Need help with jpg.

[robertoathome]

We use Access 2002.
I have a table that contains a column with hiperlinks to jpgs stored on a disk on the network.
I created a form that shows all the fields in the table including the pictures. If I use and OLE object with bmp images all is good. The problem is that I can't use bmp. If I try to display a jpg on an image box/ bound or unbound object box / text box or.... nothing comes up. I put a bmp on the first 2 records and they work; as move to the 3rd the other records no jpg is displayed. I'm not good with coding but I read somewhere that I can create a macro or code that can be a workaround.
Can any1 help me? It's really getting to me :) I've been reading for days on the syntax to use in coding but my company doesn't want me to spend all this time "training myself"
Can I use a macro or do I need to write a code?
If the workaround needs a button to open the pic in the form, ..not the ideal but...so be it.

Table name= table1
Form name= ViewRecord
field name= Photo

Thanks!

[Denburt]

O.K. it's bee a while since I fooled with linked pics but you should be able to use the on Current event of the form.

Private Sub Form_Current()
Rename the folowing Me!myLink with the field that hold the links
If Not IsNull(Me!MyLink) And Len(Me!MyLink) > 0 Then
'Me!Image0 should be the image placeholder also replace the myLink as stated above
Me!Image0.Picture = Me!MyLink
End If
End Sub

This should get you close any issues let me know.

[robertoathome]

OK, something is working but not all.

1) I'm assuming I need to use the first type of image holder, not the unbound or bound OLE object.
2) when I run the code I get an error that says"

"run-time error '2220':
Microsoft access can't open the file 'T:\SHARED FILES\Administr ation\Masters and Molds Photos\040F-6F222032-TL.JPG#T:\SHARE D FILES\Administr ation\Masters and Molds Photos\040F-6F222032-TL.JPG#.

Basically it dubles the link I have in the table1 and adds a ' # ' after it.
If I go to debud I get the following highlighted
' Image42.Picture = Photo'

Here's the code I used:

Private Sub Form_Current()
If Not IsNull(Photo) And Len(Photo) > 0 Then
Image42.Picture = Photo
End If
End Sub

THANKS!

O.K. it's bee a while since I fooled with linked pics but you should be able to use the on Current event of the form.

Private Sub Form_Current()
Rename the folowing Me!myLink with the field that hold the links
If Not IsNull(Me!MyLink) And Len(Me!MyLink) > 0 Then
'Me!Image0 should be the image placeholder also replace the myLink as stated above
Me!Image0.Picture = Me!MyLink
End If
End Sub

This should get you close any issues let me know.

[robertoathome]

In doubt of the syntax I aslo added the " Me!" :

Private Sub Form_Current()
If Not IsNull(Me!Photo ) And Len(Me!Photo) > 0 Then
Me!Image42.Pict ure = Me!Photo
End If
End Sub

"Photo" is the column in my table that holds the links to the pictures.

[Denburt]

??? How large are your pics?

I tried tried it with some windows sample pics and hadd no problem what so ever. Did you change any of the properties on the control? I simply placed the control added a dummy pic as it asked for 1 but when it opens it opens whith my first pic from the bd.
I tried it using the field photo and didn't have an issue but I don't like using any 1 word for a fieldname since you never know when or if MS has designated it as a property, method, or function.

MyPhoto as Text Field
C:\Documents and Settings\All Users\Documents \My Pictures\Sample Pictures\Blue Hills.jpg
C:\Documents and Settings\All Users\Documents \My Pictures\Sample Pictures\Sunset .jpg
C:\Documents and Settings\All Users\Documents \My Pictures\Sample Pictures\Water Lilies.jpg
C:\Documents and Settings\All Users\Documents \My Pictures\Sample Pictures\Winter .jpg

Each one in a separate record and it went smoothly.

The following site is using a class module for his example and maybe this might be a resolution for you. At this point i don't know what else I can offer.

Image Sample DB

[Denburt]

UGH Just reread your Post you are using Hyoperlink as a field to hold your pics that explains it!

If you have to use a hyperlink field then you will need to parse it.

In your current code before you set the picture.

Ytou will need to use something to the effect of:

instr(Me!'Photo,"#")
if instr>0 then
strPhoto = left(me!Photo,instr)
End if

A Hyperlink field stores the link and the text you want displayed. If you don't enter text to display it will display the link twice using # as a separator, as you posted.

