试图将数据从输入页复制到已转置的输出页。第一次在图纸之间进行复印,不是很有经验。我已经尝试了许多方法来查看前面类似的stackoverflow报告来执行copyTo行代码。 错误“缺少;语句前。”(第11行,文件“代码”) 第11行是最后一行,inRng.copyTo ...
function fetchData() {
//Browser.msgBox("Test Button");
var inData = ("FUNGRAPH9");
var outData = ("Data");
var ss = SpreadsheetApp.getActiveSpreadsheet();
var input = ss.getSheetByName(inData);
var output = ss.getSheetByName(outData);
//test getting data for 1st chart
inRng = input.getRange(10,3,1,7);
outRng = output.getRange(6,21,12,21);
inRng.copyTo(outData,6,21,12,21).CopyPasteType.transposed);
}
Tanaike回答后的修订代码。现在可以使用了。谢谢!
function fetchData() {
//Browser.msgBox("Test Button");
var inData = ("FUNGRAPH9");
var outData = ("Data");
var ss = SpreadsheetApp.getActiveSpreadsheet();
var input = ss.getSheetByName(inData);
var output = ss.getSheetByName(outData);
//test getting data for 1st chart
inRng = input.getRange(10,3,1,7);
outRng = output.getRange("U6");
inRng.copyTo(outRng, SpreadsheetApp.CopyPasteType.PASTE_NORMAL, true);
}
答案 0 :(得分:0)
我使用了Tanaike的答案(针对我的更正案例进行了修改),并且有效,更正了代码:
function fetchData() {
//Browser.msgBox("Test Button");
var inData = ("FUNGRAPH9");
var outData = ("Data");
var ss = SpreadsheetApp.getActiveSpreadsheet();
var input = ss.getSheetByName(inData);
var output = ss.getSheetByName(outData);
//test getting data for 1st chart
inRng = input.getRange(10,3,1,7);
outRng = output.getRange("U6");
inRng.copyTo(outRng, SpreadsheetApp.CopyPasteType.PASTE_NORMAL, true);
}