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  1. #1
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    Join Date
    Sep 2004

    c++ model for ListView question


    I implemented standard solution with QAbstractListModel derived class used as a model for qml ListView, it works great. Next step is to separate model from actual data a bit, instead of returning QVariant<QString> ( for example ) from data function I want to return my custom type like this: QVariant<MyType>... then in qml in delegate I want to use it, for example:
    // assuming there is data model role named myrole, model data function is returning QVariant<MyType>
      source: myrole.filename()

    Unfortunately such construction does not work, in application output window there is a message that myrole.filename() is undefined, the same happens when MyType has property filename and in qml it is used like "source: myrole.filename". However is there is just " source: myrole" then there is a message "cannot convert MyType to QUrl" which seems to be true

    How I should proceed with that? Is such approach possible at all?

    Thanks in advance


    PS What about pointers, my idea is to return QVariant<MyType*> and actually return a pointer to derived type and then in qml check what type actually is and construct logic basing on that
    Last edited by sebaw; 2011-09-16 at 12:05.

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