[robertoathome]

BETTER news !
I don't "have to" use hyperlinks but it is easier than changing all my database to text.
With simple text the code in the "Photo" field actually now works! I had to change the allowable number of characterd in the field since the links are long.

I tried to add your second piece of code u gaver me to 'parse the link' but I get a syntax error no matter where I place the code. Below is where I 'think' you wanted me to insert it.

Private Sub Form_Current()
instr(Me!'Photo ,"#")
if instr>0 then
strPhoto = left(Me!Photo,i nstr)
End If
If Not IsNull(Me!Photo ) And Len(Me!Photo) > 0 Then
Me!Image42.Pict ure = Me!Photo
End If
End Sub

The debugger stops at the very first line and makes the new code part red.

I appreciate your help!

[robertoathome]

BETTER news !
I don't "have to" use hyperlinks but it is easier than changing all my database to text.
With simple text the code in the "Photo" field actually now works! I had to change the allowable number of characterd in the field since the links are long.

I tried to add your second piece of code u gaver me to 'parse the link' but I get a syntax error no matter where I place the code. Below is where I 'think' you wanted me to insert it.

Private Sub Form_Current()
instr(Me!'Photo ,"#")
if instr>0 then
strPhoto = left(Me!Photo,i nstr)
End If
If Not IsNull(Me!Photo ) And Len(Me!Photo) > 0 Then
Me!Image42.Pict ure = Me!Photo
End If
End Sub

The debugger stops at the very first line and makes the new code part red.

I appreciate your help!

UGH Just reread your Post you are using Hyoperlink as a field to hold your pics that explains it!

If you have to use a hyperlink field then you will need to parse it.

In your current code before you set the picture.

Ytou will need to use something to the effect of:

instr(Me!'Photo,"#")
if instr>0 then
strPhoto = left(me!Photo,instr)
End if

A Hyperlink field stores the link and the text you want displayed. If you don't enter text to display it will display the link twice using # as a separator, as you posted.

[Denburt]

Glad you got it working :) sorry for my sloppy coding I just looked at it and it WAS bad. My appologies it should have looked like the following. I did test this one btw

Private Sub Form_Current()
'Always declare your variables
Dim inInt as integer
Dim strPhoto as string
inInt = InStr(Me!Photo, "#")
If inInt > 0 Then
Debug.Print Left("MyPhoto.html#photo.html", inInt - 1)
strPhoto = Left(Me!Photo, inInt)
End If
If Not IsNull(Me!Photo) And Len(Me!Photo) > 0 Then
Me!Image42.Picture = Me!Photo
End If
End Sub

[robertoathome]

I'm sorry to bug you one more time but now at least the problem seems almost solved.
I still get this error:
Run-tme error '2220'
Microsoft Access can't open the file '#C:\Documents and Settings\ELIAS\ Desktop\new\200 2\040F-6F182424.jpg#'

The debug highlights:
Me!Image42.Pict ure = Me!Photo

It looks like the '#' still gets added in the string. :(

I don't get it. I tried links that work with using just text so they are working. If I click them I can open the pics. I also used local pics, in the same folder at the database. This is what I have as the first 2 links:
C:\Documents and Settings\ELIAS\ Desktop\new\200 2\040F-6F182424.jpg
or
T:\SHARED FILES\Administr ation\Masters and Molds Photos\040F-6F18D24.JPG

The pics are only 200kb

I tried to put nothing in the "text to display" and no changes.
Thanks!!

=============== ==

Glad you got it working :) sorry for my sloppy coding I just looked at it and it WAS bad. My appologies it should have looked like the following. I did test this one btw

Private Sub Form_Current()
'Always declare your variables
Dim inInt as integer
Dim strPhoto as string
inInt = InStr(Me!Photo, "#")
If inInt > 0 Then
Debug.Print Left("MyPhoto.html#photo.html", inInt - 1)
strPhoto = Left(Me!Photo, inInt)
End If
If Not IsNull(Me!Photo) And Len(Me!Photo) > 0 Then
Me!Image42.Picture = Me!Photo
End If
End Sub

[robertoathome]

Now as my brain turns ( this thing is obsessing me! ) I thought maybe importing the data put something funky in the field so I created a fresh columb in the table an put a fresh link.
Same problem which alternates with the link getting doubled.

