Re: DataObjects Formbuilder

Alexander Petri wrote:

1. i defined: var $fieldLabels=array(..), but the form doesnt show it -> what did i do wrong?

It's $fb_fieldLabels. You're missing the prefix.

2. i have also a second table called crm_lander (with crm_laender_id) and i wrote the file crm.links.ini like this: (in the same folder where crm.ini is) [crm_firmen] crm_laender_id = crm_laender:crm_laender_id so i ought to get a select box with some countries but there is nothing at all! a lookup in the html code said thad crm_laender_id is a hidden field

From what I've seen, you're setting the linkDisplayFields to the same field names that contain the IDs of your foreign keys. I'm sure you don't want to display these to your end-users :-) Seriously, if crm_laender contains a field that holds the name of the country, supposedly named "name", then try this setting within *crm_laender* (!!!): var $fb_linkDisplayFields = array('name'); Whenever the table crm_laender is queried to get a list of countries for a selectbox, then FormBuilder will now try to use the content of the field "name" to display each option. CU Markus