2 Correct answers
Uwe, this is pretty nasty, puh – thank you very much for pointing this out!
So for completeness sake:
$.writeln( findPagesOfLastCells() );
function findPagesOfLastCells() {
// https://community.adobe.com/t5/indesign-discussions/javascript-property-to-tell-if-text-is-in-a-continued-table/td-p/6379093
var aDoc = app.activeDocument,
allTables = aDoc.stories.everyItem().tables,
thePages = [];
if( allTables.length > 0 ) {
for ( var i = 0; i < allTables.length; i++ ) {
Hi @tobias.wantzen , Uwe used the table storyOffset property answering another post this morning, and it seems like it might work here. You can get a table‘s trailing insertionPoint to find which page it ends on. Something like this:
var et = app.activeDocument.stories.everyItem().tables.everyItem().getElements();
alert(findPagesOfLastCells(et))
/**
* Gets the parent page of a table’s end
* @ param a collection of tables
* @ return an array of page names
*
*/
function findPagesOfLastCell
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Replies
Hi @tobias.wantzen , Uwe used the table storyOffset property answering another post this morning, and it seems like it might work here. You can get a table‘s trailing insertionPoint to find which page it ends on. Something like this:
var et = app.activeDocument.stories.everyItem().tables.everyItem().getElements();
alert(findPagesOfLastCells(et))
/**
* Gets the parent page of a table’s end
* @ param a collection of tables
* @ return an array of page names
*
*/
function findPagesOfLastCells(t){
var a = new Array()
var ti, psi;
for (var i = 0; i < t.length; i++){
ti = t[i].storyOffset.index;
//the parent page of the insertion point at the end of the table
psi = t[i].parent.parentStory.insertionPoints[ti+1].parentTextFrames[0].parentPage
a.push(psi.name)
};
return a
}
Hi Rob,
I never thought of inspecting the insertion point after the character that is the actual table, because that character is positioned in a different text frame if you have two threaded text frames where one table is running through.
Test this with:
characters[0].parentTextFrames.length // Returns 1
characters[0].parentTextFrames[0].parentPage.name
// Returns the page name of the page at the beginning of the table.
But well, let's test if this still true when there is more text after that table.
Alright. Tested just this and your algorithm is still working as expected:
Above: A table with one header row, one footer row and two body rows that is running through two text frames on different pages.
Regards,
Uwe Laubender
( Adobe Community Professional )
I think it works because the storyOffset insertion point is to the left of the table—not above it— and the next insertion point is to the right of the table—not below it.
I can see that if I select the insertionPoints:
The storyOffset selected
var et = app.activeDocument.stories.everyItem().tables.everyItem().getElements();
var ti = et[0].storyOffset.index;
var psi = et[0].parent.parentStory.insertionPoints[ti]
psi.select()
And the next insertionPoint selected:
var et = app.activeDocument.stories.everyItem().tables.everyItem().getElements();
var ti = et[0].storyOffset.index;
var psi = et[0].parent.parentStory.insertionPoints[ti+1]
psi.select()
I don‘t think we have to worry about parentTextFrames being longer than 1 because the insertionPoint can only be in one frame—if the selection was something like a paragraph it might crossover multiple frames.
I don’t see that, in my last example the insertionPoint stays with the table even when the return after is forced to the next page via a table space after:
Also, if I make a textframe and insert a single table and no additional text I get 1 character and 2 insertion points no matter how much space after is set in the table options:
//a selected textframe with one table inserted with any amount of space after
var s=app.activeDocument.selection[0];
$.writeln(s.parentStory.tables.length) //returns 1
$.writeln(s.parentStory.characters.length) //returns 1
$.writeln(s.parentStory.insertionPoints.length) //returns 2, before and after the table
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