recursive query

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От Joel Fradkin
Тема recursive query
Дата
Msg-id 003201c7efcb$af931500$837ba8c0@jfradkin64
обсуждение исходный текст
Список pgsql-sql
<div class="Section1"><p class="MsoNormal"><font face="Arial" size="2"><span style="font-size:10.0pt;
font-family:Arial">Hi,</span></font><p class="MsoNormal"><font face="Arial" size="2"><span style="font-size:10.0pt;
font-family:Arial">I am trying to do a hierarchy design. We currently have it hard coded to 4 levels and the location
tablehas an id field for each level where the data has a location id.</span></font><p class="MsoNormal"><font
face="Arial"size="2"><span style="font-size:10.0pt; 
font-family:Arial">This allows easy access by level (to restrict who see what) and easy to arrogate for reporting by
division,region, district, and location.</span></font><p class="MsoNormal"><font face="Arial" size="2"><span
style="font-size:10.0pt;
font-family:Arial">I am trying to make it more flexible some companies don’t use 4 levels some want more, some have
differentsets of users and would like their own hierarchy.</span></font><p class="MsoNormal"><font face="Arial"
size="2"><spanstyle="font-size:10.0pt; 
font-family:Arial">My idea is to have a virtual hierarchy (all share the location record which is the lowest level and
aseparate file that will be linked with a table that has the locationid and the hierarchy id to enable multiple
hierarchy).</span></font><pclass="MsoNormal"><font face="Arial" size="2"><span style="font-size:10.0pt; 
font-family:Arial"> I did a bit of research and found some code to do recursive sql (but it was IBM and I am not sure I
evengot it right for postgres).</span></font><p class="MsoNormal"><font face="Arial" size="2"><span
style="font-size:10.0pt;
font-family:Arial"> </span></font><p class="MsoNormal" style="text-autospace:none"><font face="Courier New"
size="2"><spanstyle="font-size:10.0pt;font-family:"Courier New"">create or replace temp view test (clientnum,id
,parentid,descr,level) as (</span></font><p class="MsoNormal" style="text-autospace:none"><font face="Courier New"
size="2"><spanstyle="font-size:10.0pt;font-family:"Courier New"">select   h.clientnum,h.id ,h.parentid,h.descr,0 as
level</span></font><pclass="MsoNormal" style="text-autospace:none"><font face="Courier New" size="2"><span
style="font-size:10.0pt;font-family:"CourierNew""> from tblhyerarchy h where parentid =0 </span></font><p
class="MsoNormal"style="text-autospace:none"><font face="Courier New" size="2"><span
style="font-size:10.0pt;font-family:"CourierNew"">union all</span></font><p class="MsoNormal"
style="text-autospace:none"><fontface="Courier New" size="2"><span style="font-size:10.0pt;font-family:"Courier
New"">select  h1.clientnum,h1.id ,h1.parentid,h1.descr,h.level +1 as level</span></font><p class="MsoNormal"
style="text-autospace:none"><fontface="Courier New" size="2"><span style="font-size:10.0pt;font-family:"Courier
New""> fromtblhyerarchy h1</span></font><p class="MsoNormal" style="text-autospace:none"><font face="Courier New"
size="2"><spanstyle="font-size:10.0pt;font-family:"Courier New"">inner join test h on h1.parentid =h.id
</span></font><pclass="MsoNormal" style="text-autospace:none"><font face="Courier New" size="2"><span
style="font-size:10.0pt;font-family:"CourierNew"">where h.level < 5</span></font><p class="MsoNormal"
style="text-autospace:none"><fontface="Courier New" size="2"><span style="font-size:10.0pt;font-family:"Courier
New"">);</span></font><pclass="MsoNormal" style="text-autospace:none"><font face="Courier New" size="2"><span
style="font-size:10.0pt;font-family:"CourierNew"">select * from test</span></font><p class="MsoNormal"
style="text-autospace:none"><fontface="Courier New" size="2"><span style="font-size:10.0pt;font-family:"Courier
