错误的值Google Apps脚本错误

Question

我在每次运行代码后都收到一条错误消息，指出“第35行的值不正确”。代码实际上一直运行，但是接下来会创建一个名为“undefined”的第二个文件，并且似乎是此错误的一部分。我的代码中没有任何地方可以创建多个电子表格，我们只是让它创建一个。如果我能找到解决这个问题的方法，将不胜感激。下面的代码：

function ssCreation(spreadsheetName, spreadsheetId1, sourceSheetName1, 
    spreadsheetId2, sourceSheetName2, 
    spreadsheetId3, sourceSheetName3, 
    spreadsheetId4, sourceSheetName4, 
    spreadsheetId5, sourceSheetName5, 
    spreadsheetId6, sourceSheetName6, 
    spreadsheetId7, sourceSheetName7, 
    spreadsheetId8, sourceSheetName8, 
    spreadsheetId9, sourceSheetName9, 
    spreadsheetId10, sourceSheetName10, 
    spreadsheetId11, sourceSheetName11) { 

    var create = SpreadsheetApp.create(spreadsheetName).getId(); 
    var renameSheet = SpreadsheetApp.openById(create).renameActiveSheet('1'); 
    for (var i = 1; i < 14; i++) { 
     var sheetName = 1 + i; 
     var insertTheSheet = SpreadsheetApp.openById(create).insertSheet(i); 
    } 

    var ss1 = SpreadsheetApp.openById(spreadsheetId1); 
    var ssd1 = SpreadsheetApp.openById(create); 
    var sourceSheet1 = ss1.getSheetByName(sourceSheetName1); 
    var sourceData1 = sourceSheet1.getDataRange().getValues(); 
    var destSheet1 = ssd1.getSheetByName("1"); 
    destSheet1.getRange(destSheet1.getLastRow() + 1, 1, sourceData1.length, sourceData1[0].length).setValues(sourceData1); 

    var ss2 = SpreadsheetApp.openById(spreadsheetId2); 
    var ssd2 = SpreadsheetApp.openById(create); 
    var sourceSheet2 = ss2.getSheetByName(sourceSheetName2); 
    var sourceData2 = sourceSheet2.getDataRange().getValues(); 
    var destSheet2 = ssd2.getSheetByName("Sheet2"); 
    destSheet2.getRange(destSheet2.getLastRow() + 1, 1, sourceData2.length, sourceData2[0].length).setValues(sourceData2); 

    var ss3 = SpreadsheetApp.openById(spreadsheetId3); 
    var ssd3 = SpreadsheetApp.openById(create); 
    var sourceSheet3 = ss3.getSheetByName(sourceSheetName3); 
    var sourceData3 = sourceSheet3.getDataRange().getValues(); 
    var destSheet3 = ssd3.getSheetByName("Sheet3"); 
    destSheet3.getRange(destSheet3.getLastRow() + 1, 1, sourceData3.length, sourceData3[0].length).setValues(sourceData3); 

    var ss4 = SpreadsheetApp.openById(spreadsheetId4); 
    var ssd4 = SpreadsheetApp.openById(create); 
    var sourceSheet4 = ss4.getSheetByName(sourceSheetName4); 
    var sourceData4 = sourceSheet4.getDataRange().getValues(); 
    var destSheet4 = ssd4.getSheetByName("Sheet4"); 
    destSheet4.getRange(destSheet4.getLastRow() + 1, 1, sourceData4.length, sourceData4[0].length).setValues(sourceData4); 

    var ss5 = SpreadsheetApp.openById(spreadsheetId5); 
    var ssd5 = SpreadsheetApp.openById(create); 
    var sourceSheet5 = ss5.getSheetByName(sourceSheetName5); 
    var sourceData5 = sourceSheet5.getDataRange().getValues(); 
    var destSheet5 = ssd5.getSheetByName("Sheet5"); 
    destSheet5.getRange(destSheet5.getLastRow() + 1, 1, sourceData5.length, sourceData5[0].length).setValues(sourceData5); 

    var ss6 = SpreadsheetApp.openById(spreadsheetId6); 
    var ssd6 = SpreadsheetApp.openById(create); 
    var sourceSheet6 = ss6.getSheetByName(sourceSheetName6); 
    var sourceData6 = sourceSheet6.getDataRange().getValues(); 
    var destSheet6 = ssd6.getSheetByName("Sheet6"); 
    destSheet6.getRange(destSheet6.getLastRow() + 1, 1, sourceData6.length, sourceData6[0].length).setValues(sourceData6); 

    var ss7 = SpreadsheetApp.openById(spreadsheetId7); 
    var ssd7 = SpreadsheetApp.openById(create); 
    var sourceSheet7 = ss7.getSheetByName(sourceSheetName7); 
    var sourceData7 = sourceSheet7.getDataRange().getValues(); 
    var destSheet7 = ssd7.getSheetByName("Sheet7"); 
    destSheet7.getRange(destSheet7.getLastRow() + 1, 1, sourceData7.length, sourceData7[0].length).setValues(sourceData7); 

