jQuery Grid Plugin - jqGrid - Topic: populate photo from a db to a grid

phicarre on populate photo from a db to a grid

Wed, 25 Nov 2009 09:00:57 +0200

For those who are looking for the/one solution:

afterInsertRow: function(rowid, aData, rowelem){
            $('#' + rowid + ' td:first-child + td + td + td').
                html('<img src =' + '”' + aData.photo + '”' +' height=”75″ width=”75″ />');
         },

- aData.photo is the path of the photo on the server

- "photo" is the colName

- the photos are stored in the fourth column in my solution

If you have a better and elegant solution, tell me it

phicarre on populate photo from a db to a grid

Mon, 16 Nov 2009 16:15:19 +0200

You are right: I should test ...

But, but I have in the cell the path to the photo and not the photo itself !

markw65 on populate photo from a db to a grid

Mon, 16 Nov 2009 08:32:06 +0200

src='/photos/ : javascript will search for this folder on your machine !

No, it wont. Try it.

Mark

phicarre on populate photo from a db to a grid

Sat, 14 Nov 2009 12:21:11 +0200

The server is one machine, the clients are thier own machine. You could be one client ! the photo are not in your machine but in my machine.

src='/photos/ : javascript will search for this folder on your machine !

Rumen[Trirand] on populate photo from a db to a grid

Sat, 14 Nov 2009 09:20:41 +0200

I am not sure I understand the scenario. What is the problem if the picture is in a directory on the server?

phicarre on populate photo from a db to a grid

Sat, 14 Nov 2009 04:59:32 +0200

No because the directory is on the server !

Rumen[Trirand] on populate photo from a db to a grid

Sat, 14 Nov 2009 02:06:31 +0200

In this case the best idea is to form the HTML of the image tag directly, instead of using formatters (and I believe there is no image formatter currently yet).

So just form the image tag, e.g.

”<img src='/photos/” . $row['Photo']." />"

phicarre on populate photo from a db to a grid

Fri, 13 Nov 2009 18:05:07 +0200

I have on the server a directory with photos.

I have in my mysql database records with a field 'photo'. the field contains the name of the photo.

I want to populate a grid with these photos with the json method.

I tried this:

while ...

$responce->rows[$i]['id']=$row['Id_Resident'];
$responce->rows[$i]['cell']=array($row['Nom'],$row['Prenom'],$row['Sexe'],"/photos/".$row['Photo']);
$i++;

echo json_encode($responce);

But the result is a link to the photo AND not the photo !

Same result if i do:

formatter:'showlink', formatoptions:{baseLinkUrl:'/photos/'},