mysql_query():

[SoulAssassin]

What's wrong with this code?
Keep on getting Error:
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in C:\apache2triad\htdocs\callforw.php on line 117
And even though I get this message, my DB still updates correctly.
It just doesn't echo my successfull message.

 $conn= mysql_connect( "localhost", "root", "password" )
or die( "Err:conn" );
$conn = mysql_select_db( "snow")
or die( "Err:conn1" );
 $sql = mysql_query("UPDATE request set forwmail='$forwmail' WHERE qid='$qid'"); 
$rs = mysql_query($sql, $conn); //Line 117

						    if($rs){ 

				echo "forwmail updated";
					}
					else
					{
				echo "Couldn't update forwmail with $forwmail<br/>";
					}

Dumb struck

[abc123]

The problem is that you have set $conn as an sql connection resource but immediatly after you have made it a db resource. The mysqlquery() function requires a connection resource but you are passing a db resource.

so...

$db = mysql_select_db( "snow") or die( "Err:conn1" );
 

Just change that one line above to use $db instead of $conn and you shouldn't get that error message.

[Blulagoon]

$sql = mysql_query("UPDATE request set forwmail='$forwmail' WHERE qid='$qid'"); 
$rs = mysql_query($sql, $conn); //Line 117 

Additionally, $sql is not an sql query string, it is the result of an sql query performed by mysql_query on the string UPDATE request etc etc. Because that line doesn't reference the connection, explains why the update happens.

HTH - Blu

edit: line beginning $rs seems actually somewhat superfluous ......

[SoulAssassin]

Thanks guys, sometimes I'm just blind.
It's sorted.

[SoulAssassin]

Thought it worked, well it did for a while.
It still does update my DB, but still echo's my else message.
I need the confirmation:
"Entry Archived"
Can anyone see the mistake?

$conn= mysql_connect( "localhost", "root", "password" );
$db = mysql_select_db( "snow", $conn);
$sql = mysql_query("UPDATE request1 set forwmail='$forwmail' WHERE qid='$qid'");
$rs = mysql_query($sql, $db);

if($rs)
{						   	 
echo ("<table align=\"center\" width='50%' border=1 bgcolor=\"#FFFFFF\">");
echo ("<tr>");
echo ("<th><font size=\"2\">Entry Archived</font></th>");
echo ("</tr>");
echo ("<table>");
}
else
{
echo ("<table align=\"center\" width='50%' border=1 bgcolor=\"#FFFFFF\">");
echo ("<tr>");
echo ("<th><font size=\"2\">Couldn't archive with entry $forwmail </font></th>");
echo ("</tr>");
echo ("<table>");
}

[abc123]

$sql = "UPDATE request1 set forwmail='$forwmail' WHERE qid='$qid'"; 

See if it works with the line above.

[SoulAssassin]

Thanks, but it also doesn't work.
I think my problem must be somewhere else in my code.

[abc123]

Okay... try changing this line as well as the other one I suggested. Remember that it requires the connection resource and the the database resource.

$rs = mysql_query($conn, $db); 

Hope it works this time. 🙂

[SoulAssassin]

Cool thanks..
It works !

[laserlight]

It works !

It does? That's rather strange, since mysql_query($conn, $db) does not make sense for the contextual clues that $conn and $db provide.

Often you can leave out the resource link identifier obtained from mysql_connect(), since the last open database connection will be assumed.

mysql_connect("localhost", "root", "password");
mysql_select_db("database");
$sql = "SELECT * FROM tablename";
$result = mysql_query($sql);

[SoulAssassin]

Well, this is how it worked

$conn= mysql_connect( "localhost", "root", "password" );
$db = mysql_select_db( "snow", $conn);
$sql = mysql_query("UPDATE request1 set forwmail='$forwmail' WHERE qid='$qid'");
$rs = mysql_query($sql, $db);

[laserlight]

SoulAssassin wrote:

Well, this is how it worked

$conn= mysql_connect( "localhost", "root", "password" );
$db = mysql_select_db( "snow", $conn);
$sql = mysql_query("UPDATE request1 set forwmail='$forwmail' WHERE qid='$qid'");
$rs = mysql_query($sql, $db);

Fascinating. The above code should fail spectacularly.

1. $sql is a result resource. It cannot be used in the second mysql_query() call, which expects a string as the first parameter.

2. $db is a boolean value. It cannot be used in the second mysql_query() call, which expects a resource link identifier as the second parameter (if one is provided at all).

[Blulagoon]

@

It is a mystery to me why the script works, except that he does the query with the line before the $rs one.

Going back to his code

$sql = mysql_query("UPDATE request1 set forwmail='$forwmail' WHERE qid='$qid'"); 
$rs = mysql_query($sql, $db); 

if($rs) 

He would be better off to delete the line beginning $rs as it doesn't achieve anything even if it doesn't actually error, and replace if($rs) with if($sql)

Blu

[SoulAssassin]

Blulagoon wrote:

@

It is a mystery to me why the script works, except that he does the query with the line before the $rs one.

Going back to his code

$sql = mysql_query("UPDATE request1 set forwmail='$forwmail' WHERE qid='$qid'"); 
$rs = mysql_query($sql, $db); 

if($rs) 

He would be better off to delete the line beginning $rs as it doesn't achieve anything even if it doesn't actually error, and replace if($rs) with if($sql)

Blu

Thanks, the first real constructive solution.
I did what yolu said, now it works perfectly.
Previously it did update, but never echoed the if, but that is fixed with:

$conn= mysql_connect( "localhost", "root", "password" );
$db = mysql_select_db( "snow", $conn);
$sql = mysql_query("UPDATE request1 set forwmail='$forwmail' WHERE qid='$qid'");


if($sql)
           {

I'll take the sarcasm, as long as it resolves challenges.
Peace

[laserlight]

It is a mystery to me why the script works, except that he does the query with the line before the $rs one.

Ah, I should have guessed that only one SQL statement is actually needed. It looked like some attempt to do a second query based on the results on the first one.

[SoulAssassin]

sweet