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sql server (alternate) : Advance ORDER BY
I'm wondering if there's any advance order by 'function'/workaround, which is reasonably efficient for MS SQL Server, without resorting to some third party indexing/search engine to achieve the following. The mechanism is to record each instance of a pattern match and order by rows with most matches first (DESC). Simplistic match but that's a separate issue. Sample: create table tmp (col varchar(50)); insert into tmp values ('a barking dog'); insert into tmp values ('a dog and cat fights over dog food'); insert into tmp values ('lovable dog is not barking dog=nice dog'); The goal for the Sample is to return resultsets in the following order: lovable dog is not barking dog=nice dog -- 3 matches a dog and cat fights over dog food -- 2 matches a barking dog -- 1 match
Thank you. Gert-Jan's solution also works.
Can you explain why? [quoted text, click to view] "Simon Hayes" <sql@hayes.ch> wrote in message news:<3f5b822c_4@news.bluewin.ch>...
> "Simon Hayes" <sql@hayes.ch> wrote in message > news:3f5b7da7$1_3@news.bluewin.ch... > > > > "Doug Baroter" <qwert12345@boxfrog.com> wrote in message > > news:fc254714.0309071031.7f6b9826@posting.google.com... > > > Hi, > > > > > > I'm wondering if there's any advance order by 'function'/workaround, > > > which is reasonably efficient for MS SQL Server, without resorting to > > > some third party indexing/search engine to achieve the following. > > > > > > The mechanism is to record each instance of a pattern match and order > > > by rows with most matches first (DESC). Simplistic match but that's a > > > separate issue. > > > > > > Sample: > > > create table tmp (col varchar(50)); > > > insert into tmp > > > values ('a barking dog'); > > > insert into tmp > > > values ('a dog and cat fights over dog food'); > > > insert into tmp > > > values ('lovable dog is not barking dog=nice dog'); > > > > > > The goal for the Sample is to return resultsets in the following > > > order: > > > lovable dog is not barking dog=nice dog -- 3 matches > > > a dog and cat fights over dog food -- 2 matches > > > a barking dog -- 1 match > > > > > > Thanks. > > > > Here's one possibility: > > > > select col > > from tmp > > order by len(replace(col, 'dog', '')) desc > > > > Simon > > > > > > Sorry - I posted that a bit too quickly. It should be this - the division by > 3 is because your search term has 3 characters, so you can count the number > of replacements made this way: > > select col > from tmp > order by (len(col) - len(replace(col, 'dog', ''))) / 3 desc >
[quoted text, click to view] "Doug Baroter" <qwert12345@boxfrog.com> wrote in message news:fc254714.0309071031.7f6b9826@posting.google.com... > Hi, > > I'm wondering if there's any advance order by 'function'/workaround, > which is reasonably efficient for MS SQL Server, without resorting to > some third party indexing/search engine to achieve the following. > > The mechanism is to record each instance of a pattern match and order > by rows with most matches first (DESC). Simplistic match but that's a > separate issue. > > Sample: > create table tmp (col varchar(50)); > insert into tmp > values ('a barking dog'); > insert into tmp > values ('a dog and cat fights over dog food'); > insert into tmp > values ('lovable dog is not barking dog=nice dog'); > > The goal for the Sample is to return resultsets in the following > order: > lovable dog is not barking dog=nice dog -- 3 matches > a dog and cat fights over dog food -- 2 matches > a barking dog -- 1 match > > Thanks. Here's one possibility: select col from tmp order by len(replace(col, 'dog', '')) desc Simon
A slight correction on Simon's solution (which will return the longest
string, regardless of number of matches) select col from tmp order by len(col) - len(replace(col, 'dog', '')) desc Gert-Jan [quoted text, click to view] Doug Baroter wrote:
> > Hi, > > I'm wondering if there's any advance order by 'function'/workaround, > which is reasonably efficient for MS SQL Server, without resorting to > some third party indexing/search engine to achieve the following. > > The mechanism is to record each instance of a pattern match and order > by rows with most matches first (DESC). Simplistic match but that's a > separate issue. > > Sample: > create table tmp (col varchar(50)); > insert into tmp > values ('a barking dog'); > insert into tmp > values ('a dog and cat fights over dog food'); > insert into tmp > values ('lovable dog is not barking dog=nice dog'); > > The goal for the Sample is to return resultsets in the following > order: > lovable dog is not barking dog=nice dog -- 3 matches > a dog and cat fights over dog food -- 2 matches > a barking dog -- 1 match >
