How to display mySQL Query?

Anon

My code is the following:

$connect = mysql_connect("localhost","testguy","test");
$snamecheck = mysql_query("SELECT * FROM aol_v2 WHERE sname=$sname2");
echo "$snamecheck";

But when I run this, it returns absolutly nothing. What is wrong? And yes, the values exists in the database.

Anon

Two things: You may need to specify which database to use. after your $connect, specify the database with:

mysql_select_db("database_name",$connect);

also, do you want to display the query that has been sent or the results of the query? if you want to display the query that was sent, try this:

$connect = mysql_connect("localhost","testguy","test");

mysql_select_db("database_name",$connect);

$sendthisquery = "SELECT * FROM aol_v2 WHERE sname=" . $sname2;

$snamecheck = mysql_query($sendthisquery);

echo $sendthisquery;

Hope this helps! 🙂

mithril

As a collary, mysql_query() returns a resource #id to the query result. If you want to display the returned records you need to use one of the mysql_fetch_*() functions to parse the resource.

-geoff

tridoc

This may help.

$connect = mysql_connect("localhost","testguy","test");

$dblinkid = mysql_select_db("aol_v2", $connect);

$snamecheck = mysql_query("SELECT * FROM aol_v2 WHERE sname=$sname2");

while ($row = mysql_fetch_row($snamecheck))

echo "$snamecheck";

tridoc

Should be something like this

$connect = mysql_connect("localhost","testguy","test");

$dblinkid = mysql_select_db("aol_v2", $connect);

$snamecheck = mysql_query("SELECT * FROM aol_v2 WHERE sname=$sname2");

while ($row = mysql_fetch_row($snamecheck))

echo "$row";

[deleted]

Read this:
http://www.hvt-automation.nl/yapf/faq/index.php?cmd=viewitem&itemid=12

tridoc

Great Page wish I had seen that a couple of months ago

[deleted]

lol thanks, and sorry I didn't create it sooner :-)

andykal

Firstly, I assume that you are passing the $sname2 variable approproately - I bet you are, that's why you have not included that on the code sent.

My suggestion is to include the qoutes on the '$sname2' variable in the query.

$connect = mysql_connect("localhost","testguy","test");
$snamecheck = mysql_query("SELECT * FROM aol_v2 WHERE sname='$sname2'");
echo "$snamecheck";

-Andy-

Anon

all your informatino helped me very much and now my database is running nicely ! =D