MYSQL get rows not axisting in other table

Thank you for your reply. I further investigated more in this yesterday and got this solution:

SELECT * FROM species WHERE stockimg <> 'noimage.png' AND stockimg <> '' AND suid NOT IN (SELECT suid FROM printpipe WHERE memb_guid = '$guid') ORDER BY name_sc, name_de

this worked for me. "noimage.png" is an image file showing a dummy image in case no corresponding image has already been uploaded. To be sure also empty fields are handled i have also exclaimed them. So only fields with images are taken. And they needed to be compared with the other table printpipe where suid is also used and so can be compared. So now if the user has the species already in his printpipe (cart) the species is not shown in the other list (catalogue) anymore to be chosen for the shopping cart for prints

You should use LEFT JOIN for this which will take all the records from the table of the left hand side of the JOIN. Then you can use a WHERE clause to filter out anything but the species not in printpipe. It'll make the code a lot simpler, like so:

SELECT S.*
FROM species S
LEFT JOIN printpipe P
    ON S.suid = P.suid
WHERE 
    S.memb_guid = '$guid' -- Member GUID
    AND S.stockimg <> '' -- Not empty
    AND S.stockimg <> 'noimage.png' -- Not noimage.png
    AND P.suid IS NULL -- Doesn't exist in printpipe
ORDER BY S.name_sc, S.name_de

This assumes the memb_guid field is part of the species table (otherwise if it's part of the printpipe table it's not possible to get records that don't exist in printpipe for a specific value in printpipe).

Also your post says "having empty field "stockimg"" but your query does not empty field stockimg. I assumed your post had a typo and you really meant not empty.

Note for anyone who cares about security in addition to the logical and syntactical answer to this question, using inline variables as part of a SQL string (such as in OP's example) are susceptible to SQL injections. There are alternative ways to leverage variables as part of the query instead.