<>oracle中关于null排序的问题</P>1 X$ x" m& h: @* C# u9 S' R, E
<>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> ( p8 {, j6 U- d, e! I<>问题解决:<BR>方案1:<BR>可以使用复杂的使用sql:</P>, J# j- w4 z5 c* O, {% O+ ~9 [
<>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>7 v' ?0 a* c, U: i; I2 [3 u
<>and ZBRL is null</P>+ v' f" ~( z' z
<>) 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>5 p1 |1 F7 C d; g
<>and ZBRL is null</P> : N+ L. J7 ~1 Y8 K<>)) 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>, z0 W) c$ X4 x! h; O! e. Y, G
<>and ZBRL is not null order by ZBRL <BR>) b<BR>)<BR>order by my_sys_rownum desc</P>: F& B7 a2 S. F* T4 d6 ~4 Y
<>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>