DonkeyGallery Help

[infomaniaco]

Hello everyone, I am new here and I would like to explain to you a problem that is concerning me regarding Donkey Gallery (http....). Donkey Gallery is a set of ajax, php and css which cooperate in order to create a beautiful, fast and clean picture gallery wich can be easily inserted into websites. Donkey Gallery is a set of ajax, php and css which cooperate in order to create a beautiful, fast and clean picture gallery which is easy to insert into websites. . As you can see it will be enough to charge your own pictures into the /images cartel and open an empty class. example: (

class id="gallery-id"

) this is a html page which hosts good picture presentations by DonkeyGallery. My problem is: my database contains information for every single picture and I would like it to appear along with the picture. How can I do that? source: https://help.github.com/articles/github-flavored-markdown Donkey gallery download: https://github.com/Gix075/donkeyGallery

[IT] Salve a tutti, sono nuovo quì, e vorrei esporvi un problema che mi affligge riguardo Donkey Gallery (http://factory.brainleaf.eu/donkeyGallery/) DonkeyGallery è un insieme di ajax, php e css che cooperano per avere una galleria di immagini bella,veloce,pulita, e sopratutto facile da implementare nei propri siti web. Infatti basta caricare le proprie immagini nella cartella /images e aprire un div vuoto

esempio:(

class id="gallery-id"

)

Nella pagina in html per avere una buona presentazione delle immagini. Questo è il mio problema: ho delle informazioni contenute nel mio database per ogni immagine, e vorrei e si potessero visualizzare insieme all'immagine, come potrei farlo?

fonte:https://help.github.com/articles/github-flavored-markdown Donkey gallery download: https://github.com/Gix075/donkeyGallery

[Gix075]

Hi, this is impossible to do with donkeyGallery at now. Probably in the next future I will add a new feature that allows a json data loading instead the actual folder file listing. The only thing you can do is to write a php script that calls your db and returns a json response with the data you want to add on your images, then, with some javascript code, you can append the new data on each image.

[infomaniaco]

thanks for your answer, I'm thinking to calls my DB and take the info modifing this function :

galleryGenerate($galleryPath,$forceGenerate,$thumbSizes,$elementId,$root) { $md5check = md5check($galleryPath,$root); $galleryImages = imageList($root.$galleryPath); $galleryNumb = count($galleryImages); for ($i=0; $i<$galleryNumb; $i++) { if (($forceGenerate == 1) or (md5check == "changed")) { thumbGenerate($galleryImages[$i],$galleryPath,$thumbSizes,$root); }
$alt = $i+1; //call DB and take info of "galleryImages[$i]" //$code['jsonInfo'][]= "IMAGES INFO HERE" $code['jsonFancy'][] = "<a href=\"".$galleryPath.$galleryImages[$i]."\"><img src=\"".$galleryPath."thumb/thumb-".$galleryImages[$i]."\" alt=\"Image ".$alt."\" />"; $code['jsonNoFancy'][] = "<img src=\"".$galleryPath."thumb/thumb-".$galleryImages[$i]."\" alt=\"Image ".$alt."\" />"; $galleryJsArray[] = " { 'href' : '".$galleryPath.$galleryImages[$i]."' } "; } return $code; }

Is that possible? or am I doing it wrong?

[Gix075]

Yes i think is possible, but remember that this function is used to generate a json response for the jquery plugin, so (to obtain a new markup with new data) you need to modify the jquery.donkeyGallery.js file too .

[infomaniaco]

Well, so I must use a new markup for my new information from my DB or can I use that one?

---

PHP galleryFunction.php ==> function galleryGenerate

---

function galleryGenerate($galleryPath,$forceGenerate,$thumbSizes,$elementId,$root) { $md5check = md5check($galleryPath,$root); $galleryImages = imageList($root.$galleryPath); $galleryNumb = count($galleryImages); for ($i=0; $i<$galleryNumb; $i++) { if (($forceGenerate == 1) or (md5check == "changed")) { thumbGenerate($galleryImages[$i],$galleryPath,$thumbSizes,$root); }
$alt = $i+1;

                //HERE I OBTAIN THE INFO (STRING) BY USING THE NAME OF  IMAGE
       $info= infoDB($galleryImages[$i]); 
               $code['jsonInfo'][]=$info; //HERE I WANT TO PASS THIS INFO

$code['jsonFancy'][] = "<a href=\"".$galleryPath.$galleryImages[$i]."\"><img src=\"".$galleryPath."thumb/thumb-".$galleryImages[$i]."\" alt=\"Image ".$alt."\" />"; $code['jsonNoFancy'][] = "<img src=\"".$galleryPath."thumb/thumb-".$galleryImages[$i]."\" alt=\"Image ".$alt."\" />"; $galleryJsArray[] = " { 'href' : '".$galleryPath.$galleryImages[$i]."' } "; } mysqli_close($connessione); return $code; }

---

//this is a piece of JQUERY.DONKEYGALLERY.JS

---

switch (settings.fancybox.active) { case false: var count = Object.keys(data.jsonNoFancy).length; for (var i=0; i<count; i++){ markup += "

"+ fCssOpen + data.jsonNoFancy[i] + fCssClose +"

"; //markup += "

"+ fCssOpen + data.jsonInfo[i] + fCssClose +"

"; // markup += "

"+ fCssOpen + data.jsonNoFancy[i] + fCssClose +"

" +data.jsonInfo[i] ; }

---

I really need help to continue with clarity, my target is to view the image with description saved in a DB with the name of the image (puoi rispondere anche in italiano)

[Gix075]

I think for .... { markup += ""+ fCssOpen + data.jsonNoFancy[i] + data.jsonInfo[i] + fCssClose +""; }