CODEWITHSUNDEEP
qtqmlqtvirtualkeyboard
Rui Sebastião
Rui Sebastião

Reputation: 883

Hide key from Qt Virtual keyboard

Is there a way to hide the language selection key from the virtual keyboard without use a custom layout?

enter image description here

Upvotes: 5

Views: 3663

Answers (3)

Sagar Sutar
Sagar Sutar

Reputation: 11

I was able to hide the hideKeyboard key with this trick. I basically tried to get the reference of the emoji key and thereby was able to disable the next key which is hideKeyboard key.

function disableKey(parent, objectText) 
{

   var obj = null
   if (parent === null)
        return null
   var children = parent.children       
   for (var i = 0; i < children.length; i++) {
       obj = children[i]
       if (obj.text === objectText && obj.toString().substring(0, 7) === "BaseKey") {
           console.log("Disabling symbols. " + obj.text)
           obj.enabled = false
       }
       else if(obj.displayText === "HWR"){
            console.log("Disabling Handwriting mode button." + obj.displayText + " " + objectText)
            obj.visible = false
       }          
       else if(obj.text === ":-)" && obj.toString().substring(0, 7) === "BaseKey"){
           console.log("Disabling hidekeyboard key." + obj.text)
           children[i+1].visible = false
       } 
   obj = disableKey(obj, objectText)
       if (obj)
           break
   }
   return obj
}

Upvotes: 1

Rui Sebasti&#227;o
Rui Sebasti&#227;o

Reputation: 883

I was able to hide the language key with a workaround:

    property var keyboardLayout: inputPanel.keyboard.layout


    function findChildByProperty(parent, propertyName, propertyValue, compareCb) {
        var obj = null
        if (parent === null)
            return null
        var children = parent.children
        for (var i = 0; i < children.length; i++) {
            obj = children[i]
            if (obj.hasOwnProperty(propertyName)) {
                if (compareCb !== null) {
                    if (compareCb(obj[propertyName], propertyValue))
                        break
                } else if (obj[propertyName] === propertyValue) {
                    break
                }
            }
            obj = findChildByProperty(obj, propertyName, propertyValue, compareCb)
            if (obj)
                break
        }
        return obj
    }



    onKeyboardLayoutChanged: {
        if(keyboardLayout!=""){
            var ChangeLanguageKey= findChildByProperty(inputPanel.keyboard, "objectName", "changeLanguageKey", null)
            if(ChangeLanguageKey){
                ChangeLanguageKey.visible=false
            }
        }
    }


    InputPanel {
        id: inputPanel
        z: 99
        y: parent.height
        anchors.left: parent.left
        anchors.right: parent.right




        states: State {
            name: "visible"

            when: inputPanel.active
            PropertyChanges {
                target: inputPanel
                y: parent.height - inputPanel.height
            }
        }
        transitions: Transition {
            from: ""
            to: "visible"
            reversible: true
            ParallelAnimation {
                NumberAnimation {
                    properties: "y"
                    duration: 400
                    easing.type: Easing.InOutBack
                }
            }
        }





        CustomComponents.AutoScroller {

            id:autoscroller

            panelY: inputPanel.y
        }


    }

enter image description here

This only works in version 5.9 where the objectname property is defined with "changeLanguageKey", for previous versions set the property in the source code and recompile.

Upvotes: 3

Mitch
Mitch

Reputation: 24386

No, not without using a custom layout.

You can always modify the layouts that come with the keyboard though.

Upvotes: 1

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