counting images in a DB (should be easy)

gat

This is a very easy one, im sure.....but i dont know how to go about it.

basically i want text that says.

"there are currently ( ) images in this section."

and i would like the number of files to come in the brackets.

please can someone help???

ir4z0r

so I guess, you have a table with images then?

$query = myqsl_query("SELECT COUNT(*) AS number FROM images");
list($number) = mysql_fetch_array($query);

echo "there are " . $number . " images in the table";

gat

thanks for the reply,

i have had a fiddle, and im still getting the message.

Fatal error: Call to undefined function: myqsl_query() in c:\program files\apache group\apache\htdocs\ky\walls.php on line 195

Here is the code.

(

$conn= @mysql_connect("localhost","GWorsnup","bracknell")
                                       or die("Err:Conn");

$rs= @mysql_select_db("kriminalyouth", $conn)
                                       or die("Err:Db");
									   /////////////////////////////////////////////////

 (line 195 where i get error) $query = myqsl_query ("SELECT COUNT(*) AS number FROM walls_berks") ;

 list($count) = mysql_fetch_array($query);

echo "there are " . $count . " images in the table";?>

hope this makes it more clear.

the db = kriminalyouth
the table i want to count is 'walls_berks'

many many thanks

ir4z0r

uuhh, typo, please correct the function name to mysql_query

the s and q have been mixed up

gat

ok, that got it on to the next line.

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\program files\apache group\apache\htdocs\ky\walls.php on line 197

197 = list($count) = mysql_fetch_array($query);

???????

ir4z0r

that's strange

usually this msg appears, when the query doesn't return anything
but the query should ALWAYS return the number, hence 1 entry

have you tried to execute the query in your mysql console?

(go to command prompt, navigate to mysql/bin dir, type "mysql" then
at the "mysql>" prompt, enter "use your_db;" and the query: "SELECT ...;")

gat

ok, i did the query in the console.

and it worked fine.

it returned

number :
---------- :

84 :

so we know the query is ok.

could it be to do with my php version??
im using 4.3.1

current error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\program files\apache group\apache\htdocs\ky\walls.php on line 197

ir4z0r

not really, 4.3.1 is okay

have you removed "(line 195 where i get error)" from your source?

I've done this a thousand times and it looks ok

insert after the query:
echo "Rows: " . mysql_num_rows($query);

this should always return 1

gat

damn damn damn,

i have taken the code and put it in a page called 'countertest.php'
untill i get it working.

when i insert either lines

echo "Rows: " . mysql_num_rows($query);

list($count) = mysql_fetch_array($query);

i still get the same error.....

One last time, ill paste the whole page code in

<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>

<?php
$conn= @mysql_connect("localhost","GWorsnup","bracknell")
or die("Err:Conn");

$rs= @mysql_select_db("kriminalyouth", $conn)
or die("Err😃b");
$query = mysql_query ( "SELECT COUNT (*)AS number FROM walls_berks") ;

list($count) = mysql_fetch_array($query);

echo "there are " . $count . " images in the table";
?>

</body>

</html>

error = Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\program files\apache group\apache\htdocs\ky\countertest.php on line 18

18 = list($count) = mysql_fetch_array($query);

ahhhh i hate this.....but i apreciate the help. 😉

ir4z0r

take away the space between "COUNT" and "(*)" and insert one
before "AS"

the rest is okay

hint: while testing, remove the @'s cause they suppress error msgs

gat

hahaha, thats all it was.

the worst thing is, yo ugave it to me right inthe first place....lol

ah well, just one of those programming thing i guess.

thanks a lot IR4Z0R....

eternalprophet

also, out of habit, i remove my username and password from the code i copy and paste anywhere 🙂

ir4z0r

look at the error msg, it's a windows machine, hence it's highly probable
a local system for testing, so no security problem I guess :p