I am trying to store the result of an mysql query output in an array..
for testing purpose i wrote the following code:

$conn = mysql_connect("localhost","name","pwd");
mysql_select_db("media",$conn);
$sql = "select images from site";
//mysql_query($sql, $conn);
$result = mysql_query($sql, $conn);
$number_of_rows = mysql_num_rows($result);
$count = mysql_fetch_array($result,MYSQL_ASSOC);
print ("<br>");
echo "<b>Number of contacts = $number_of_rows<br></b>";
print_r($count);

but when i execute this query i get the following output:
Number of contacts = 6582
Array ( [images] => ----,3642 )

My question is why is the number of rows of indexed array more than
associative array ??

thecoolone wrote:
Your code doesn't make any sense.

mysql_fetch_array() returns one row from the result set. From your
SELECT statement this would be one element of the column "image".

I don't see anywhere you are getting the entire result set into your
array, so there's no way to count the number of elements in it.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex (AT) attglobal (DOT) net
==================

Jerry Stuckle wrote:
ok i will explain what i am trying to do here.
I have a table that contains a list of files(images) downloaded which
is stored as ids.
And another table that stores the name of the file.
Therefore i am first trying to store all the id's in an associative
array (say $count), and if the id is present in the first query(i.e.
from table that contain all ids) then increment the count of that id in
the array $count.

how would you go about doing this??
the reason i had used mysql_fetch_array is so that i can get an
associative array.

thecoolone wrote:
No problem using mysql_fetch_array - but that fetches ONE ROW into an
array. If you want to get all of the rows, you need to continue issuing
mysql_fetch_array calls until it returns false.

But if all you want is a count of id's in an array, let MySQL do it for
you. Check out the COUNT SQL function, GROUP BY clause and JOIN clauses.

I don't know your table structure or data, but something like:

SELECT id, COUNT(id) FROM table1
JOIN table2 ON table1.id = table2.id
GROUP BY id

Or something similar.

But without knowing exactly your table layout, a little sample data and
the results you want, it's hard to tell exactly what you need.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex (AT) attglobal (DOT) net
==================

Jerry Stuckle wrote:
i tried that as follows:
while ($count = mysql_fetch_array($result,MYSQL_ASSOC))
{
print_r($count);
print "<br>"; }

but i want to be able to store the all results of mysql_fetch_array in
one single array.
i thought of using a for loop like:
$number_of_rows = mysql_num_rows($result);
for ($i=0;$i<$number_of_rows;$i++)
{
print_r($count[$i]);
print "<br>";
global $num;
$num=array();
$num=$count;
}

but it isint working the way i expected.

the problem here is that some rows have id's separated by commas.
and so doing a count(id) wont work here.

i tried sending u a mail with details of what i need but it failed to
deliver.
if u cud email me then i can send the details to you.

many thanks,

thecoolone wrote:
No, because you never put anything in $count, so printing it won't show
anything. You need to fetch the data from the mysql result before you
can us it.

Also every time through your loop you're reinitializing $num.

That's a broken database design then. Do a search on "database
normalization". You should probably have a third table which links the
two current tables together.

Sorry, but I also don't discuss these things in email. It's better to
keep it in the newsgroup where others can make comments, also.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex (AT) attglobal (DOT) net
==================

Jerry Stuckle wrote:
Then how do i loop in such a way that it appends the value of
mysql_fetch_array to $count

its not a broken db design. There is a reason for that, which i had
wrote in a mail i had sent you.
the reason being each id's indicate the files that the user has choose
to download.
i want to be able to scan through each row of $count array and if the
id is present in that row
then increment the count for that id a $counter array (which is
associative). If my thinking is correct
i will have to do an in_array comparision for each id in the first
table against the $count array and then
increment $counter right??

thank you.

thecoolone wrote:
See the examples at
http://www.php.net/manual/en/functio...etch-assoc.php and
http://www.php.net/manual/en/functio...etch-array.php. They have
some good examples.

Your first try has the right idea - but you overwrite $count each time
when you fetch the new row. Additionally, the array in $count only
represents a single row. If you want multiple elements, you need an
array of arrays, i.e.

$count = array();
while ($temp = mysql_fetch_array($result), MYSQL_ASSSOC) {
$count [] = $temp;
....

This creates the $count array then appends each row read into it.

To start, it is a violation of first normal form (having more than one
value in a column). I have *never* found a reason this is "better" than
the correct way of doing things - with a link table. And I've been
doing RDB design for over 20 years, starting with DB2 on mainframes.
And yes, when I say never, I do mean *never*.

Read up on database normalization. You will find a correct database
design will make this job a lot easier (plus give you more options). In
fact, you'll be able to get your count entirely in SQL without the need
for PHP or any other language.

Also, as I said before - IMHO newsgroup threads should remain in the
newsgroup. You can explain why you have such a design here. I tend to
delete email related to newsgroup messages.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex (AT) attglobal (DOT) net
==================

thecoolone wrote:
It IS a broken DB design. The reason you gave is not a reason for having the
design that you have. The reason why you have the broken design is because
you do not know how to design a database. Read up on normalisation.

You don't even need any programming to get the answer that you want. With a
properly designed database a single SQL query will give the required result.