Skip to Main Content

APEX

Announcement

For appeals, questions and feedback about Oracle Forums, please email oracle-forums-moderators_us@oracle.com. Technical questions should be asked in the appropriate category. Thank you!

Display an image on an interactive grid

J.colOct 25 2018 — edited Oct 31 2018

Hello everyone,

I would like to do the following :

Create a file browse item P1_FIC, which insert a file into APEX_APPLICATION_TEMP_FILES table.

A button UPLOAD does a submit of the page, so upload the file.

On the same page, an IG displays the lines into apex_application_temp_files, and if the file mimtype is image, the image is displayed on the last column.

I tried two things, both didn't work.

1) put the BLOB_CONTENT column in display image type (unsupported), but I got this error :

Data type BLOB can not be converted to VARCHAR2!

2) Use this SQL piece :

select id, application_id, name, filename, mime_type, created_on, CASE WHEN mime_type like 'image%'

    THEN '<img src="'||apex_util.get_blob_file_src('BLOB_CONTENT',id)||'" />'

    ELSE

    '<a href="'||apex_util.get_blob_file_src('BLOB_CONTENT', id)||'">Download</a>'

    end blob_content

from apex_application_temp_files;

But I don't know what the first parameter of apex_util.get_blob_file_src should be.
I saw it should be a field, but why ? In my case, I don't want it to be a field.

Any help please ?

Comments
Post Details
Added on Oct 25 2018
9 comments
4,885 views