sholud be easy to solve

[trilbyfish]

hi

 

i have the sql query

 

SELECT COUNT( * ) 
FROM `bookingtable` 
WHERE `eight` = '-' 

 

and the output is 2.

 

how do i echo the output using php? I know it is probably right in front of me, but i cant find the solution.

 

The php code i have at the moment is

<?php

require_once ('../project/connect.php'); // Connect to the db.

$query = "SELECT COUNT(*) FROM `bookingtable` WHERE `eight` = '-'";		
	$result = @mysql_query ($query); // Run the query.
	$row = mysql_fetch_array ($result, MYSQL_NUM); // Return a record, if applicable.



echo ($row['count(*)']);






?>

 

The output of this is a blank page.

 

If anyone can tell me what i need to echo, or if there are any problems above that, i would be grateful.

 

Thanks in advance

[revraz]

Check this out and scroll down to the "As" part

 

http://www.techonthenet.com/sql/count.php

[moselkady]

You cannot use count( as an array index. You can try:

 

echo $row[0];

 

Another way is to assign a name to COUNT(*) in your query and use that name as array index:

 

$query = "SELECT COUNT(*) AS mycount FROM `bookingtable` WHERE `eight` = '-'";
...
...
echo $row['mycount'];

[sasa]

 

echo ($row['COUNT(*)']);

 

[trilbyfish]

Ok thanks for the help.

 

The only thing that seems to work for me is

 echo $row[0];

 

Is there a way to do this for multiple columns? ive tried changing the number in the $row[0] but that just echoes nothing.

[moselkady]

 

If you select more than one field, the array $row will have their values in the order you put in SELECT statement

 

<?php

......
$query = "SELECT field1, field2, field3 FROM `bookingtable` WHERE `eight` = '-'";		
$result = @mysql_query ($query); // Run the query.
$row = mysql_fetch_array ($result, MYSQL_NUM); // Return a record, if applicable.
echo $row[0];   // field1
echo $row[1];   // field2
echo $row[2];   // field3
?>

 

Another thing you code do is user print_r($row) to print all the values in the array and check its structure