    toggleState(id = "variable")
                    updatePickerInput(session = session,
                                        inputId = "variable", label = NULL,  
                                        choices = c("Cylinders" = "cyl",
                                                      "Transmission" = "am",
                                                      "Gears" = "gear"),
                                        selected = "Cylinders")
                }
            })
  }
)

任何用于解决此问题的指针都将受到高度赞赏。

shiny

shinyapps

shinywidgets

回答 1

Stack Overflow用户

发布于 2020-03-21 02:02:57

您可以使用updateSelectInput (不要忘记在function(input, output, session)中添加session )。另外，当checkboxInput是TRUE或FALSE时，您需要详细说明情况。

下列守则应能发挥作用：

library(shiny)
library(shinyjs)

shinyApp(
  ui = fluidPage(
    useShinyjs(),
    checkboxInput("checkbox", label = "Choice a value", value = FALSE),
    selectInput("variable", "Variable:",
                c("Cylinders" = "cyl",
                  "Transmission" = "am",
                  "Gears" = "gear"),
                selected = "gear")
  ),
  server = function(input, output, session) {
    observe({
      if(input$checkbox == "TRUE"){
        toggleState(id = "variable")
      }

      else if(input$checkbox == "FALSE"){
        toggleState(id = "variable")
        updateSelectInput(session = session,
                          inputId = "variable",
                          choices = c("Cylinders" = "cyl",
                                      "Transmission" = "am",
                                      "Gears" = "gear"),
                          selected = "cyl")
      }

    })
  }
)