[PHP] Re: multiple search
After executing the exact script I get the following (rough)error:
Warning: Undefined offset: 1 in :/Inetpub/co2_busters_mk2
The script(Which is being executed about 25m away) soon cause my computer to
fill up up it's memory(96mb used to about 256)
That is why I couldn't give you the message(IE crashed).
It is looping forever for some reason(I've never filled a client's computer
that fast before!)
How can I fix it?
Or if you cant tell me what does the error mean?
--
JJ Harrison
[EMAIL PROTECTED]
www.tececo.com
Richard Lynch [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]...
My question is what can I do if someone searches for php script for
search?
I would want to use OR in my SQL statment. The problem is how do I write
the
query so that it will detect the number of words sent then put enough ORs
and LIKE '%$search[1]% in?
The LIKE operator will cheerfully return 1 or 0, which you can add up,
or
in your case, weight and add up in your SELECT:
$query = select 0 ;
# The 0 is a kind of 'yeast' to start off our summation of matches.
$words = explode(' ', $search);
while (list(,$word) = $words){
$word = trim($word);
if ($word){
$query .= + article_keyword.weight * article_keyword.keyword like
'%$word%' ;
}
}
$query .= as score ;
$query .= from article_keyword, article_data ;
$query .= where score 0 ;
$query .=and article_data.aid = article_keyword.aid ;
$query .= order by score desc ;
echo $query, BR\n;
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[PHP] Re: multiple search
My question is what can I do if someone searches for php script for search?
I would want to use OR in my SQL statment. The problem is how do I write the
query so that it will detect the number of words sent then put enough ORs
and LIKE '%$search[1]% in?
The LIKE operator will cheerfully return 1 or 0, which you can add up, or
in your case, weight and add up in your SELECT:
$query = select 0 ;
# The 0 is a kind of 'yeast' to start off our summation of matches.
$words = explode(' ', $search);
while (list(,$word) = $words){
$word = trim($word);
if ($word){
$query .= + article_keyword.weight * article_keyword.keyword like
'%$word%' ;
}
}
$query .= as score ;
$query .= from article_keyword, article_data ;
$query .= where score 0 ;
$query .=and article_data.aid = article_keyword.aid ;
$query .= order by score desc ;
echo $query, BR\n;
--
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[PHP] Re: Multiple search options...
Now I can do one or two of these seperate but can anyone suggest a logic
to
take to allow someone to say: I'd like to search for Computer Programmers
between 1 and 3 years experience in Kitchener and that have these
keywords.
But the next person may come and say I want Computer Programmers with 2
years experience in any city and any keywords...
MySQL:
create table programmer(
programmer_id int(11) auto_increment not null primary key,
category int(4),
experience int(2),
location varchar(255),
name text,
email text
);
create table category(
category_id int(11) auto_increment not null primary key,
category varchar(255)
);
insert into category(category) values('Computer Programmer');
insert into programmer(category, experience, location, name, email)
values(1, 5, 'Chicago', 'Richard Lynch', '[EMAIL PROTECTED]');
create table skill(
skill_id int(11) auto_increment not null primary key,
skill varchar(255)
);
insert into skill(skill) values('PHP');
insert into skill(skill) values('MySQL');
insert into skill(skill) values('PostgreSQL');
create table programmer_skill(
programmer_id int(11),
skill_id int(11)
);
insert into programmer_skill(programmer_id, skill_id) values(1, 1);
insert into programmer_skill(programmer_id, skill_id) values(1, 2);
insert into programmer_skill(programmer_id, skill_id) values(1, 3);
PHP:
$query = select id, name, email, (0 ;
if (isset($category)){
$query .= + category = $category ;
}
if (isset($experience)){
$query .= + experience between $experience[0] and $experience[1] ;
}
if (isset($location)){
$query .= + location = '$location' ;
}
if (isset($skills)){
$query .= + count(skills.id) ;
}
$query .= ) as score ;
$query .= from programmer, programmer_skill ;
$query .= where programmer.programmer_id = programmer_skill.programmer_id
;
$query .=and score 0 ;
$query .= order by score desc ;
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