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기본 CRM 2011 Opportunities by Sales Stage 깔때기 차트를 사용하면 각 단계에 대한 데이터가 없더라도 범례의 모든 단계를 표시하려면 어떻게해야합니까? 이것은 단순한 문제처럼 보이지만 나는 혼란 스럽습니다.CRM 2011 차트 - 범례에 모든 값을 표시하는 방법?
차트에는 가져 오기 및 가져 오기로 검색 한 데이터 만 표시되며 각 스테이지에 대해 적어도 하나 이상의 레코드가있을 필요는 없습니다. 또한 깔때기 형 차트는 하나 이상의 시리즈를 가질 수 없습니다.
이것은 내가 생각해 봤지만 깔때기 형 차트로 작동하지 않습니다. 레이다 챠트로 스와핑을하면 레이아웃에 의한 무언가가이 유형의 데이터에는 쓸모가 없습니다.
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<name>Opportunities by Sales Stage</name>
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<attribute groupby="true" alias="groupby_column" name="salesstagecode">
<filter type="or">
<condition attribute="salesstagecode" operator="eq" value="1" />
</filter>
</attribute>
<attribute alias="aggregate_column_2" name="opportunityid" aggregate="count">
<filter type="or">
<condition attribute="salesstagecode" operator="eq" value="200000" />
</filter>
</attribute>
<attribute alias="aggregate_column_3" name="opportunityid" aggregate="count">
<filter type="or">
<condition attribute="salesstagecode" operator="eq" value="200001" />
</filter>
</attribute>
<attribute alias="aggregate_column_4" name="opportunityid" aggregate="count" >
<filter type="or">
<condition attribute="salesstagecode" operator="eq" value="200002" />
</filter>
</attribute>
<attribute alias="aggregate_column_5" name="opportunityid" aggregate="count">
<filter type="or">
<condition attribute="salesstagecode" operator="eq" value="100000000" />
</filter>
</attribute>
<attribute alias="aggregate_column_6" name="opportunityid" aggregate="count">
<filter type="or">
<condition attribute="salesstagecode" operator="eq" value="100000001" />
</filter>
</attribute>
<attribute alias="aggregate_column_7" name="opportunityid" aggregate="count" >
<filter type="or">
<condition attribute="salesstagecode" operator="eq" value="100000002" />
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</attribute>
<attribute alias="aggregate_column_8" name="opportunityid" aggregate="count">
<filter type="or">
<condition attribute="salesstagecode" operator="eq" value="100000003" />
</filter>
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<category>
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<measure alias="aggregate_column_3" />
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<measure alias="aggregate_column_7" />
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<measurecollection>
<measure alias="aggregate_column_8" />
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누구나 더 좋은 방법이 있습니까?