New""> </span></font><pclass="MsoNormal" style="text-autospace:none"><font face="Courier New" size="2"><span
style="font-size:10.0pt;font-family:"CourierNew"">but get a </span></font><p class="MsoNormal"
style="text-autospace:none"><fontface="Courier New" size="2"><span style="font-size:10.0pt;font-family:"Courier
New"">ERROR:infinite recursion detected in rules for relation "test" SQL state: 42P17</span></font><p class="MsoNormal"
style="text-autospace:none"><fontface="Courier New" size="2"><span style="font-size:10.0pt;font-family:"Courier
New""> </span></font><pclass="MsoNormal" style="text-autospace:none"><font face="Courier New" size="2"><span
style="font-size:10.0pt;font-family:"CourierNew"">I am basing on a table created with</span></font><p class="MsoNormal"
style="text-autospace:none"><fontface="Courier New" size="2"><span style="font-size:10.0pt;font-family:"Courier
New"">CREATETABLE tblhyerarchy</span></font><p class="MsoNormal" style="text-autospace:none"><font face="Courier New"
size="2"><spanstyle="font-size:10.0pt;font-family:"Courier New"">(</span></font><p class="MsoNormal"
style="text-autospace:none"><fontface="Courier New" size="2"><span style="font-size:10.0pt;font-family:"Courier New""> 
clientnumcharacter varying(16) NOT NULL,</span></font><p class="MsoNormal" style="text-autospace:none"><font
face="CourierNew" size="2"><span style="font-size:10.0pt;font-family:"Courier New"">  id integer NOT
NULL,</span></font><pclass="MsoNormal" style="text-autospace:none"><font face="Courier New" size="2"><span
style="font-size:10.0pt;font-family:"CourierNew"">  parentid integer NOT NULL,</span></font><p class="MsoNormal"
style="text-autospace:none"><fontface="Courier New" size="2"><span style="font-size:10.0pt;font-family:"Courier New""> 
descrcharacter varying(250),</span></font><p class="MsoNormal" style="text-autospace:none"><font face="Courier New"
size="2"><spanstyle="font-size:10.0pt;font-family:"Courier New"">  CONSTRAINT pk_tblhyerarchy PRIMARY KEY (clientnum,
id)</span></font><pclass="MsoNormal" style="text-autospace:none"><font face="Courier New" size="2"><span
style="font-size:10.0pt;font-family:"CourierNew"">) </span></font><p class="MsoNormal"
style="text-autospace:none"><fontface="Courier New" size="2"><span style="font-size:10.0pt;font-family:"Courier
New"">WITHOIDS;</span></font><p class="MsoNormal"><font face="Arial" size="2"><span style="font-size:10.0pt; 
font-family:Arial"> </span></font><p class="MsoNormal"><font face="Arial" size="2"><span style="font-size:10.0pt;
font-family:Arial"> </span></font><p class="MsoNormal"><font face="Arial" size="2"><span style="font-size:10.0pt;
font-family:Arial">Any help is much appreciated. I am also a little confused how to do a crosstab on the
output.</span></font><pclass="MsoNormal"><font face="Arial" size="2"><span style="font-size:10.0pt; 
font-family:Arial">In the end I will want to get at a data record that has the location id and then link to the
hierarchywith each level of the hierarchy present for aggregation etc.</span></font><p class="MsoNormal"><font
face="Arial"size="2"><span style="font-size:10.0pt; 
font-family:Arial"> </span></font><p class="MsoNormal"><font color="navy" face="Arial" size="2"><span
style="font-size:10.0pt;font-family:Arial;color:navy">JoelFradkin</span></font><p class="MsoNormal"
style="margin-left:1.0in"><fontcolor="navy" face="Times New Roman" size="3"><span
style="font-size:12.0pt;color:navy"> </span></font><pclass="MsoNormal"><font color="navy" face="Arial" size="2"><span
style="font-size:
10.0pt;font-family:Arial;color:navy">Wazagua, Inc.<br /></span></font><font color="navy" face="Arial" size="2"><span
style="font-size:10.0pt;font-family:Arial;color:navy">2520 Trailmate Dr</span></font><font color="navy" face="Arial"
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