Re: [dbforms] Foreign key handling in dbforms_2.3_20040329

[Shawn]

 I got a error:
"Make sure that field DESIGNINFOOBJID is a KEY of the table
CA_REPORTSTATICINFO! Otherwise you can not use it as PARENT/CHILD LINK
argument!"
But: The field DESIGNINFOOBJID IS already declared in dbform-config.xml 
as
<field name="DESIGNINFOOBJID" fieldType="number" size="22" 
isKey="true"/>.

Any idea how to solve this problem?
Is it a key in your database too?  Otherwise just confirm that it is a key 
in both the parent/child forms.


Has anyone a xsl-template (sample)  which enable me to create such
drop-down-boxes automaticly using the devgui-generation process?
Any help is welcome!
Have not used this myself but someone posted it a while back as working 
for them...

              <field name="tipo" isForeignKey="true" fieldType="int" 
size="10">
<foreign-key  name="FK_tabela_tabela_tipos"
      foreignTable="tabela_tipos"
      displayType="select"
      foreign="id_tipo"
      visibleFields="descricao"/></field>

So i have direct acess to the foreign table, etc... and then in the xsl i 
do
something very simple

 <xsl:choose>
 <xsl:when test="@isForeignKey='true'">
       <db:dataLabel fieldName="[EMAIL PROTECTED]" >
<xsl:for-each select="foreign-key">
<xsl:variable name="temp_name" select="@name"/>
<xsl:variable name="foreign_table" select="@foreignTable" />
<xsl:variable name="visib_fields" select="@visibleFields"/>
<xsl:variable name="str_field" select="@foreign" />
<db:tableData name="{$temp_name}"  foreignTable="{$foreign_table}"
         visibleFields="{$visib_fields}" storeField="{$str_field}"/>
</xsl:for-each>
</db:dataLabel>
</xsl:when>
<xsl:otherwise>
<db:label fieldName="[EMAIL PROTECTED]" />
</xsl:otherwise>
</xsl:choose>

--
Shawn
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