从电子表格中的单元格值解析字符串和对象

I got spreadsheet including annoying values and struggle to this. Required data is in each cell of column A. Value of a cell is {"p0":70,"u3":71,"s7110":40},t45,{"t78":60,"s3310":15},p37,p36,{"p29":44,"s8110":95},p85,p14,{"s2710":41},u47. Number of such values is about 1000. I have to parse these values. I need strings and objects from such values. I can ignore order of parsed value. I cannot parse manually. So I decided to use script.

我得到了电子表格，包括恼人的价值观和挣扎。所需数据在列a的每个单元格中。单元格的值为{“p0”:70，“u3”:71，“s7110”:40}，t45，{“t78”:60，“s3310”:15}，p37,p36，{“p29”:44，“s8110”:95}，p85, p85,p14。这些值的数量大约是1000。我必须解析这些值。我需要这些值中的字符串和对象。我可以忽略解析值的顺序。我不能手动解析。所以我决定使用脚本。

I tried to parse using "split".

我尝试使用“split”来解析。

var sheet = SpreadsheetApp.getActiveSheet();
var values = sheet.getRange("A1:A" + sheet.getLastRow()).getValues();
var result = [];
for (var i = 0; i < values.length; i++)
{
  result.push(values[i][0].split(","));
}

I got t45,p37,p36,p85,p14,u47 as string. But "split" also splits all objects. For example, {"p0":70,"u3":71,"s7110":40} is split to {"p0":70, "u3":71, "s7110":40}. Is there way to solve this?

得到t45 p37 p36 p85 p14 u47作为字符串。但是“分割”也分割所有的对象。例如,{“p0”:70年,“u3”:71年,“s7110”:40 }是分裂{“p0”:70年,“u3”:71年,“s7110”:40 }。有办法解决这个问题吗?

Sample values are this. Each line is in cell A1, A2, A3, A4, A5.

这个示例值。每一行都在A1 A2 A3 A4 A5中。

{"p0":70,"u3":71,"s7110":40},t45,{"t78":60,"s3310":15},p37,p36,{"p29":44,"s8110":95},p85,p14,{"s2710":41},u47
s6610,{"t25":70,"u8":43,"p35":86},u85,u74,{"s7710":83},{"p70":70,"u67":84},{"u71":43,"s1210":73},{"u45":84,"s710":15},{"u14":79,"p22":45},p31
u73,u12,{"t51":98,"u57":96},u31,p41,s1110,s6610,p55,{"t57":71,"s7510":83,"u62":17},u73
t50,{"t83":22,"p18":76},{"p47":12,"s8710":18,"u11":35},{"t14":74,"u72":51},{"p74":21,"t77":77},{"u62":84,"s3010":11},p81,u36,p67,{"t79":12,"u2":70,"s6010":98}
{"u54":51,"t31":31},t56,s4110,{"s3110":84,"t25":92,"p80":19},s3210,{"p65":54,"s8510":45},{"t73":78,"s6210":11},{"s2110":98,"p11":16},{"p61":55,"t88":75},p38

Thank you so much for your time. And I'm sorry for my immature question.

非常感谢您的时间。我为我不成熟的问题感到抱歉。

2 个解决方案

#1

B1:

=ARRAYFORMULA(REGEXEXTRACT(A1:A5&",",REGEXREPLACE(REGEXREPLACE(A1:A5&",","{.*?}","($0)"),"([A-Za-z]\d+),","($1),")))

We are enclosing all objects and strings except the commas , with () and then extracting them later.

除了逗号、with()之外，我们将所有对象和字符串都封装在一起，然后稍后提取它们。

EDIT:An easier anchor: The , commas to split by are NOT followed by "

编辑:一个更简单的锚:分隔的逗号后面不跟着"

=ARRAYFORMULA(SUBSTITUTE(SPLIT(SUBSTITUTE(A1:A5,","&CHAR(34),"????"),","),"????",","&CHAR(34)))


=ARRAYFORMULA(split(REGEXREPLACE(A1:A5,"(,)([^"&CHAR(34)&"])","????$2"),"????"))

#2

You can use RegExp to replace characters with something unique that can be found without affecting anything else.

您可以使用RegExp将字符替换为可以找到的惟一字符，而不影响任何其他内容。

function myFunction() {
  var L,newArray,thisElement;

  var myStrng = '{"p0":70,"u3":71,"s7110":40},t45,{"t78":60,"s3310":15},p37,p36,{"p29":44,"s8110":95},p85,p14,{"s2710":41},u47 \
s6610,{"t25":70,"u8":43,"p35":86},u85,u74,{"s7710":83},{"p70":70,"u67":84},{"u71":43,"s1210":73},{"u45":84,"s710":15},{"u14":79,"p22":45},p31 \
u73,u12,{"t51":98,"u57":96},u31,p41,s1110,s6610,p55,{"t57":71,"s7510":83,"u62":17},u73 \
t50,{"t83":22,"p18":76},{"p47":12,"s8710":18,"u11":35},{"t14":74,"u72":51},{"p74":21,"t77":77},{"u62":84,"s3010":11},p81,u36,p67,{"t79":12,"u2":70,"s6010":98} \
{"u54":51,"t31":31},t56,s4110,{"s3110":84,"t25":92,"p80":19},s3210,{"p65":54,"s8510":45},{"t73":78,"s6210":11},{"s2110":98,"p11":16},{"p61":55,"t88":75},p38';

  var re = new RegExp("\},","g");
  var parsedObj = myStrng.replace(re,"}zq^");//Replace all }, characters with }zq^

  //Logger.log(parsedObj)

  parsedObj = parsedObj.replace(/,\{/g,"zq^{");//Replace all ,{ characters with zq^{

  //Logger.log(parsedObj)
  parsedObj = parsedObj.replace(/\}\{/g,"}zq^{");//Replace all back to back brackets
  parsedObj = parsedObj.replace(/\} \{/g,"}zq^{");//Replace all back to back brackets with a space between

  parsedObj = parsedObj.split("zq^");//split on zq^

  L = parsedObj.length;

  newArray = [];

  for (var i=0;i<L;i++) {
    thisElement = parsedObj[i];
    //Logger.log('thisElement: ' + thisElement)

    if (thisElement.indexOf("{") !== -1) {
      newArray.push(thisElement);
      continue;
    }

    if (thisElement.indexOf(",") !== -1) {
      thisElement = thisElement.split(",");

      for (var j =0;j<thisElement.length;j++) {
        newArray.push(thisElement[j]);
      }
      continue;
    }

    if (thisElement.indexOf(" ") !== -1) {
      thisElement = thisElement.split(" ");

      for (var j =0;j<thisElement.length;j++) {
        newArray.push(thisElement[j]);
      }
      continue;
    }

    newArray.push(thisElement);
  }

  L = newArray.length;

  for (var i=0;i<L;i++) {
    Logger.log(newArray[i])

  }
}

#1

B1:

=ARRAYFORMULA(REGEXEXTRACT(A1:A5&",",REGEXREPLACE(REGEXREPLACE(A1:A5&",","{.*?}","($0)"),"([A-Za-z]\d+),","($1),")))