can somebody help?

DXL

for some reason, my MySQL select wont gather and output what i was hoping it would say

This is my coding. Be aware, since im only a beginner with MySQL, this has some notes, to help me remember.

<?php

$user = "root";
$pass = "";
$database = "test";

######################Connect##########################################
$link = @mysql_connect( "localhost", $user, $pass );
if ( ! $link )
{
die( "Could not connect to MySQL ERROR: ".mysql_error() );
}

$link2 = @mysql_pconnect( "localhost", $user, $pass );
if ( ! $link2 )
{
die( "Could not connect to MySQL ERROR: ".mysql_error() );
}
######################Select Database##################################
@mysql_select_db( $database )
or die ( "Could not open $database ERROR: ".mysql_error() );

######################Create Table#####################################
$xR = "CREATE TABLE res ( title VARCHAR(30), author VARCHAR(40), domain VARCHAR(250), desc VARCHAR(100) )";

$createxR = mysql_query( $xR, $link );

######################Add Data to Table################################
$add_A = "INSERT INTO res( title, author, domain, desc ) values( 'Darth' 'DXL' 'http://www.gamingw.net/' 'Darth is back' )";

mysql_query( $add_A );

######################Obtaining the Data###############################
$select = "SELECT author FROM res WHERE title = 'Darth'";
$found = mysql_query( $select );
print("Thank you for submitting your resource, $found");

for some reason, it will not display "DXL" after the comma. help 🙁 😕

iNFERiON

I'm pretty new to mySQL too, so I can't guarantee it'll work, but try this:

print("Thank you for submitting your resource, " $found);

DXL

Parse error: parse error, unexpected T_VARIABLE in c:\program files\easyphp1-7\www\mysql1\testing.php on line 40

nope, doesn't work.

i thought that sidescape in the print function was only for other functions anyway...

iNFERiON

Hmm, maybe you should try echo instead of print.

echo "Thank you for submitting your resource, " $found ".";

DXL

Parse error: parse error, unexpected T_VARIABLE, expecting ',' or ';' in c:\program files\easyphp1-7\www\mysql1\testing.php on line 40

didnt work

iNFERiON

My mistake, should be:

echo "Thank you for submitting your resource, " . $found . ".";

helz

in your sql query where you add data into the database:

######################Add Data to Table################################
$add_A = "INSERT INTO res( title, author, domain, desc ) values( 'Darth' 'DXL' 'http://www.gamingw.net/' 'Darth is back' )";

you forgot to add commas in between the values for each field, that is why you are getting the error:

######################Add Data to Table################################ 
$add_A = "INSERT INTO res( title, author, domain, desc ) values( 'Darth'[B],[/b] 'DXL'[b],[/b] 'http://www.gamingw.net/'[b],[/b] 'Darth is back' )";

the reason why you are not getting the authors name after the comma is that you are just requesting information from the database and not putting it into a form that you can use, you need to add this after the $found = mysql_query($select):

$found1 = mysql_fetch_row($found);
$found = $found1[0];

DXL

ok, i changed it into this:

$add_A = "INSERT INTO res( title, author, domain, desc ) values( 'Darth', 'DXL', 'http://www.gamingw.net/', 'Darth is back' )";

mysql_query( $add_A );

######################Obtaining the Data###############################
$select = "SELECT author FROM res WHERE title = 'Darth'";
$found = mysql_query( $select );
echo "Thank you for submitting your resource, ";
print("$found");
print(".");

what it displayed was "Thank you for submitting your resource, ."

umm...can u help me a little longer?

helz

you forgot the end of my post:

$found1 = mysql_fetch_row($found);
$found = $found1[0];

add that below:
$found = mysql_query( $select );

so it becomes:
$found = mysql_query($select);
$found1 = mysql_fetch_row($found);
$found = $found1[0];

then try it

DXL

######################Obtaining the Data###############################
$select = "SELECT author FROM res WHERE title = 'Darth'";
$found = mysql_query( $select );
$found1 = mysql_fetch_row($found);
$found = $found1[0];
echo "Thank you for submitting your resource, ";
print("$found");
print(".");

this came up with the error:

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in c:\program files\easyphp1-7\www\mysql1\testing.php on line 40
Thank you for submitting your resource, .

what now?

drawmack

what you have in found is a pointer to the query itself, you need to pull the data from that try this

<?php
    echo "Thank you for submitting " . mysql_result($found,0,0) . "<br />\n";
?>

DXL

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in c:\program files\easyphp1-7\www\mysql1\testing.php on line 40

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in c:\program files\easyphp1-7\www\mysql1\testing.php on line 42
Thank you for submitting
.

any other ideas?

mparker

Hey I am willing to help, but could you please tell me how you are getting the data to insert into the database? Is it coming off of an HTML form, or are you just inserting static data?

helz

i have never known the reason why sometimes it doesn't work, as in the past when i have tried mysql_num_rows i have had the same problem...
usually it only affects num_rows and not any other query

if you want to try a replacement for the num_rows you can try this:

$numofrow = mysql_query("SELECT COUNT(*) FROM res") or die(mysql_error());

not sure whether you need to get it from the resource with a fetch_array or not

DXL

mparker-im using just plain data, if u have read above, u would have noticed. im not working MySQL from forms yet, since im a starter with it

helz-i dont wanna find the row number, i just want it to pick DXL from the database

helz

sorry thats me getting confused 🙂

change:
mysql_fetch_row to mysql_fetch_array
and see if that works

DXL

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\program files\easyphp1-7\www\mysql1\testing.php on line 40
Thank you for submitting your resource, .

ummm...this is really starting to get on my nerves

helz

try this:

$found = mysql_query("SELECT author FROM res WHERE title = 'Darth'") or die(mysql_error()); 
$found1 = mysql_fetch_array($found);
$found = $found1[0];
echo "Thank you for submitting your resource, ".$found.".";

mparker

Okay bud, this is what you need to do... Use this script to retreive the data you had inserted with your last script...

// Declare Database Connection Variables // 
$user = "yourUserName"; 
$pass = "yourPassword"; 
$database = "test";
$host = "localhost"; 

// Establish A Connection //
$connection = @mysql_connect($host,$user,$pass) or die("Could not connect to MySQL<br><br>ERROR:'" . mysql_error() . "'");
$db = mysql_select_db($database,$connection) or die ("Could not open database"); 

// Obtaining the Data //
$SQL = "SELECT title FROM res WHERE title = 'Darth'"; 
$result = @mysql_query($SQL, $connection) or die("Could not execute query.");
$nums = @mysql_num_rows($result); 

// If a record was found Display It //		
if($nums) {
	$data = mysql_fetch_array($result);
	echo $data['title'];
	} else {
		echo "Sorry no records were found...";
		}

Remember to change the variables at the top and Let me know how it goes...

mparker

Did you try that script out?