[SOLVED] my script will not connect to the database :@--must be error in code

runnerjp · May 9, 2007

<?

$conn = mysql_connect("localhost", "username", "password");

$db = @mysql_select_db("runnerse_profiles", "images", $conn);

$sql = "SELECT * FROM $tbl_images WHERE image_id = $auth[member_id] AND default_pic = \"yes\"";

$result = @mysql_query($sql,$connection) or die("Couldn't execute query.");

$PICSnum=mysql_num_rows($result);

while ($row = mysql_fetch_array($result)) {

$url = $row['url'];

$image = $row['image'];

}

if(empty($url)) {$newurl="$imgdir";}else{$newurl="$userdir/$url";}

if(empty($image))$image="pic.gif";

$showimg="/$newurl/$image";

?>

function openpopup(popurl){

var winpops=window.open(popurl,"","width=610px,height=625px,resizable")

}

</script>

<a href="javascript:openpopup('showpic.php?img=<?echo $image ?>')"><IMG SRC="../../thumbs/phpThumb.php?src=<?echo $showimg ?>&w=150" border="0"></a>

<?

if($PICSnum > "0") {

?>

<br><a href="<?echo $PHP_SELF ?>?action=pics">My Pictures >></a>

<? }

if($profile_id == $auth[member_id]){

echo "<br><a href=\"$userurl/upload.php\">upload picture</a>";

} else {

if($profile_id != "00000001")

echo "<br><a href=\"/message.php?mail_to=$profile_id&p_name=$p_name\">

<img src=\"$imgurl/friend.gif\" border=\"0\" alt=\"Send Friend Request!\"></a>";

}

?>

btw the script is retreving users images from the data base... like an avator

only one · May 9, 2007

$conn = mysql_connect("localhost", "username", "password");

$db = @mysql_select_db("runnerse_profiles", "images");

you dont need to put $conn iin the mysql_selecet_db() function..

runnerjp · May 9, 2007

thanks... sill mistake there lol but its still not connecting

runnerjp · May 9, 2007

if it help...all im doing is going into the table images and getting the users image they have uploaded

the table looking like this

CREATE TABLE `images` (
  `image_id` varchar( NOT NULL default '',
  `directory` varchar(150) NOT NULL default '',
  `image` varchar(60) NOT NULL default '',
  `url` varchar(150) NOT NULL default '',
  `displayname` varchar(25) NOT NULL default '',
  `default_pic` enum('yes','no') NOT NULL default 'no'
) TYPE=MyISAM;

i displayed the structure as its easyer for u guys 2 see then

the image id is the same is the users id

Nameless12 · May 9, 2007

$conn = mysql_connect("localhost", "username", "password");

$db = @mysql_select_db("runnerse_profiles", "images");

you dont need to put $conn iin the mysql_selecet_db() function..

$conn is needed so the mysql_* functions know what connection to use, if you do not do this you will have problems handling multiple db servers.

Why it is optional to pass the connection to mysql_* functions it is not optional in other database apis.

The reason for the connection problems is you are using two strings followed by $conn in your mysql_select_db, it should be

mysql_select_db($name, $conn) //$conn being optional

taith · May 9, 2007

$conn = mysql_connect("localhost", "username", "password");

$db = @mysql_select_db("runnerse_profiles", "images");

you dont need to put $conn iin the mysql_selecet_db() function..

$conn is needed so the mysql_* functions know what connection to use, if you do not do this you will have problems handling multiple db servers.

Why it is optional to pass the connection to mysql_* functions it is not optional in other database apis.

The reason for the connection problems is you are using two strings followed by $conn in your mysql_select_db, it should be

mysql_select_db($name, $conn) //$conn being optional

usually i'd agree... but unless your doing a server of massive proportions... having 2 databases powering one page... is not really necessary...

runnerjp · May 9, 2007

ahhh lol ok so what do i do then if u dnt agree taith

taith · May 9, 2007

i assume your getting a username/password error?

Nameless12 · May 9, 2007

$conn = mysql_connect("localhost", "username", "password");

$db = @mysql_select_db("runnerse_profiles", "images");

you dont need to put $conn iin the mysql_selecet_db() function..