    var ss8 = SpreadsheetApp.openById(spreadsheetId8); 
    var ssd8 = SpreadsheetApp.openById(create); 
    var sourceSheet8 = ss8.getSheetByName(sourceSheetName8); 
    var sourceData8 = sourceSheet8.getDataRange().getValues(); 
    var destSheet8 = ssd8.getSheetByName("Sheet8"); 
    destSheet8.getRange(destSheet8.getLastRow() + 1, 1, sourceData8.length, sourceData8[0].length).setValues(sourceData8); 

    var ss9 = SpreadsheetApp.openById(spreadsheetId9); 
    var ssd9 = SpreadsheetApp.openById(create); 
    var sourceSheet9 = ss9.getSheetByName(sourceSheetName9); 
    var sourceData9 = sourceSheet9.getDataRange().getValues(); 
    var destSheet9 = ssd9.getSheetByName("Sheet9"); 
    destSheet9.getRange(destSheet9.getLastRow() + 1, 1, sourceData9.length, sourceData9[0].length).setValues(sourceData9); 

    var ss10 = SpreadsheetApp.openById(spreadsheetId10); 
    var ssd10 = SpreadsheetApp.openById(create); 
    var sourceSheet10 = ss10.getSheetByName(sourceSheetName10); 
    var sourceData10 = sourceSheet10.getDataRange().getValues(); 
    var destSheet10 = ssd10.getSheetByName("Sheet10"); 
    destSheet10.getRange(destSheet10.getLastRow() + 1, 1, sourceData10.length, sourceData10[0].length).setValues(sourceData10); 

    var ss11 = SpreadsheetApp.openById(spreadsheetId11); 
    var ssd11 = SpreadsheetApp.openById(create); 
    var sourceSheet11 = ss11.getSheetByName(sourceSheetName11); 
    var sourceData11 = sourceSheet11.getDataRange().getValues(); 
    var destSheet11 = ssd11.getSheetByName("Sheet11"); 
    destSheet11.getRange(destSheet11.getLastRow() + 1, 1, sourceData11.length, sourceData11[0].length).setValues(sourceData11); 

} 

Answer 1

，而不是通过23个参数的函数来创建一个新的电子表格，然后从其他11个电子表格复制数据，为什么不种田了副本的其他功能&然后将源表信息到功能？这样，您只需在每个复制操作的代码部分进行调试，而不是在当前版本中有11个复制。下面我汇总了一个这样做的方法：重复6行数据复制代码现在在copyData()函数中有5行。

我们创建目标电子表格&仅在我们将电子表格对象传递给copyData()时引用它。数组sheetsList[]是一种key =>值对列表。因此，要从源文件复制到目标文件，我们在sheetsList[]中为每个关键字=>值对调用copyData()一次。我们通过“spreadsheetId Ñ”（键），“sourceSheetName Ñ'（值）copyData()与destinationSpreadsheet对象&片材在目标电子表格将从源接受的数据的名称一起。

希望这有助于

function copyData(sourceSpreadsheetID, sourceSheetName, destSpreadsheet, destSheetName){ 
    // sourceSpreadsheetID - ID of the source spreadsheet (string) 
    // sourceSheetName - name of the sheet in the source spreadsheet (string) 
    // destSpreadsheet - the destination spreadsheet object 
    // destSheetName - name of the destination sheet (string) 
    var srcSS = SpreadsheetApp.openById(sourceSpreadsheetID); 
    var sourceSheet = srcSS.getSheetByName(sourceSheetName); 
    var sourceData = sourceSheet.getDataRange().getValues(); 
    var destSheet = destSpreadsheet.getSheetByName(destSheetName); 
    destSheet.getRange(destSheet.getLastRow() + 1, 1, sourceData.length, sourceData[0].length).setValues(sourceData); 
} 

function doDataCopy(){ 
    // create the destination spreadsheet & store it as a Spreadsheet object that we can pass to copyData() 
    var destinationSpreadsheet = SpreadsheetApp.create("a spreadsheet"); 
    // the source spreadsheets & sheets as sort of (key, value) pairs 
    var sheetsList = ["spreadsheetId1", "sourceSheetName1", 
        "spreadsheetId2", "sourceSheetName2", 
        "spreadsheetId3", "sourceSheetName3", 
        "spreadsheetId4", "sourceSheetName4", 
        "spreadsheetId5", "sourceSheetName5", 
        "spreadsheetId6", "sourceSheetName6", 
        "spreadsheetId7", "sourceSheetName7", 
        "spreadsheetId8", "sourceSheetName8", 
        "spreadsheetId9", "sourceSheetName9", 
        "spreadsheetId10", "sourceSheetName10", 
        "spreadsheetId11", "sourceSheetName11" 
        ]; 
    for(var i = 0, lim = (sheetsList.length/2); i < lim; ++i){ 
    destinationSpreadsheet.insertSheet("Sheet" + (i+2), i+1); // <= we have to skip the name "Sheet1" as the newly created sheet already has a "Sheet1" 
    // As we're treating sheetsList[] as (key,value) pairs, the index of each 'key' is i*2 & each 'value' is (i*2)+1 
    // Also we can pass the destinationSpreadsheet object as is & copyData() can work on it directly 
    copyData(sheetsList[i*2], sheetsList[(i*2)+1], destinationSpreadsheet, "Sheet" + (i+1)); 
    } 
}