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External
Since: Jan 02, 2011
Posts: 3
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(Msg. 1) Posted: Sun Jan 02, 2011 2:25 pm
Post subject: how to detect non-presence
Archived from groups: alt>php>sql (more info?)
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I'm building a website where I want to present information about some
items. Those items can for a longer or shorter time be present on
addresses that will be given is such is the case.
The relevant tables are of this type:
vartype, with the fields: ID(key), varnr, description
and
var, with the fields id(key), address, varnr.
Of course a join is to be made between the varnr-fileds.
The visitor can make a choice from a list generated by
$sql='SELECT DISTINCT varnr, description FROM vartype';
As the items to be presented come and go, sometimes in the list
generated are items that cannot be found in the var-table, when there
is no address to be given.
Now I wonder if there is a simple way to to present only those items
from the vartype-tabel that have a counterpart in the var-table?
I hope I made myself clear.
Any hint will be appreciated.
Susan |
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External
Since: Apr 23, 2007
Posts: 89
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(Msg. 2) Posted: Sun Jan 02, 2011 2:25 pm
Post subject: Re: how to detect non-presence [Login to view extended thread Info.]
Archived from groups: per prev. post (more info?)
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On Jan 2, 6:58 pm, Susan Boscha wrote:
> I'm building a website where I want to present information about some
> items. Those items can for a longer or shorter time be present on
> addresses that will be given is such is the case.
> The relevant tables are of this type:
> vartype, with the fields: ID(key), varnr, description
> and
> var, with the fields id(key), address, varnr.
> Of course a join is to be made between the varnr-fileds.
> The visitor can make a choice from a list generated by
> $sql='SELECT DISTINCT varnr, description FROM vartype';
> As the items to be presented come and go, sometimes in the list
> generated are items that cannot be found in the var-table, when there
> is no address to be given.
> Now I wonder if there is a simple way to to present only those items
> from the vartype-tabel that have a counterpart in the var-table?
>
> I hope I made myself clear.
> Any hint will be appreciated.
> Susan
That is what a JOIN does anyway! |
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External
Since: Jan 02, 2011
Posts: 3
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(Msg. 3) Posted: Sun Jan 02, 2011 5:25 pm
Post subject: Re: how to detect non-presence [Login to view extended thread Info.]
Archived from groups: per prev. post (more info?)
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Op Sun, 2 Jan 2011 11:54:45 -0800 (PST) schreef Captain Paralytic
:
>On Jan 2, 6:58 pm, Susan Boscha wrote:
>> I'm building a website where I want to present information about some
>> items. Those items can for a longer or shorter time be present on
>> addresses that will be given is such is the case.
>> The relevant tables are of this type:
>> vartype, with the fields: ID(key), varnr, description
>> and
>> var, with the fields id(key), address, varnr.
>> Of course a join is to be made between the varnr-fileds.
>> The visitor can make a choice from a list generated by
>> $sql='SELECT DISTINCT varnr, description FROM vartype';
>> As the items to be presented come and go, sometimes in the list
>> generated are items that cannot be found in the var-table, when there
>> is no address to be given.
>> Now I wonder if there is a simple way to to present only those items
>> from the vartype-tabel that have a counterpart in the var-table?
>>
>> I hope I made myself clear.
>> Any hint will be appreciated.
>> Susan
>
>That is what a JOIN does anyway!
Of course I make a join when I present the items that are there. But
how to eliminate items from the generated list where a join cannot be
made at that moment? >> Stay informed about: how to detect non-presence |
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External
Since: Apr 23, 2007
Posts: 89
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(Msg. 4) Posted: Mon Jan 03, 2011 4:34 am
Post subject: Re: how to detect non-presence [Login to view extended thread Info.]
Archived from groups: per prev. post (more info?)
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On Jan 2, 9:52 pm, Susan Boscha wrote:
> Op Sun, 2 Jan 2011 11:54:45 -0800 (PST) schreef Captain Paralytic
> :
>
>
>
>
>
> >On Jan 2, 6:58 pm, Susan Boscha wrote:
> >> I'm building a website where I want to present information about some
> >> items. Those items can for a longer or shorter time be present on
> >> addresses that will be given is such is the case.
> >> The relevant tables are of this type:
> >> vartype, with the fields: ID(key), varnr, description
> >> and
> >> var, with the fields id(key), address, varnr.
> >> Of course a join is to be made between the varnr-fileds.
> >> The visitor can make a choice from a list generated by
> >> $sql='SELECT DISTINCT varnr, description FROM vartype';
> >> As the items to be presented come and go, sometimes in the list
> >> generated are items that cannot be found in the var-table, when there
> >> is no address to be given.
> >> Now I wonder if there is a simple way to to present only those items
> >> from the vartype-tabel that have a counterpart in the var-table?
>
> >> I hope I made myself clear.
> >> Any hint will be appreciated.
> >> Susan
>
> >That is what a JOIN does anyway!
>
> Of course I make a join when I present the items that are there. But
> how to eliminate items from the generated list where a join cannot be
> made at that moment?
As I said, "That is what a JOIN does anyway!"
The result set of a JOIN (or more correctly an INNER JOIN) consists of
only those records where a successful match is made. >> Stay informed about: how to detect non-presence |
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External
Since: Jan 02, 2011
Posts: 3
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(Msg. 5) Posted: Mon Jan 03, 2011 5:25 pm
Post subject: Re: how to detect non-presence [Login to view extended thread Info.]
Archived from groups: per prev. post (more info?)
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Op Mon, 3 Jan 2011 04:34:02 -0800 (PST) schreef Captain Paralytic
:
>On Jan 2, 9:52 pm, Susan Boscha wrote:
>> Op Sun, 2 Jan 2011 11:54:45 -0800 (PST) schreef Captain Paralytic
>> :
>>
>>
>>
>>
>>
>> >On Jan 2, 6:58 pm, Susan Boscha wrote:
>> >> I'm building a website where I want to present information about some
>> >> items. Those items can for a longer or shorter time be present on
>> >> addresses that will be given is such is the case.
>> >> The relevant tables are of this type:
>> >> vartype, with the fields: ID(key), varnr, description
>> >> and
>> >> var, with the fields id(key), address, varnr.
>> >> Of course a join is to be made between the varnr-fileds.
>> >> The visitor can make a choice from a list generated by
>> >> $sql='SELECT DISTINCT varnr, description FROM vartype';
>> >> As the items to be presented come and go, sometimes in the list
>> >> generated are items that cannot be found in the var-table, when there
>> >> is no address to be given.
>> >> Now I wonder if there is a simple way to to present only those items
>> >> from the vartype-tabel that have a counterpart in the var-table?
>>
>> >> I hope I made myself clear.
>> >> Any hint will be appreciated.
>> >> Susan
>>
>> >That is what a JOIN does anyway!
>>
>> Of course I make a join when I present the items that are there. But
>> how to eliminate items from the generated list where a join cannot be
>> made at that moment?
>
>As I said, "That is what a JOIN does anyway!"
>
>The result set of a JOIN (or more correctly an INNER JOIN) consists of
>only those records where a successful match is made.
Aha, I see, and now I'm feeling a bit stupid!
Thank you anyway. >> Stay informed about: how to detect non-presence |
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