Proper Syntax

[tcorbeil]

I am having trouble with this code:

 

$database = "oilers"

$query="SELECT * FROM '$database' WHERE Page = '$pagepath'";

 

Even though I know I have a table in my database called oilers, it doesn't seem to work..

I'm thinking it is just my syntax using the variable $database.. can someone let me know if there is something wrong?

 

Thanks everyone.

 

T.

[ultrus]

Hmmm. Try this:

 

$query="SELECT * FROM '" . $database . "' WHERE Page = '" . $pagepath . "'";

 

Variables inside of single quotes have been giving me troubles today.

 

With my frustration today anyway, echo('Are you happy?: $happy'); would output:

Are you happy?: $happy

 

With double quotes - echo("Are you happy?: $happy");

Are you happy?: Yes!

 

[btherl]

Table names shouldn't be inside ' , instead they should be inside ` , or inside no quotes at all.

[ultrus]

Ah yes. My code above is bunk.

 

$query="SELECT * FROM $database WHERE Page = '" . $pagepath . "'";

[tcorbeil]

Ok i tried:

 

$database = "Tim_123";

$query="SELECT * FROM $database WHERE Page = '".$pagepath."'";

 

...still no go..  I can't think of anything else to try.. seems I tried every different syntax variations and no go..

 

Any other ideas before I pull my hair out?

 

T.

[ultrus]

what is the error you are getting? If you pull your hair out, I recommend using this php class:

http://www.phpclasses.org/browse/package/790.html

 

It helps on my rush projects.

[tcorbeil]

Ok maybe I'm on the wrong track...

 

I'm getting this error:

 

Warning: mysql_numrows(): supplied argument is not a valid MySQL result resource in /home/tcorbeil/public_html/Entertainment/Tim_123/Tim_123.php on line 897

 

with this code:

 

 

$database = "Tim_123";

$query="SELECT * FROM $database WHERE Page = '".$pagepath."'";

$result=mysql_query($query);

$num=mysql_numrows($result);

 

it seems the error would have to do with the numrow..  I use $num=mysql_numrows($result); to count the amount of occurences in the search...

 

T.

[Lytheum]

Should be

$result = mysql_num_rows($result);

I do believe.

[hitman6003]

change:

 

$result=mysql_query($query);

 

to

 

$result=mysql_query($query) or die(mysql_error());

 

And see what the error is.

[trq]

Firstly, get in the habbit of checking your queries actually work before trying to use them, the code you posted is extremely error prone.

 

<?php

 // connect to db.

 $database = "Tim_123";
 $query = "SELECT * FROM $database WHERE Page = '$pagepath';";
 if ($result = mysql_query($query)) {
   if (mysql_num_rows($result)) {
     while ($row = mysql_fetch_assoc($result)) {
       print_r($row);
     }
   } else {
     echo "No results found.";
   }
 } else {
   echo "Query failed. $query<br />" . mysql_error();
 }

?>

Secondly... where is $pagepath defined?