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C
CamiloU
Inspiring
February 21, 2024
Answered

A couple of scripts that worked in earlier versions of ID

  • February 21, 2024
  • 2 replies
  • 332 views

 

1. One is a script wrote by Manan Joshi that indeed inserts a row at the end of the table not in the character style indicated in the code but in the style of the row preceding it. 

 

 table.rows[-1].cells.everyItem().texts.everyItem().appliedCharacterStyle = “myriad regular”

 

**********

Manam wrote another version that gives this error.

 

app.documents[0].stories.everyItem().tables.everyItem().rows.add(LocationOptions.AT_END).cells.everyItem().appliedCellStyle = "StyleName"

https://community.adobe.com/t5/indesign-discussions/it-is-possible-to-add-one-row-or-a-line-of-text-at-the-end-of-a-table-in-a-document-with-many-table/m-p/11727702

/////////////////////

 

2. The another one, by Jump_Over, gives an error 21: “undefined is not an object”.

app.documents[0].stories.everyItem().tables.everyItem().rows.add(LocationOptions.AT_END).cells.everyItem().appliedCellStyle = “EndRow”

 

https://community.adobe.com/t5/indesign-discussions/selecting-last-row-of-table/m-p/5151572/highlight/true

 

 

 

    This topic has been closed for replies.
    Correct answer Peter Kahrel

    Camilo -- You'll have to split the long statement:

    rows = app.documents[0].stories.everyItem().
  tables.everyItem().rows.add (LocationOptions.AT_END);

for (i = 0; i < rows.length; i++) {
  rows[i].cells.everyItem().appliedCellStyle = "c";
}

    .add() creates an array of rows, and therefore cells.everyItem() doesn't work.

    2 replies

    Peter Kahrel
    Community Expert
    Peter KahrelCommunity ExpertCorrect answer
    Community Expert
    February 21, 2024

    Camilo -- You'll have to split the long statement:

    rows = app.documents[0].stories.everyItem().
  tables.everyItem().rows.add (LocationOptions.AT_END);

for (i = 0; i < rows.length; i++) {
  rows[i].cells.everyItem().appliedCellStyle = "c";
}

    .add() creates an array of rows, and therefore cells.everyItem() doesn't work.

    C
    CamiloUAuthor
    Inspiring
    February 22, 2024

    Peter, thanks for these splendid solutions.
    The strength of this forum lies in people like you.
    Best regards.

    C.

      brian_p_dts
      Community Expert
      brian_p_dtsCommunity Expert
      Community Expert
      February 21, 2024

      For the latter two scripts, do you have Cell Styles (not Character Styles) with the correct name? 

      C
      CamiloUAuthor
      Inspiring
      February 21, 2024

      Brian,
      for 

      app.documents[0].stories.everyItem().tables.everyItem().rows.add(LocationOptions.AT_END).cells.everyItem().appliedCellStyle = "StyleName"

      the error  persists. Cell Style is adequately named. In the previous example used other names for cell styles but were changed in other tests.

       

      and

      app.documents[0].stories.everyItem().tables.everyItem().rows.add(LocationOptions.AT_END).cells.everyItem().appliedCellStyle = "EndRow"

      the same. Cell styles names checked.

       

      Thanks.


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