The I put an "NA" in the 'text to display' field of the hyperlink and the error now is:

Run-tme error '2220'
Microsoft Access can't open the file 'NA#C:\Documents and Settings\ELIAS\ Desktop\new\200 2\040F-6F182424.jpg#'

[Denburt]

O.K.

Debug.Print Left("MyPhoto.html#photo.html", inInt - 1)
strPhoto = Left(Me!Photo, inInt-1)

The debug.print statement in the VBA window will show you the results in the immediate window. In the VBA window go to the View Menu then click on immediate window (Ctrl G) you should see it show up.

I had that in while testing and if you look closely the debug statement had a minus 1 after the variable inInt that will remove the trailing # symbol. The above should resolve your issue. Good luck let me know how it works.

[robertoathome]

Well, I swear, if some1 asked me now to rate my intelligence I'd be very crititcal....an d confused :)

I'm still getting the same error.

I opened the 'immediate window' and the first time it showed a few lines with " "myphoto" or similar.
Now nothing, only return lines, empty.
I added the - before the 1 :

Private Sub Form_Current()
'Always declare your variables
Dim inInt As Integer
Dim strPhoto As String
inInt = InStr(Me!Photo, "#")
If inInt > 0 Then

Debug.Print Left("MyPhoto.html#photo.html", inInt - 1)
strPhoto = Left(Me!Photo, inInt - 1)

End If
If Not IsNull(Me!Photo) And Len(Me!Photo) > 0 Then
Me!Image42.Picture = Me!Photo
End If
End Sub

Also noticed that you have

Debug.Print Left("MyPhoto.h tml#photo.html" , inInt - 1)

Shouldn't it be Me!Photo and capital P after # ?

How about the space, on the first one you have
inInt - 1

on the following line
inInt-1

I just I'm just reaching :)

If I'm understanding correctly the code is only trying ot take the initial #. There is one also at the end of the hyperlink.

The time you're taking is immensly appreciated.

[Denburt]

The code I supplied will pick up the first part of the link, I take it that this would be the text to display so if that isn't the link my bad. Slight adjustment needs to be made.

The I put an "NA" in the 'text to display' field of the hyperlink and the error now is:
Run-tme error '2220'
Microsoft Access can't open the file 'NA#C:\Document s and Settings\ELIAS\ Desktop\new\200 2\040F-6F182424.jpg#'

Also my debug statement was just parsing a string I wasn't trying to tie it into your form or anything.
Debug.Print Left("MyPhoto.h tml#photo.html" , inInt - 1)

Dim inInt As Integer
Dim strPhoto As String
inInt = InStr(Me!Photo, "#")
If inInt > 0 Then
    strPhoto = Mid(Me!Photo, inInt + 1, Len(Me!Photo) - inInt - 1)
End If
If Not IsNull(Me!Photo) And Len(Me!Photo) > 0 Then
    Me!Image42.Picture = Me!Photo
End If

[robertoathome]

Here I am again.
Same problem.

The # is still there. The code is reading from the "text to display" which is ok to me.

When I insert a new link in the table the code reads it twice and puts it twice in the len part. If I point the mouse on the last Len statement the tooltip shows a count of 142 instead of 71. Plus is still puts the ' # ' . This only happens when I insert a new link in my table.

What I also don't understand is why if I look at

strPhoto = Mid(Me!Photo, inInt + 1, Len(Me!Photo) - inInt - 1)

and point the mouse on strPhoto, it shows the correct link without the # but if I scroll down to the last line of code

Me!Image42.Picture = Me!Photo

and point the mouse to Me!Photo it shows the link with the #.


Aren't you happy to have decided to help me? :)

Current code:

Private Sub Form_Current()
Dim inInt As Integer
Dim strPhoto As String
inInt = InStr(Me!Photo, "#")
If inInt > 0 Then
    strPhoto = Mid(Me!Photo, inInt + 1, Len(Me!Photo) - inInt - 1)
End If
If Not IsNull(Me!Photo) And Len(Me!Photo) > 0 Then
    Me!Image42.Picture = Me!Photo
End If
End Sub