$conn is needed so the mysql_* functions know what connection to use, if you do not do this you will have problems handling multiple db servers.

Why it is optional to pass the connection to mysql_* functions it is not optional in other database apis.

The reason for the connection problems is you are using two strings followed by $conn in your mysql_select_db, it should be

mysql_select_db($name, $conn) //$conn being optional

usually i'd agree... but unless your doing a server of massive proportions... having 2 databases powering one page... is not really necessary...

I never said having two databases was ideal, But to restrict your self to one when you could otherwise have support for multiple is just stupid. Also in other database apis it is not an optional parameter so I think it is best to do to be consistent.

Multiple mysql databases like you said is most probably pointless like you say in relation to mysql but in there are times multiple databases come in handy.

runnerjp · May 9, 2007

no get "Couldn't select database." which is weird as that is the database name

taith · May 9, 2007

$conn = mysql_connect("localhost", "username", "password") or die(mysql_error());

$db = mysql_select_db("runnerse_profiles", "images") or die(mysql_error());

runnerjp · May 9, 2007

no still just get "Couldn't select database."

jitesh · May 9, 2007

<?php

$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
    die('Not connected : ' . mysql_error());
}

// make foo the current db
$db_selected = mysql_select_db('foo', $link);
if (!$db_selected) {
    die ('Can\'t use foo : ' . mysql_error());
}
?>

taith · May 9, 2007

no still just get "Couldn't select database."

meaning...

a) you have a typo

b) the database doesnt exist yet

c) your mysql is pooched

runnerjp · May 9, 2007

i have looked at it... yet all looks correct to me

runnerjp · May 9, 2007

should i start from scratch...is there an easyer way of doing this???...does it have to be done by db

taith · May 9, 2007

technically... no... you can switch a database out for straight files... but databases make everyones lives SO EASY!

only one · May 9, 2007

could be probleme with the host.. it might not be localhost (if your using a webhost)

per1os · May 9, 2007

One thing is for sure, the @'s just make the problem 10x harder to diagnose. It is never good to use those to surpress an error.

One option is to create 2 connections to the database and use 1 connection for 1 database and the 2nd for the other IE:

<?php
$conn1 = mysql_connect("localhost", "username", "password");
$conn2 = mysql_connect("localhost", "username", "password");
mysql_select_db("db1", $conn1);
mysql_select_db("db2", $conn2);

$query = mysql_query("SELECT * FROM TABLE", $conn2);

Unsure if it works and probably is not very proficient and may cause mysql max user problems down the line.

If at all possible and it is "unnescary" to use 2 database I would combine them into one.

I am sure there is someone who has expierence with this, but usually when more than 1 database is used the 2nd one is hosted on a seperate server.

Just incase the mysql_select_db() function still won't work with multiple database connections (as has happened to me before).

$dbh1 = mysql_pconnect($host,$user,$pass);

$dbh2 = mysql_pconnect($host,$user,$pass);

You could do this...

mysql_query("USE database1",$dbh1);

mysql_query("Use database2",$dbh2);

This does the same thing as the mysql_select_db() function...

or this...

You don't even have to select the database for each connection.

mysql_query("SELECT * FROM database1.table",$dbh1);

mysql_query("SELECT * FROM database2.table",$dbh2);

You could also try that too.

runnerjp · May 9, 2007

what i have done is added it all into 1 database!

