Hello, if a record has a "photo" field and a "name" field, is there a way that the "photo" field will be filled with a picture that i name with the same info as the "name" field. Thanks!

Huh?

Where's the path field?

maybe im not explain it the right way. Let me put an example.

Lets say i have "file.fp7" and a folder with 3 pictures: "James.jpg", "Jhon.jpg" and "Peter.jpg".

In file.fp7, there is a record with two fields "Name" and "Photo", i want that if in "Name" i put "James", automatically the photo container will be "James.jpg". Tell me if i explained myself better. Thanks!

Where is the folder with 3 pictures... Is it in the same folder as file.fp7?

yes, it would be in the same folder.

create a calculation result container and try:

"image:" & "yourFolderName/" & nameField & ".jpg"

image:yourFolderName/James.jpg

hello! i tried it and it didnt work. Is there another way to do it?

sorry! i did it and it did work. But how about if i want the picts embedded, how would i do that?! thanks!

It does work, See Attached.

imageExample.zip

Embedded? Only if you actually use the insert picture script step...

Go To Field[YourImageField]

Insert Picture[]

Try this:

Set variable [$path; value: Substitute ( Get(FilePath);["file" ; "image"];[Get ( FileName ) & ".fp7" ; yourDB::name & ".jpg" ])]

Go to field [yourDB::photo]

Insert Picture["$path"]

You can make the "photo" field a button that fires this script

or

you can use EventScript plugin and an auto-enter calculation for field "name" that will fire this script