Re: [PHP-DB] populating select and searching to criteria
First:
Use the SQL distinct command.. it's explained here:
http://www.mysql.com/doc/en/SELECT.html
That will get you unique values.
Then just loop through the result:
while ($d = mysql_fetch_assoc($return)) {
?option value=?=$d['value']??=$d['name']?/option
?
}
Second:
get the value from the other table into a variable, during each iteration of
the loop check to see if they match, if so add 'selected'.
while ($d = mysql_fetch_assoc($return)) {
?option value=?=$d['value']??
if ($other_table_value == $d['value'])
echo selected;
??=$d['name']?/option
?
}
Third:
If you're using MySQL, check this out:
http://www.mysql.com/doc/en/Fulltext_Search.html
If you're using something else, it's more complex I think. Someone may have an
elegant solution, but I would do something like:
// Get total number of keywords:
$numofkeywords = count(str_replace ( , , $keywords));
// Split up your search words:
$search = explode( , $keywords);
$total_matched = 0;
// loop through the array of search terms and get number of returns.
foreach ($search as $searchword) {
$total_matched += count(str_replace($searchword, $searchword, $keywords));
}
// echo out the result in percent. (to one decimal place even!)
echo Percent Matched: .round(($total_matched/$numofkeywords)*100), 1). %;
I'm sure there's a better way to get a word count from a string, but it's
late.. I didn't try the above code, so I'm just guessing it works.. I think
it's the right idea though.
-Micah
On Mon October 27 2003 4:54 pm, Shannon Doyle wrote:
Hi People,
I have a three part question here. Well actually its 3 separate
questions.
First.
How would I populate a select menu from a mysql databse with the entries
from a particular field, but only showing those that are unique, ie not
showing more than 1 of the same value.
Second.
How would I then have one of those options selected depending on what it
contained in another table?
Third.
I need to understand how the following would be achieved.
In the database there is a text field that contains a number of
'keywords' I need to be able to search through those individual keywords
and return results based on a % match from the original search. Ie if I
search for 'secretary, typing, customer relations' I need to return all
records from the database that has one or all of those words contained
in the text field and display them as a % match of that search, ie if 1
match then 33% 2, 66% and so on.
Any help on the above three would be fantastic.
Cheers,
Shannon
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RE: [PHP-DB] populating select and searching to criteria
Thanks Micah,
while ($d = mysql_fetch_assoc($return)) {
?option value=?=$d['value']??=$d['name']?/option - I get
a parse error on this line.
while ($d = mysql_fetch_assoc($return)) {
?option value=?=$d['value']??
if ($other_table_value == $d['value'])
echo selected;
??=$d['name']?/option
?
}
Ok cool, I am assuming that this will be OK once I work out the parse
error on the above option.
Third:
If you're using MySQL, check this out:
http://www.mysql.com/doc/en/Fulltext_Search.html
If you're using something else, it's more complex I think. Someone may
have an
elegant solution, but I would do something like:
// Get total number of keywords:
$numofkeywords = count(str_replace ( , , $keywords));
// Split up your search words:
$search = explode( , $keywords);
$total_matched = 0;
// loop through the array of search terms and get number of returns.
foreach ($search as $searchword) {
$total_matched += count(str_replace($searchword, $searchword,
$keywords));
}
// echo out the result in percent. (to one decimal place even!)
echo Percent Matched: .round(($total_matched/$numofkeywords)*100),
1). %;
This only partly resolves my problem, I am not looking for a word count,
I am looking to return those records that have one or more of the
keywords in them, and display a percentage result next to each record
that matches.
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To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] populating select and searching to criteria
This may be a dumb comment, but those variable don't make any sense, you'll
have to replace them with stuff that relates to the query and the database.
-Micah
On Mon October 27 2003 8:34 pm, Shannon Doyle wrote:
Thanks Micah,
while ($d = mysql_fetch_assoc($return)) {
?option value=?=$d['value']??=$d['name']?/option - I get
a parse error on this line.
while ($d = mysql_fetch_assoc($return)) {
?option value=?=$d['value']??
if ($other_table_value == $d['value'])
echo selected;
??=$d['name']?/option
?
}
Ok cool, I am assuming that this will be OK once I work out the parse
error on the above option.
Third:
If you're using MySQL, check this out:
http://www.mysql.com/doc/en/Fulltext_Search.html
If you're using something else, it's more complex I think. Someone may
have an
elegant solution, but I would do something like:
// Get total number of keywords:
$numofkeywords = count(str_replace ( , , $keywords));
// Split up your search words:
$search = explode( , $keywords);
$total_matched = 0;
// loop through the array of search terms and get number of returns.
foreach ($search as $searchword) {
$total_matched += count(str_replace($searchword, $searchword,
$keywords));
}
// echo out the result in percent. (to one decimal place even!)
echo Percent Matched: .round(($total_matched/$numofkeywords)*100),
1). %;
This only partly resolves my problem, I am not looking for a word count,
I am looking to return those records that have one or more of the
keywords in them, and display a percentage result next to each record
that matches.
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
