Re: How do I get the number

[Igor Korot <ikorot01 gmail com>]

Hi, Chris,
On Mon, Oct 8, 2018 at 7:28 PM Chris Moller <moller mollerware com> wrote:
> /
> |
> |-----> disk1
>     |
>     |
>     |-------X-----------> folder1
>     |--------------------> folder2
> |-----> disk2
>     |
>     |--------------------> folder3
>     |--------------------> folder4
>     |--------------------> subdir
>                               |
>                               |--------------------> folder5
>                               |--------------------> folder6
>
>
>
>
> From that, folder1 would have an indices value of [0, 0], folder2 would be inidices[0, 1], folder3 [1,0], 
> folder4 [1,1]

So folder1 will have row 0, column 0?

But that's wrong, because folder1 is at row 2 column 2.

> By extension, folder5 would be indices[1,2,0] and folder6 [1, 2, 1]

And folder5 should be at row 8 column 3.

And what is the coordinate be for disk1?
It should be row 1, column 1, right?

So as you can see I'm trying to get a value of the column number.
On top of that I will need to get a column 1 if I click where the "X" is
on the diagram...

Thank you.

> On 08/10/2018 18:25, Igor Korot wrote:
>
> Ok.
> Now let's get back.
>
> Do you know how I can get the column from the screenshot I sent?
>
> On Mon, Oct 8, 2018, 5:04 PM Chris Moller <moller mollerware com> wrote:
> > Sorry, I misunderstood you.
> >
> > Actually, my daughter speaks pretty good Russian--she was a competitive figure skater for a lot of years 
> > and both her principal coaches were Russian.   The preferred Russian over English so my daughter learned 
> > Russian.  Anyway, she translated the songs years ago just to practise.
> >
> > On 08/10/2018 17:15, Igor Korot wrote:
> >
> > On Mon, Oct 8, 2018 at 4:09 PM Chris Moller <moller mollerware com> wrote:
> >
> > The ones by the Soviet Army?  No.  They're from a recording made in London in about 1959.  Once upon a 
> > time, back in the 60s, I had an LP version that had some translations printed on the liner, but that LP 
> > vanished decades ago.  Other than Tipperary, Annie Laurie, and Oh no, John!, they're in Russian.
> >
> > I understand.
> > And that's why I offer to translate them to English if you are interested.
> >
> > Thank you.
> >
> > On 08/10/2018 16:13, Igor Korot wrote:
> >
> > Chris,
> >
> > On Mon, Oct 8, 2018 at 3:08 PM Chris Moller <moller mollerware com> wrote:
> >
> > Here's a more complex tree, some of  of the files in my music collection.  (It's not exactly how GTK would 
> > show it, but close enough.)
> >
> > Do you want a translation for some those songs?
> >
> > ── misc
> > │   ├── PiSymphony
> > │   │   ├── Pi-and-e.ogg
> > │   │   └── The-Circle.ogg
> > │   └── SovietArmy
> > │       ├── 01 - Song of Youth.mp3
> > │       ├── 02 - A birch tree in a field did stand.mp3
> > │       ├── 03 - Far Away.mp3
> > │       ├── 04 - Song of the Volga Boatmen.mp3
> > │       ├── 05 - You are always beautiful.mp3
> > │       ├── 06 - Along Peter's Street.mp3
> > │       ├── 07 - Tipperary (sung in English).mp3
> > │       ├── 08 - Ah! Lovely Night.mp3
> > │       ├── 09 - Kamarinskaya [instrumental].mp3
> > │       ├── 10 - Annie Laurie (sung in English).mp3
> > │       ├── 11 - Song of the Plains (Meadowland).mp3
> > │       ├── 12 - Kalinka.mp3
> > │       ├── 13 - Bandura (sung in Ukrainian).mp3
> > │       ├── 14 - Oh no, John! (sung in English).mp3
> > │       ├── 15 - Snowflakes.mp3
> > │       ├── 16 - Ukrainian Poem.mp3
> > │       └── 17 - Soldiers' Chorus (from The Decembrists).mp3
> > └── synthesiser
> >     ├── Carlos
> >     │   ├── Aurora_Borealis.ogg
