JSON + PHP Help

[emeryc]

Hi, this is probably really simple, but i can't get it to work, i need to do a query in PHP / MYSQL based off of the $_GET[] from the URL and then echo it to JSON,

this is what i have

<?Php
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("ipod") or die(mysql_error());

$query = "SELECT * FROM users WHERE name='". $_GET['name']. "'"; 
$result = mysql_query($query) or die(mysql_error());
$rows = mysql_fetch_array($result) or die(mysql_error());
echo json_encode($rows);
?>

[sunfighter]

Your code is good. You maybe using a $_GET['name'] that is wrong. echo out the $query string to see what it is.

[emeryc]

Ok, can't right now, upgrading database etc ... will let you know

[emeryc]

<?Php
mysql_connect("localhost", "root", "e12250") or die(mysql_error());
mysql_select_db("ipod") or die(mysql_error());
$dsn = "mysql:host=localhost;dbname=ipod";
$username = "root";
$password = "e12250";
$query = "SELECT * FROM users WHERE name='". $_GET['name']. "'"; 
echo $query;
$result = mysql_query($query) or die(mysql_error());
$rows = mysql_fetch_array($result) or die(mysql_error());
echo json_encode($rows);
?>

I did that and it prints nothing ...

[sunfighter]

So what did the echo $query; print out?
Did you get a name for $_GET['name'] and in your DB is there a name with this 'name'? You can use phpMyAdmin to find out.

[emeryc]

Originally Posted by sunfighter

So what did the echo $query; print out?
Did you get a name for $_GET['name'] and in your DB is there a name with this 'name'? You can use phpMyAdmin to find out.

yes, it is there, there is data there, and it is in the url

[emeryc]

Originally Posted by emeryc

yes, it is there, there is data there, and it is in the url

whoops, it doesn't print the query$, it is a blank screen

[emeryc]

Are you still helping?

[sunfighter]

Your code is good. You maybe using a $_GET['name'] that is wrong. echo out the $query string to see what it is.

<?Php
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("ipod") or die(mysql_error());

$query = "SELECT * FROM users WHERE name='". $_GET['name']. "'"; 
echo $query;
$result = mysql_query($query) or die(mysql_error());
$rows = mysql_fetch_array($result) or die(mysql_error());
echo json_encode($rows);
?>