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Hi Igor, you actually need to fetch existing columns from the tree view one by one, until you find a match, something like the code below. Regards, Miroslav int current = 0; GtkTreeViewColumn *cur_column = gtk_tree_view_get_column ( while (cur_column != NULL) { if (clicked_column == cur_column) { break; //found it } current ++; cur_column = gtk_tree_view_get_column ( } if (cur_column != NULL) { // found the matching column, index stored in "current" } On 10/9/2018 3:00 AM, Igor Korot via gtk-list wrote: Hi, list, Does anybody know how do I get the information? Thank you. ---------- Forwarded message --------- From: Chris Moller <moller@xxxxxxxxxxxxxx> Date: Mon, Oct 8, 2018 at 8:02 PM Subject: Re: How do I get the number To: Igor Korot <ikorot01@xxxxxxxxx> To be honest, I'm not sure how GTK would do it. Maybe the thing to do is write a bit of code and see what happens--I've never needed to to know the column number. On 08/10/2018 08:38, Igor Korot wrote: Hi, Chris, On Mon, Oct 8, 2018 at 7:28 PM Chris Moller <moller@xxxxxxxxxxxxxx> 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@xxxxxxxxxxxxxx> 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@xxxxxxxxxxxxxx> 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@xxxxxxxxxxxxxx> 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@xxxxxxxxxxxxxx> 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@xxxxxxxxxxxxxx> 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. _______________________________________________ gtk-list mailing list gtk-list@xxxxxxxxx https://mail.gnome.org/mailman/listinfo/gtk-list _______________________________________________ gtk-list mailing list gtk-list@xxxxxxxxx https://mail.gnome.org/mailman/listinfo/gtk-list _______________________________________________ gtk-list mailing list gtk-list@xxxxxxxxx https://mail.gnome.org/mailman/listinfo/gtk-list |
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