> >     │   ├── Midnight_Sun.ogg
> >     │   └── SonicSeasonings.ogg
> >     ├── Hyman
> >     │   ├── Evening_Thoughts.ogg
> >     │   ├── Four_Duets_in_Odd_Meter.ogg
> >     │   ├── Give_It_Up_or_Turn_It_Loose.ogg
> >     │   ├── Improvisation_in_Fourths.ogg
> >     │   ├── Kolumbo.ogg
> >     │   ├── Tap_Dance_in_the_Memory_Banks.ogg
> >     │   ├── The_Legend_of_Johnny_Pot.ogg
> >     │   ├── The_Minotaur.ogg
> >     │   ├── The_Moog_and_Me.ogg
> >     │   ├── Time_Is_Tight.ogg
> >     │   ├── Topless_Dancers_of_Corfu.ogg
> >     │   └── Total_Bells_and_Tony.ogg
> >     ├── Kitaro
> >     │   ├── Silk_Road_II.ogg
> >     │   └── Silk_Road_I.ogg
> >     ├── Tomita
> >     │   ├── Compilation
> >
> >
> > On 08/10/2018 14:48, Igor Korot wrote:
> >
> > On Mon, Oct 8, 2018 at 12:11 PM Chris Moller <moller mollerware com> wrote:
> >
> > Hi, Igor,
> >
> > The row-activated handler calls:
> >
> > void
> > user_function (GtkTreeView       *tree_view,
> >                GtkTreePath       *path,
> >                GtkTreeViewColumn *column,
> >                gpointer           user_data)
> > {
> >
> > }
> >
> > so what you need is :
> >
> > gint *indices =  gtk_tree_path_get_indices (path);
> >
> > which will return an integer vector ow the row, column, sub-column, subsub-column... depending on the 
> > topgraphy of your tree.  (You can get the depth of the tree--thus the number of entries in the 
> > vector--with:
> >
> > gint len = gtk_tree_path_get_depth (path);
> >
> > )
> >
> > If it's a simple two-dimensional tree, the column number will be indices[1], but you  might want to make 
> > sure that the tree depth is at least two before doing that or you might be pointing past the end of the 
> > indices array.
> >
> > BTW, what do you mean by saying: "simple 2-dimensional tree"?
> > Is there going to be n-dimensional tree? How it will be represented on
> > the screen? Do you have a screenshot?
> >
> > Thank you.
> >
> > By the way, don't try to free the indices array--it points to a GTK internal structure.
> >
> > HTH,
> > Chris
> >
> >
> >
> > On 08/10/2018 12:52, Igor Korot wrote:
> >
> > Hi, Chris,
> >
> > On Sun, Oct 7, 2018 at 3:34 PM Chris Moller <moller mollerware com> wrote:
> >
> > Take a look at
> >
> > gint * gtk_tree_path_get_indices (GtkTreePath *path);
> >
> > with the  "GtkTreePath  *path"  parameter you get from the row-activated callback.
> >
> > https://developer.gnome.org/gtk3/stable/GtkTreeModel.html#gtk-tree-path-get-indices
> >
> > I tried to do:
> >
> > int column = gtk_tree_path_get_indices( column );
> >
> > but got an error:
> >
> > Can't convert GtkTreeViewColumjn to GtkTreePath
> >
> > The "row-activated" callback is defined as:
> >
> > void user_function(GtkTreeView *tree_view, GtkTreePath *path,
> > GtkTreeViewColumn *column, gpointer user_data);
> >
> > And so the types are unrelated.
> >
> > What am I doling wrong?
> >
> > Thank you.
> >
> >
> >
> >
> > On 07/10/2018 20:04, Igor Korot via gtk-list wrote:
> >
> > Hi, ALL,
> > For the GtkTreeView there is a "row-activated" signal:
> > https://developer.gnome.org/gtk3/stable/GtkTreeView.html#GtkTreeView-row-activated.
> >
> > That function receives as a parameter
> >
> > column the GtkTreeViewColumn in which the activation occurred.
> >
> > The problem is (at least as I see it) is that it is not a simple integer
> > for the column number.
> >
> > Is there a simple way to get a column number on which I activated
> > the row from this signal?
> >
> > Thank you.
> > _______________________________________________
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> >
> >
> >
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