[quoted text, click to view] "Simon Hayes" <sql@hayes.ch> wrote in message news:3f5b7da7$1_3@news.bluewin.ch... > > "Doug Baroter" <qwert12345@boxfrog.com> wrote in message > news:fc254714.0309071031.7f6b9826@posting.google.com... > > Hi, > > > > I'm wondering if there's any advance order by 'function'/workaround, > > which is reasonably efficient for MS SQL Server, without resorting to > > some third party indexing/search engine to achieve the following. > > > > The mechanism is to record each instance of a pattern match and order > > by rows with most matches first (DESC). Simplistic match but that's a > > separate issue. > > > > Sample: > > create table tmp (col varchar(50)); > > insert into tmp > > values ('a barking dog'); > > insert into tmp > > values ('a dog and cat fights over dog food'); > > insert into tmp > > values ('lovable dog is not barking dog=nice dog'); > > > > The goal for the Sample is to return resultsets in the following > > order: > > lovable dog is not barking dog=nice dog -- 3 matches > > a dog and cat fights over dog food -- 2 matches > > a barking dog -- 1 match > > > > Thanks. > > Here's one possibility: > > select col > from tmp > order by len(replace(col, 'dog', '')) desc > > Simon > > Sorry - I posted that a bit too quickly. It should be this - the division by 3 is because your search term has 3 characters, so you can count the number of replacements made this way: select col from tmp order by (len(col) - len(replace(col, 'dog', ''))) / 3 desc Simon
[quoted text, click to view]
>> Thank you. Gert-Jan's solution also works. Can you explain why? << Take the expression used in the ORDER BY clause and add it in the SELECT list. The answer then becomes obvious. -- - Anith ( Please reply to newsgroups only )
Of course it does :-)
It works because for the ORDER BY clause you do not need the actual number of occurrences. You just need them sorted. In that respect "1 barking", "2 dog and cat", "3 lovable dog" is the same as "3 barking", "6 dog and cat", "9 lovable dog". Gert-Jan [quoted text, click to view] Doug Baroter wrote:
> > Thank you. Gert-Jan's solution also works. > Can you explain why? > > "Simon Hayes" <sql@hayes.ch> wrote in message news:<3f5b822c_4@news.bluewin.ch>... > > "Simon Hayes" <sql@hayes.ch> wrote in message > > news:3f5b7da7$1_3@news.bluewin.ch... > > > > > > "Doug Baroter" <qwert12345@boxfrog.com> wrote in message > > > news:fc254714.0309071031.7f6b9826@posting.google.com... > > > > Hi, > > > > > > > > I'm wondering if there's any advance order by 'function'/workaround, > > > > which is reasonably efficient for MS SQL Server, without resorting to > > > > some third party indexing/search engine to achieve the following. > > > > > > > > The mechanism is to record each instance of a pattern match and order > > > > by rows with most matches first (DESC). Simplistic match but that's a > > > > separate issue. > > > > > > > > Sample: > > > > create table tmp (col varchar(50)); > > > > insert into tmp > > > > values ('a barking dog'); > > > > insert into tmp > > > > values ('a dog and cat fights over dog food'); > > > > insert into tmp > > > > values ('lovable dog is not barking dog=nice dog'); > > > > > > > > The goal for the Sample is to return resultsets in the following > > > > order: > > > > lovable dog is not barking dog=nice dog -- 3 matches > > > > a dog and cat fights over dog food -- 2 matches > > > > a barking dog -- 1 match > > > > > > > > Thanks. > > > > > > Here's one possibility: > > > > > > select col > > > from tmp > > > order by len(replace(col, 'dog', '')) desc > > > > > > Simon > > > > > > > > > > Sorry - I posted that a bit too quickly. It should be this - the division by > > 3 is because your search term has 3 characters, so you can count the number > > of replacements made this way: > > > > select col > > from tmp > > order by (len(col) - len(replace(col, 'dog', ''))) / 3 desc > >
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