我想将数据从一个电子表格复制到另一个电子表格中。但 1.数据在几张纸上 2.只有某些列(只是为了加快速度,不需要所有列)
一开始我使用了以下脚本:
function CopyDataToNewFile() {
var sss = SpreadsheetApp.openById('0AjN7uZG....'); // sss = source spreadsheet
var ss = sss.getSheetByName('Monthly'); // ss = source sheet
//Get full range of data
var SRange = ss.getDataRange();
//get A1 notation identifying the range
var A1Range = SRange.getA1Notation();
//get the data values in range
var SData = SRange.getValues();
var tss = SpreadsheetApp.openById('8AjN7u....'); // tss = target spreadsheet
var ts = tss.getSheetByName('RAWData'); // ts = target sheet
//set the target range to the values of the source data
ts.getRange(A1Range).setValues(SData);
}
我从user2741获取了这个,发布了here:
..这个适用于一张纸。 但现在我不确定如何从更多工作表中获取数据并实际对其进行排序。有没有人对如何做到这一点有好主意?
谢谢大家, 的Sascha
如果所有内容都发生在同一个电子表格中,则此处可以使用
SpreadsheetApp.getActiveSpreadsheet().getSheetByName("SummarySheet").clear();
mergeSheet2("SummarySheet", "123");
mergeSheet2("SummarySheet", "345");
mergeSheet2("SummarySheet", "748");
mergeSheet2("SummarySheet", "293");
function mergeSheet2(targetSheetName, sourceSheetName) {
var ss = SpreadsheetApp.getActiveSpreadsheet();
var sourceSheet = ss.getSheetByName(sourceSheetName);
var lastRow = sourceSheet.getLastRow();
var lastCol = sourceSheet.getLastColumn();
var source = sourceSheet.getRange(1,1,lastRow,lastCol); // Error after some iterations: The coordinates or dimensions of the range are invalid. (line 11, file "copy-m.js") -> might be empty spreadsheet
var destSheet = ss.getSheetByName(targetSheetName);
var destRange = destSheet.getRange(destSheet.getLastRow()+1,1);
source.copyTo(destRange, {contentsOnly: true});
}
但由于某种原因,我在使用不同的电子表格时无法运行它 错误:“目标范围和来源范围必须位于同一电子表格中”
我尝试过:
SpreadsheetApp.getActiveSpreadsheet().getSheetByName("SummarySheet").clear();
SpreadsheetApp.openById('ID');
mergeSheet2("SummarySheet", "123");
function mergeSheet2(targetSheetName, sourceSheetName) {
var ss = SpreadsheetApp.openById('ID');
var ssd = SpreadsheetApp.openById('ID');
var sourceSheet = ss.getSheetByName(sourceSheetName);
var lastRow = sourceSheet.getLastRow();
var lastCol = sourceSheet.getLastColumn();
var source = sourceSheet.getRange(1,1,lastRow,lastCol); // Error after some iterations: The coordinates or dimensions of the range are invalid. (line 11, file "copy-m.js") -> might be empty spreadsheet
var destSheet = ssd.getSheetByName(targetSheetName);
var destRange = destSheet.getRange(destSheet.getLastRow()+1,1);
source.copyTo(destRange, {contentsOnly: true});
有什么想法吗?
答案 0 :(得分:2)
解决方案比您尝试的更简单,因为您只想复制值并且知道copyTo
仅适用于同一个电子表格,只需从源获取值并将其粘贴到目标中,如下所示:
function copyFromSourceToTarget(targetSheetName, sourceSheetName) {
var ss = SpreadsheetApp.openById('ID');
var ssd = SpreadsheetApp.openById('ID');
var sourceSheet = ss.getSheetByName(sourceSheetName);
var sourceData = sourceSheet.getDataRange().getValues();
var destSheet = ssd.getSheetByName(targetSheetName);
destSheet.getRange(destSheet.getLastRow()+1,1,sourceData.length,sourceData[0].length).setValues(sourceData);
}
为了使其更灵活,我建议使用4个参数并再次简化代码:
function copyFromSourceToTarget(targetSheetName,targetSsId, sourceSheetName,sourceSsId) {
var ss = SpreadsheetApp.openById(sourceSsId)getSheetByName(sourceSheetName);
var ssd = SpreadsheetApp.openById(targetSsId).getSheetByName(targetSheetName);
var sourceData = ss.getDataRange().getValues();
ssd.getRange(ssd.getLastRow()+1,1,sourceData.length,sourceData[0].length).setValues(sourceData);
}