Like the title says, I am only getting the first result, and getting it twice.
example:
<?php
$con = mysql_connect("localhost","USER","PASS");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("DB_NAME", $con);
$sql = 'SELECT ptext FROM provisions WHERE pnumber =
"119441a" OR pnumber = "119441b"';
$data = mysql_query( $sql, $con );
$info = mysql_fetch_array( $data );
$arr = array_values($info);
print_r($arr);
?>I only get the (a) value - twice:
Array ( [0] => (a) Product support documentation provided to end-users shall be made available in alternate formats upon request, at no additional charge. [1] => (a) Product support documentation provided to end-users shall be made available in alternate formats upon request, at no additional charge. )The first result twice.
Same result for SELECT DISTINCT
I have a shadow of the db on localhost, SQL on localhost works fine.
On localhost running the query in phpmyadmin I get the two text fields, (a) and (b)..
decibel.places posted this at 15:45 — 1st September 2008.
He has: 1,550 posts
Joined: Jun 2008
mysql_fetch_array($data)only returns one row from the db at a time...
so you need a "while" loop to loop through all the rows in the db..
<?php
$con = mysql_connect("localhost","communit_508dev","commUNIT");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("communit_508prov", $con);
$sql = 'SELECT ptext FROM provisions';
$data = mysql_query( $sql, $con );
while($row = mysql_fetch_array($data))
{
$arr .= $row['ptext'] . "<br />";
}
print_r($arr);
mysql_close($con);
?>