select and count

clowngirl

This works and I can echo $grid and all three user_id shows up, but I also want to echo how many it found and in this case it is three.

Can anyone help me out here...

 $query = "SELECT * FROM members where girl = '0' and user_id = '$user2'";
 $result = mysql_query($query) or die( "ERROR: " . mysql_error() . "\n");	    

 while ($myrow = mysql_fetch_array($result)) {
 $gird=$myrow["user_id"];
echo"$gird<BR>"; 
}

So result should be like this :
1
2
3

( 3 ) user_id found

harmor

Try

echo count($grid);

clowngirl

Nope, still showing one for the count and it should show 3

harmor

Change
$gird=$myrow["user_id"];

To
$gird .= $myrow["user_id"];

krrose27

Off topic-- to harmor----
do u know where their is a good tut on how to make an IMAGE uploader?

harmor

Go through these krrose
http://www.pixel2life.com/tutorials/PHP_Coding/File_Upload_and_Manipulation/
Do not thank me or comment, just click the link and leave clowngirl's thread alone.

As for clowngirl.
I would drop the variable $grid and just use the array $myrow["user_id"]
If you are echoing values from the database then using the variable form will give you problems.
It'll take a while to explain so maybe you have discovered what I'm trying to say with this script if you in fact tried to echo $grid.

clowngirl

Ok now I get

111

And I should get :

If I echo $gird without:
echo count($gird);

I get the user id's:

1
2
3

When I echo :
echo count($gird);
I get :
111
With the change you just made..

If I echo both I get :
1
112
1123
1

clowngirl

I tried that and still get :
111 when it should read 3

What I need is the count variable of 3, I actully do not need to echo it.
I am using echo for testing reasons to make sure what ever I come up with works correct..

harmor

Try this code

Edit:
I just read your post above this one.

$query = "SELECT * FROM members where girl = '0' and user_id = '$user2'";
 $result = mysql_query($query) or die( "ERROR: " . mysql_error() . "\n");	

$myrow = mysql_num_rows($result)

echo $myrow;

clowngirl

I am getting this with that echo:
Parse error: parse error, unexpected T_ECHO

harmor

I am sorry, I'm not focused.

put a semicolon after
$myrow = mysql_num_rows($result)

So it'll be
$myrow = mysql_num_rows($result);

clowngirl

Now I am getting :
1110

Why am I not able to show count, but I am able to echo all three user id's

harmor

Yikes,

I had that problem about 3 months ago but I forgot how I solved it.
I guess I have to say "I'm Stumped", sorry.

clowngirl

dang, well now i go to bed and think on this in morning..

If you or anyone can figure this out, please post 🙂

jc94062

Hmmm, if one thing fails try another...

You know at one stage it was reading three (111 = 1.1.1 right?) so just do += there if you need to prehaps?

Heres some code that should work, without doing a fetch and a count.
Take a look and let me know, I'll try to pop on from work later 🙂
(Functions may be off as I've been working from OOP so long I've forgotten more than ->fetch 😛)

$sql = "select * from members";
$sql .= " where girl='0' and user_id='".$user2."'";
$query = mysql_query($sql);
$totCount=0;
while ($rows = mysql_fetch_array($query))
{
	$totCount++;
	echo $rows['user_id']."<br>";

//alternate
$gird[] = $rows["user_id"];
$out .= $gird."<br>";
}
echo "( ".$totCount." ) user_id found";

echo "<br><br>.$out;
echo "COUNT TEST: ( ".count($gird)." ) user_id found";

You can only count() on an array... and it needs to be outputted after the loop of course

Roger_Ramjet

At a guess, but an educated one:

You are trying to get the count after you have read a row from the result, and echoing it within the get row loop.

If you get a row from the resultset, then you move the resultset cursor and the count changes. Use the code like this

$query = "SELECT * FROM members where girl = '0' and user_id = '$user2'";
$result = mysql_query($query) or die( "ERROR: " . mysql_error() . "\n");    

$num_rows = mysql_num_rows($result);

while ($row = mysql_fetch_array($result)) {
   echo $row['user_id'] . '<br />';
}

echo $num_rows . ' user ids found';

clowngirl

With that I get :
Parse error: parse error, unexpected T_STRING, expecting ',' or ';' in

On this line :
echo "COUNT TEST: ( ".count($gird)." ) user_id found";

clowngirl

Oh ok Roger I will try yours and see how that goes now..

clowngirl

nope still showing just the user ids and nothing for count

Roger_Ramjet

EH?? Are you sure you copied my code correctly? mysql_num_rows will return the row count of the resultset. So if you are getting 3 rows returned, your 3 ids, then you really must be getting mysql_num_rows = 3. I've never heard of it not working.

Please post the code, exactly as you used it, and the output it produced.