 <?
		$conn = mysql_connect("localhost", "", "") or die(mysql_error());
            $db = mysql_select_db("runnerse_profiles") or die(mysql_error());
		$sql = "SELECT * FROM $tbl_images WHERE image_id = $auth[member_id] AND default_pic = \"yes\"";
		$result = @mysql_query($sql,$connection) or die("Couldn't execute query.");
		$PICSnum=mysql_num_rows($result);
		while ($row = mysql_fetch_array($result)) {
		$url = $row['url'];
		$image = $row['image'];
		}
		if(empty($url)) {$newurl="$imgdir";}else{$newurl="$userdir/$url";}
		if(empty($image))$image="pic.gif";
		$showimg="/$newurl/$image";
            ?>
            <script type="text/javascript">
		function openpopup(popurl){
		var winpops=window.open(popurl,"","width=610px,height=625px,resizable")
		}
            </script>
            <p align="center">
             <a href="javascript:openpopup('showpic.php?img=<?echo $image ?>')"><IMG SRC="../../thumbs/phpThumb.php?src=<?echo $showimg ?>&w=150" border="0"></a>
             <?
             if($PICSnum > "0") {
             ?>
             <br><a href="<?echo $PHP_SELF ?>?action=pics">My Pictures >></a>
             <? }
                        if($profile_id == $auth[member_id]){
                        echo "<br><a href=\"$userurl/upload.php\">upload picture</a>";
                        } else {
                        if($profile_id != "00000001")
					echo "<br><a href=\"/message.php?mail_to=$profile_id&p_name=$p_name\">
					<img src=\"$imgurl/mail.gif\" border=\"0\" alt =\"Send Message!\"></a> 
            <a href=\"$siteurl/friends.php?action=invite&profile_id=$profile_id\">
            <img src=\"$imgurl/friend.gif\" border=\"0\" alt=\"Send Friend Request!\"></a>";
                        }
                        ?>

but it still wont connect to that database??...thing is i have other scripts connecting to it

runnerjp · May 9, 2007

still getting not connecting to DB

per1os · May 9, 2007

<?

$conn = mysql_connect("localhost", "", "") or die(mysql_error());

$db = mysql_select_db("runnerse_profiles") or die(mysql_error());

$sql = "SELECT * FROM $tbl_images WHERE image_id = $auth[member_id] AND default_pic = \"yes\"";

$result = @mysql_query($sql,$connection) or die("Couldn't execute query.");

You are using $connection, which should be $conn and do not use the @ signs, they do no good. Be consistent with referecing variable names.

runnerjp · May 9, 2007

now i have this and guess wat...still cannot connect lol duin my head in this

 <?
		$conn = mysql_connect("localhost", "", "") or die("Couldn't connect.");
            $db = mysql_select_db(runnerse_profiles) or die("Couldn't select database.");
		$sql = "SELECT * FROM $tbl_images WHERE image_id = $auth[member_id] AND default_pic = \"yes\"";
		$result = mysql_query($sql,$conn) or die("Couldn't execute query.");
		$PICSnum=mysql_num_rows($result);
		while ($row = mysql_fetch_array($result)) {
		$url = $row['url'];
		$image = $row['image'];
		}
		if(empty($url)) {$newurl="$imgdir";}else{$newurl="$userdir/$url";}
		if(empty($image))$image="pic.gif";
		$showimg="/$newurl/$image";
            ?>
            <script type="text/javascript">
		function openpopup(popurl){
		var winpops=window.open(popurl,"","width=610px,height=625px,resizable")
		}
            </script>
            <p align="center">
             <a href="javascript:openpopup('showpic.php?img=<?echo $image ?>')"><IMG SRC="../../thumbs/phpThumb.php?src=<?echo $showimg ?>&w=150" border="0"></a>
             <?
             if($PICSnum > "0") {
             ?>
             <br><a href="<?echo $PHP_SELF ?>?action=pics">My Pictures >></a>
             <? }
                        if($profile_id == $auth[member_id]){
                        echo "<br><a href=\"$userurl/upload.php\">upload picture</a>";
                        } else {
                        if($profile_id != "00000001")
					echo "<br><a href=\"/message.php?mail_to=$profile_id&p_name=$p_name\">
					<img src=\"$imgurl/mail.gif\" border=\"0\" alt =\"Send Message!\"></a> 
            <a href=\"$siteurl/friends.php?action=invite&profile_id=$profile_id\">
            <img src=\"$imgurl/friend.gif\" border=\"0\" alt=\"Send Friend Request!\"></a>";
                        }
                        ?>

runnerjp · May 10, 2007

there error is in here $conn = mysql_connect("localhost", "", "") or die("Couldn't connect.");

$db = mysql_select_db(runnerse_profiles) or die("Couldn't select database.");

$sql = "SELECT * FROM $tbl_images WHERE image_id = $auth[member_id] AND default_pic = \"yes\"";

$result = mysql_query($sql,$conn) or die("Couldn't execute query.");

or is it bad database>

runnerjp · May 10, 2007

"bump"

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[SOLVED] my script will not connect to the database :@--must be error in code

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