我投赞成票:

=INDEX(A2:A,COUNTA(A2:A),1)

因此,对于OP的需要:

=DAYS360(A2,INDEX(A2:A,COUNTA(A2:A),1))

If the column expanded only by contiguously added dates as in my case - I used just MAX function to get last date.

最后公式是:

=DAYS360(A2; MAX(A2:A)) 

这里指的是:

=indirect("A"&max(arrayformula(if(A:A<>"",row(A:A),""))))

最后是:

=DAYS360(A2,indirect("A"&max(arrayformula(if(A:A<>"",row(A:A),"")))))

这里工作的另一个方程式是这样,但我喜欢这样一来,因为它使人数变得容易,我认为我需要做更多的工作。 仅仅这样说:

=max(arrayformula(if(A:A<>"",row(A:A),"")))

我最初试图找到解决办法,解决一个电子表格问题,但找不到任何用处,只是给最后入境次数增加,因此,希望这对一个人有益。

此外,这还具有额外优势,即它能够按任何顺序为任何类型的数据服务,而且你可以在有内容的浏览器之间有空白的浏览器,并且用“评价”的公式计算电池。 它还可以处理重复的价值观。 所有这一切都与此处使用最大(G:G<>”*row(G:G)的公式非常相似,但如果把增长数推向你们之后,则会稍加容易。

Alternatively, if you want to put a script on your sheet you can make it easy on yourself if you plan on doing this a lot. Here s that scirpt:

function lastRow(sheet,column) {
  var ss = SpreadsheetApp.getActiveSpreadsheet();
  if (column == null) {
    if (sheet != null) {
       var sheet = ss.getSheetByName(sheet);
    } else {
      var sheet = ss.getActiveSheet();
    }
    return sheet.getLastRow();
  } else {
    var sheet = ss.getSheetByName(sheet);
    var lastRow = sheet.getLastRow();
    var array = sheet.getRange(column + 1 +  :  + column + lastRow).getValues();
    for (i=0;i<array.length;i++) {
      if (array[i] !=   ) {       
        var final = i + 1;
      }
    }
    if (final != null) {
      return final;
    } else {
      return 0;
    }
  }
}

如果您希望把最后一行放在你目前重新编辑的同一页上的话,你可以这样做:

=LASTROW()

or if you want the last row of a particular column from that sheet, or of a particular column from another sheet you can do the following:

=LASTROW("Sheet1","A")

And for the last row of a particular sheet in general:

=LASTROW("Sheet1")

然后获得实际数据。

=INDIRECT("A"&LASTROW())

或您可在最后两条回归线上修改上述文字(最后两条,因为为了从实际的栏目中获取实际价值,你必须填写表格和栏目),并将变数改为:

return sheet.getRange(column + final).getValue();

and

return sheet.getRange(column + lastRow).getValue();

One benefit of this script is that you can choose if you want to include equations that evaluate to "". If no arguments are added equations evaluating to "" will be counted, but if you specify a sheet and column they will now be counted. Also, there s a lot of flexibility if you re willing to use variations of the script.

Probably overkill, but all possible.

这为我工作。 谷歌中A栏的最后价值:

=index(A:A,max(row(A:A)*(A:A<>"")))

(如果存在的话,也会出现空白段)

This seems like the simplest solution that I ve found to retrieve the last value in an ever-expanding column:

=INDEX(A:A,COUNTA(A:A),1)

严格将最后一批非致命的囚室放在一栏中,这项工作应当做到......

=LOOKUP(2^99, A2:A)

达到最后价值的公式是什么:

=index(G:G;max((G:G<>"")*row(G:G)))

这将是你最初任务的最后公式:

=DAYS360(G10;index(G:G;max((G:G<>"")*row(G:G))))

假设你最初的日期是10国集团。

我走了一条不同的路线。 自2006年以来 我知道,我只想一把东西加进一行/栏,我先点一下有数据的领域。 页: 1

=COUNT(A5:A34)

So, let s say that returned 21. A5 is 4 rows down, so I need to get the 21st position from the 4th row down. I can do this using inderect, like so:

=INDIRECT("A"&COUNT(A5:A34)+4)

它寻找有数据的浏览量,并以指数变数将我送回一米。

for a row:

=ARRAYFORMULA(INDIRECT("A"&MAX(IF(A:A<>"", ROW(A:A), ))))

for a column:

=ARRAYFORMULA(INDIRECT(ADDRESS(1, MAX(IF(1:1<>"", COLUMN(1:1), )), 4)))

这将使最后一个单元的内容:

=indirect("A"&max(ARRAYFORMULA(row(a:a)*--(a:a<>""))))

这将使最后一个单元的地址:

="A"&max(ARRAYFORMULA(row(a:a)*--(a:a<>"")))

这将使最后一间牢房的囚室:

=max(ARRAYFORMULA(row(a:a)*--(a:a<>"")))

Maybe you d prefer a script. This script is way shorter than the huge one posted above by someone else:

供稿编辑,并拯救这一文字:

function getLastRow(range){
  while(range.length>0 && range[range.length-1][0]==  ) range.pop();
  return range.length;
}

其中一项行动是,你刚刚需要在一个牢房里进入:

=getLastRow(A:A)

With the introduction of LAMBDA and REDUCE functions we can now compute the row number in a single pass through the cells (Several of the solutions above filter the range twice.) and without relying on magic text or numeric values.

=lambda(rng, 
   REDUCE(0, rng, lambda(maxrow, cell, if(isblank(cell),maxrow,row(cell)) ) )
)(A:A)

单装成像使用一样的纳米功能

=LAST_ROWNUM(A:A)

该系统使用间带和多科米宽的栏目(在第一行的距离范围内,再加固),而部分栏(如A20:A),则实际增长数(在幅度内未加抵消)。

This can then be combined with Index to return the value

=DAYS360(A2, Index(A1, LAST_ROWNUM(A:A)))

(In truth, though, I suspect that the OPs date values are monotonic (even if with blanks in between), and that he could get away with

=DAYS360(A2, MAX(A2:A))

This solution is identified above as relying on the dates being "contiguous" - whether that means "no blanks" or "no missing dates" I m not certain - but either stipulation is not necessary.)