<>oracle中关于null排序的问题</P>& c) |3 F/ [- l2 T. @# E% ]9 J
<>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>2 T+ |9 P4 B2 B& F7 @$ h( \- W
<>问题解决:<BR>方案1:<BR>可以使用复杂的使用sql:</P> / o8 d8 t: p$ u! l2 c' [<>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>9 `1 e* ?0 e! Z" G6 H. p+ Q* @
<>and ZBRL is null</P> 7 K# _/ B3 D' H<>) 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>2 b1 o2 f) H4 |2 i
<>and ZBRL is null</P>; \- c9 ]) d# i5 w+ n, k4 y
<>)) 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> o% d9 \1 G! H: m7 a
<>and ZBRL is not null order by ZBRL <BR>) b<BR>)<BR>order by my_sys_rownum desc</P> E# w" t5 G F l
<>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>