<>oracle中关于null排序的问题</P>+ a, B+ N, O/ ^6 O
<>问题描述:<BR>在处理一般的数据记录中,对于数字类型的字段,在oracle的排序中,默认把null值做为<BR>大于任何数字的类型,当然对于varchar2类型的字段,默认也是该处理方式,但是客户<BR>要求排序的过程中,需要把null的字段默认排在前边(从小-->大)。一般的<BR>order by xxxx,无法解决。</P>( j2 D: F: B+ q3 c6 ^4 b
<>问题解决:<BR>方案1:<BR>可以使用复杂的使用sql:</P> 4 u# W+ s S, [0 }<>select * from <BR>(select a.*,rownum as my_sys_rownum from (<BR>select deptid,nvl(BDZNAME,' '),nvl(VOLLEVEL,'0'),ZBRL,nvl(ZBTS, '0'),nvl(FZR,'0'),nvl(DEPTIDDES,' '),nvl(TEL,' '),nvl(RUNSTATEDES,' '),nvl(ADDRESS,' '),BDZID from V_BDZ where rownum<2000 </P> / ?6 n$ h4 J# X& P<>and ZBRL is null</P>) p& e* m$ d' h6 X
<>) a<BR>union<BR>select b.*,rownum+(select count(*) from (<BR>select deptid,nvl(BDZNAME,' '),nvl(VOLLEVEL,'0'),ZBRL,nvl(ZBTS, '0'),nvl(FZR,'0'),nvl(DEPTIDDES,' '),nvl(TEL,' '),nvl(RUNSTATEDES,' '),nvl(ADDRESS,' '),BDZID from V_BDZ where rownum<2000 </P>' G: I, q6 j# N4 e
<>and ZBRL is null</P>8 N: i8 r' ]( [0 r. ^
<>)) as my_sys_rownum from (<BR>select deptid,nvl(BDZNAME,' '),nvl(VOLLEVEL,'0'),ZBRL,nvl(ZBTS, '0'),nvl(FZR,'0'),nvl(DEPTIDDES,' '),nvl(TEL,' '),nvl(RUNSTATEDES,' '),nvl(ADDRESS,' '),BDZID from V_BDZ where rownum<2000 </P># n4 p' a7 P: y
<>and ZBRL is not null order by ZBRL <BR>) b<BR>)<BR>order by my_sys_rownum desc</P> 5 _6 l. }- g3 k# T6 l6 K, R/ S( ~<>方案2:<BR>可以利用oracle中可以对order by中对比较字段做设置的方式来实现:<BR> 如: ……order by nvl( aaa,'-1')</P><BR>