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Posted by Jimbus on 11/05/69 11:46
Use an associative array with the coordinates as the key. That way any biz
that has the same coords will appear under that key in the array.
while($row = mysql_fetch_assoc($result)) {
$data[ $row['coordinates'] ][] = $row['bizname'];
}
Note, 'coordinates' and 'bizname' are the db cols, your db col name will
probably differ from my example.
Also note second set of empty square braces. These create a new var under
that key, otherwise you'd just be overwriting the last bizname you saved.
"elyob" <newsprofile@gmail.com> wrote in message
news:4454f07a$1@news1.homechoice.co.uk...
> Hi,
>
> I've got a database that outputs markers on a google map project. The
> problem I am encountering is that some businesses have been located the
> same latitude and longitude, therefore markers don't show (others are
> underneath the top one). So, I need to get this MySQL output, parse it
> through an array creating an output such as
>
> (-0.28183499,51.37974167), biz1 biz2 biz3
> (-0.28301400,51.38087463), biz4
> (-0.28151301,51.38144684), biz5
> (-0.28693399,51.38250351), biz6
> (-0.28626299,51.38276672), biz7
> (-0.29334599,51.38436890), biz8 biz9 biz10
> (-0.29725999,51.38850021), biz11 biz12
> (-0.29800400,51.38896179), biz13 biz14
>
> Here's an example of my input data as it is at the moment ..
>
> (-0.28183499,51.37974167), biz1
> (-0.28183499,51.37974167), biz2
> (-0.28183499,51.37974167), biz3
> (-0.28301400,51.38087463), biz4
> (-0.28151301,51.38144684), biz5
> (-0.28693399,51.38250351), biz6
> (-0.28626299,51.38276672), biz7
> (-0.29334599,51.38436890), biz8
> (-0.29334599,51.38436890), biz9
> (-0.29334599,51.38436890), biz10
> (-0.29725999,51.38850021), biz11
> (-0.29725999,51.38850021), biz12
> (-0.29800400,51.38896179), biz13
> (-0.29800400,51.38896179), biz14
>
> I already sort the data by latitude, so they will always be next to each
> other. But I am going to be writing this using AJAX and need the fastest
> solution possible. There could be hundreds needed to be parsed in a split
> second.
>
> Thanks for any pointers.
>
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