Programming Forums > Web Development and Design > PHP

Kelvoron · Nov 26th, 2007, 3:51 PM

Hey guys I have been working on getting my selct statement to work for some time now and just can not get it to work I did manage to get it to stop returning errors but now it returns a blank page, so i went in to my DBMS and opend up the table double checked that i have every thing spelled the same as it is in there. I would be greatful if you could look at the code for me.

Quote:

<?
$host="localhost";
$username="----------";
$password="*******";
$db_name="kelvoron_testscripts";
$table_name="test1";

mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

$sql="SELECT name FROM $tabLE_name WHERE gender='male'";
$result=mysql_query($sql);

echo($result);

mysql_close();
?>

DaWei · Nov 26th, 2007, 5:38 PM

Kelvoron, how long have you been around, now? Not long enough to distinguish the differnce between code tags and quote tags? If you don't have two brain cells, try to rent one or two.

Dameon · Nov 26th, 2007, 5:39 PM

$tabLE_name and $table_name aren't the same thing. Make sure you have php error display enabled.

Kelvoron · Nov 26th, 2007, 6:19 PM

Oh my bad i changed it. it still does not work thoug thanks for cathing that for me Dameon. I checked my code against othr peoples and bought a MySQL book the code looks corect as far as i can see but i still get no return

i changed echo(results); to echo $results now it displays " Resource id #2 " X|

DaWei · Nov 26th, 2007, 6:59 PM

Check your book regarding what "resource" means. As an example, a FILE * in C does not represent the contents of a file. It is merely an access mechanism.

Dameon · Nov 26th, 2007, 8:44 PM

Any basic example of the PHP MySQL functions will show you what to do with your $result variable. You certainly don't try to convert it to a string and display it. Reread your book and the online documentation before simply asking around for the next line -- both contain working examples that you can build from with a little personal effort.

Infinite Recursion · Nov 27th, 2007, 1:28 PM

Look into the mysql_fetch_row() function, passing in your $result variable as the parameter.

Sanjay Aggarwal · Nov 29th, 2007, 12:01 AM

The Variable names are case sensitive in PHP. See the variable you have used for table name.

Sanjay Aggarwal

titaniumdecoy · Nov 29th, 2007, 12:51 AM

Quote:

Originally Posted by Sanjay Aggarwal

The Variable names are case sensitive in PHP. See the variable you have used for table name.

Sanjay Aggarwal

Thanks for repeating Dameon's post (second reply in this thread).

pegasus001 · Nov 29th, 2007, 9:37 AM

Quote:

mysql_connect("$host", "$username", "$password")or die("cannot connect");

The $host variable is a string by itself, try removing the double quotes.

Just a thought, though.

Nov 26th, 2007, 5:39 PM	#3
Dameon Troll Join Date: Apr 2005 Location: Texas Posts: 732 Rep Power: 4	Re: SELECT statement $tabLE_name and $table_name aren't the same thing. Make sure you have php error display enabled. __________________ MD5(sig) = bcef75433db02e9ad9bf81d6f7c5c270

Nov 26th, 2007, 6:19 PM	#4
Kelvoron Programmer Join Date: Aug 2007 Location: Ohio Posts: 46 Rep Power: 0	Re: SELECT statement Oh my bad i changed it. it still does not work thoug thanks for cathing that for me Dameon. I checked my code against othr peoples and bought a MySQL book the code looks corect as far as i can see but i still get no return i changed echo(results); to echo $results now it displays " Resource id #2 " X\| __________________ Good better best, never let it rest, untill your good is better and your better is the best. Last edited by Kelvoron; Nov 26th, 2007 at 6:34 PM.

Nov 26th, 2007, 6:59 PM	#5
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	Re: SELECT statement Check your book regarding what "resource" means. As an example, a FILE * in C does not represent the contents of a file. It is merely an access mechanism. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Nov 26th, 2007, 8:44 PM	#6
Dameon Troll Join Date: Apr 2005 Location: Texas Posts: 732 Rep Power: 4	Re: SELECT statement Any basic example of the PHP MySQL functions will show you what to do with your $result variable. You certainly don't try to convert it to a string and display it. Reread your book and the online documentation before simply asking around for the next line -- both contain working examples that you can build from with a little personal effort. __________________ MD5(sig) = bcef75433db02e9ad9bf81d6f7c5c270

Nov 27th, 2007, 1:28 PM	#7
Infinite Recursion Programming Guru Join Date: Jul 2004 Location: United States Posts: 3,467 Rep Power: 8	Re: SELECT statement Look into the mysql_fetch_row() function, passing in your $result variable as the parameter. __________________ http://jasonpowers.net "There are a thousand hacking at the branches of evil to one who is striking at the root."

Nov 26th, 2007, 5:38 PM	#2
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	Re: SELECT statement Kelvoron, how long have you been around, now? Not long enough to distinguish the differnce between code tags and quote tags? If you don't have two brain cells, try to rent one or two. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Nov 29th, 2007, 12:01 AM	#8
Sanjay Aggarwal Newbie Join Date: Oct 2007 Posts: 12 Rep Power: 0	Re: SELECT statement The Variable names are case sensitive in PHP. See the variable you have used for table name. Sanjay Aggarwal

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Programming Forums > Web Development and Design > PHP

SELECT statement