<>oracle中关于null排序的问题</P> 6 h$ H6 x7 |0 G, Z<>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>8 e' G, o3 e* S3 y
<>问题解决:<BR>方案1:<BR>可以使用复杂的使用sql:</P>- O1 w% C3 W9 u6 h+ i. J) [0 q0 R
<>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>( P7 |' w/ G9 Y q- S+ L- w
<>and ZBRL is null</P>5 M" v! v( D# E5 C; W
<>) 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>' U, y4 w; e& D. k
<>and ZBRL is null</P> 2 I. `6 H6 _ m<>)) 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> ! q# u. O5 l8 e+ Q<>and ZBRL is not null order by ZBRL <BR>) b<BR>)<BR>order by my_sys_rownum desc</P> % W1 B+ A- w" P% D- a- @% `<>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>