我使用反應js爲我的意見!我正在彈出一個彈出窗口,顯示用戶鍵入的數據的一些細節,彈出窗口會在點擊提交按鈕後顯示! 如何將作業狀態通過單擊提交按鈕傳遞給彈出窗口?彈出代碼是 ! (使用咖啡腳本)傳遞@states和道具彈出模型
@Jobs.TipIncrease = React.createClass
displayName: "TipIncrease"
componentDidMount: ->
$(".ui.modal").modal
# detachable: false
onApprove: =>
data =
meet_at_car: false
@disableButtonShowSpinner()
@props.saveJobValues(data, @props.handleSubmit)
onDeny: =>
data =
meet_at_car: false
@disableButtonShowSpinner()
@props.saveJobValues(@props.handleSubmit)
disableButtonShowSpinner: ->
$("#submit-done").addClass('hide')
$("#submit-button").addClass('disabled').css("padding", "1.5em")
$("#submit-spinner").removeClass('hide')
render: ->
job = @props.job
# @setState jobs: jobs
# CSS & Stylings
modalStyle =
maxWidth: "33%"
marginLeft: "-275px"
# Render
React.DOM.div
className: "ui modal"
# style: modalStyle
React.DOM.div
className: "header"
"Increase Tip peeps"
# Gogetter Tips
React.DOM.div
className: "row"
# style: borderBottom
React.DOM.div
className: "sixteen wide column"
className: "content"
React.createElement Jobs.TipBox,
# poster_tip: job.poster_tip
saveJobValues: @props.saveJobValues
React.DOM.div
className: "actions"
React.DOM.div
className: "ui red cancel inverted medium button"
style: padding: "9px 50px 7px 17px"
React.DOM.i
className: "remove icon"
"Later"
React.DOM.div
className: "ui green ok inverted medium button"
style: padding: "9px 60px 7px 17px"
React.DOM.i
className: "checkmark icon"
"Submit"
這裏(poster_tip:job.poster_tip)海報尖是不確定的。多數民衆贊成在我面臨的問題,因爲國家沒有通過!我正在展示它!
# Done button
React.DOM.div
className: "sixteen wide column"
React.createElement Jobs.Button.DoneModal
# TipIncrease at Car Modal
React.DOM.div
className: "sixteen wide column"
React.createElement Jobs.TipIncrease,
saveJobValues: @props.saveJobValues
handleSubmit: @props.handleSubmit
彈出窗口工作正常,但它沒有采取創建工作的狀態!
@Jobs.Button.DoneModal = React.createClass
displayName: "Button-DoneModal"
showModal: ->
# $(".ui.small.modal").modal("show")
$(".ui.modal").modal("show")
$('.ui.dimmer.modals').appendTo('.ui.fluid.container.pushable');
render: ->
# CSS & Stylings
buttonStyle =
backgroundColor: "#00b4ad"
color: "white"
borderRadius: "2em"
width: "100%"
padding: "1em"
fontSize: "18px"
margin: "1em auto"
# Render
React.DOM.button
type: "button"
className: "ui button"
id: "submit-button"
onClick: @showModal
style: buttonStyle
React.DOM.span
id: "submit-done"
"DONE"
React.DOM.div
className: "ui small hide active loader"
id: "submit-spinner"
我怎麼會把狀態? 錯誤是:
(Uncaught Error: Invariant Violation: setState(...): Cannot update during an existing state transition (such as within render). Render methods should be a pure function of props and state.)
請更新您粘貼的代碼中的縮進,這在coffeescript中很重要;否則很難讀取代碼邏輯。 –
@ArieShaw它幾乎正確!它的超級難以在這裏正確縮進!我收到錯誤! (未捕獲錯誤:不變違例:setState(...):無法在現有狀態轉換期間更新(例如'render'內)。渲染方法應該是道具和狀態的純函數。) – bilal
我想獲取當前「提示」狀態並將其顯示到模型中,但每次我都可以工作。提示其未定義! – bilal