replace this option.text = data[i].styleName;let keys=Object.keys(data[i]) with option.text = data[i][keys[0]].styleName我的問題是我需要從一個數組（tFields）中讀取表中的名稱和列數，以及從第二個數組（tRecords）中讀取元素數據。我已經實現了列的讀取和創建，請大家幫忙或者告訴我創建列表的問題出在哪里。這是代碼：function buildColumns() {var rc = vConnection.recCountvar fc = vConnection.fldCountfor(var i = 0; i < fc; ++i){ var str = 'import QtQuick 2.1; import QtQuick.Controls 1.0; import QtQuick.Layouts 1.0; TableViewColumn{ id: idColumn; role: ' str += '"' + vConnection.tFields[i] + '"' + '; title: ' + '"' + vConnection.tFields[i] + '"' + '; width: ' + (tblPage.width / vConnection.fldCount) +'; Component.onCompleted: console.log("YEY");}' tbl.addColumn(Qt.createQmlObject(str, tbl))}for(var j = 0; j < rc; ++j){ var str2 = 'import QtQuick 2.12; import QtQuick.Window 2.12; import QtQuick.Controls 2.4; import QtQuick.Controls.Imagine 2.3; import QtQuick.Templates 2.5; import QtQuick.Controls 1.4; ListElement { id: idColumn; ' for(var f = 0; f < fc; ++f) { str2 += vConnection.tFields[f] + ': "' + vConnection.tRecords[f + rc] + '"; ' } str2 += '}' Qt.createQmlObject(str2, lmTable)}}Component.onCompleted: { buildColumns();}TableView {id: tblanchors.fill: parentmodel: ListModel { id: lmTable}}}
如何制作動態表
桃花長相依
2022-05-26 11:12:33
