BiLe asked 3 years ago
Hello,
I am trying to show an image, but frontend only returns:
<img src=”undefined”>
The “intro-image” fields value would be for example ‘images/videos/video-id-9.jpg’.
Did I miss something?
In addition: is it possible to use crop/scale with image upload? I will need some images in a specific size.
$col = array();
$col[“title”] = “intro_image”; // caption of column
$col[“name”] = “intro_image”; // grid column name, must be exactly same as returned column-name from sql (tablefield or field-alias)
$col[“width”] = “50”;
$col[“align”] = “left”;
$col[“search”] = false;
$col[“sortable”] = false;
$col[“editable”] = true;
$col[“edittype”] = “image”; // render as textarea on edit
$col[“upload_dir”] = “images/temp”; // upload here
//$col[“editoptions”] = array(“rows”=>2, “cols”=>20); // with these attributes
//$col[“editoptions”] = array(“size”=>20); // with default display of textbox with size 20
//$col[“editrules”] = array(“required”=>true); // required:true(false), number:true(false), minValue:val, maxValue:val
$col[“formatter”] = “function(cellval,options,rowdata){ return ‘<img src=\”‘+cellval+’\” />’; }”;
$col[“formatoptions”] = array(“width”=>’160′,”height”=>’90’);
//$col[“hidden”] = true; // don’t show this column in list, but in edit/add mode
//$col[“editrules”] = array(“edithidden”=>true); // don’t show this column in list, but in edit/add mode
//$col[“export”] = true;
//$col[“template”] = “<img height=20 src=’http://mydomain.de/’>”; // make this a custom column with no relation to database
$cols[] = $col;
BiLe answered 3 years ago
Tried this, where {intro_image} is supposed to be the value in database:
$col[“default”] = “<img width=160 height=90 src=’http://www.mydomain.de/{intro_image}’ >”;
Will not work either.
Should I use the formatter or default? Difference?
Abu Ghufran Staff answered 3 years ago
Hello,
To use image in datagrid, You need 2 fields.
Referring this demo code: https://www.gridphp.com/demo/demos/editing/file-upload.phps
One for file add/edit purpose which is not shown in listing (line 96-108)
And other one for displaying purpose which is actually shown on listing (line 110-147)
In the later column, You can have this code with placeholder in default property {intro_image}.
PS: I’m adding the crop feature in todos.
BiLe answered 3 years ago
Hello,
thank you for adding the crop feature.
I have looked at the example and tried to replicate, but there still has to be something wrong with my code.
As said, I have a database column “intro_image” with value e.g. “images/videos/myvideo.jpg”. I would like to display the image inside the grid and on click link to another page.
Right now I am getting “None” as output 🙂
// intro_image
$col = array();
$col[“title”] = “intro_image”; // caption of column
$col[“name”] = “intro_image”; // database field name
$col[“show”] = array(“list”=>false,”edit”=>true,”add”=>true); // show in list, edit, add
$col[“resizable”] = true;
$col[“width”] = “50”;
$col[“align”] = “left”;
$col[“search”] = false;
$col[“sortable”] = false;
$col[“editable”] = true;
$col[“edittype”] = “file”;
$col[“upload_dir”] = “images/temp”;
$col[“editrules”][“allowedext”] = “png,jpeg,jpg”; // comma separated list of extensions
$col[“editrules”][“allowedsize”] = 5 * 640 * 360; // allowed size in bytes e.g. 3 MB = 3 x 1024 x 1024
$cols[] = $col;
// virtual column
$col = array();
$col[“title”] = “image”;
$col[“name”] = “image”;
$col[“width”] = “360”;
$col[“editable”] = true;
$col[“on_data_display”] = array(“display_image”,””);
function display_image($data) {
$upload_url = explode(“/”,$_SERVER[“REQUEST_URI”]);
array_pop($upload_url);
$upload_url = implode(“/”,$upload_url).”/”;
$file = $data[“intro_image”];
$arr = explode(“.”,$file);
$ext = $arr[count($arr)-1];
if (empty($file))
return “None”;
else if (strtolower($ext) == “doc” || strtolower($ext) == “xls” || strtolower($ext) == “docx” || strtolower($ext) == “xlsx” || strtolower($ext) == “pdf”)
return “<a href=’$upload_url/$file’ target=’_blank’>Download File</a>”;
else if (strtolower($ext) == “png” || strtolower($ext) == “jpg” || strtolower($ext) == “jpeg”)
return “<a href=’$upload_url/$file’ target=’_blank’><img height=50 src=’$upload_url/$file’></a>”;
else
return $file;
}
$col[“show”] = array(“list”=>true,”edit”=>false,”add”=>false,”view”=>true);
$col[“editoptions”][“dataInit”] = “function(o){jQuery(o).parent().html(o.value);}”;
$cols[] = $col;
BiLe answered 3 years ago
I think I found one mistake I made. I need to select all columns I want to show:
$g->select_command = “SELECT
title
FROM
pages
WHERE
app_id = 6
“;
Abu Ghufran Staff answered 3 years ago
Thanks correct. If you provide sql query, only selected fields will be there in dataset.
I’ve sent you latest build as well.
