help with my first learning database

geyids · #1 Mar 15th, 2007, 18:55

Is there something wrong with these?

CREATE TABLE personnel
(
id int NOT NULL AUTO_INCREMENT,
firstname varchar(25),
lastname varchar(20),
nick varchar(12),
email varchar(35),
salary int,
PRIMARY KEY (id),
UNIQUE id (id)
);
INSERT INTO personnel VALUES ('1','John','Lever','John', 'john@everywhere.net','75000');
INSERT INTO personnel VALUES ('2','Camilla','Anderson','Rose', 'rose@flower.com','66000');

I put the following codes in php and...

Code: Select all

<HTML>
<?php
$db = mysql_connect("localhost", "root", "");
mysql_select_db("learndb",$db);
$result = mysql_query("SELECT * FROM personnel",$db);
echo "<TABLE>";
echo"<TR><TD><B>Full Name</B><TD><B>Nick Name</B><TD><B>Salary</B></TR>";
while ($myrow = mysql_fetch_array($result))
{
echo "<TR><TD>";
echo $myrow["firstname"];
echo " ";
echo $myrow["lastname"];
echo "<TD>";
echo $myrow["nick"];
echo "<TD>";
echo $myrow["salary"];
}
echo "</TABLE>";
?>
</HTML>

...this is what I get

Warning: mysql_connect(): Access denied for user: 'root@localhost' (Using password: NO) in /home/(location)/index.php on line 10

Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in /home/(location)/index.php on line 11

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/(location)/index.php on line 12

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/(location)/index.php on line 15
Full NameNick NameSalary

Please someone help me

grahame · #2 Mar 16th, 2007, 07:08

You haven't given your MySQL password (for the MySQL account called root) as the third parameter to mysql_connect, so you are being denied access to the database from withing the PHP. The other error messages are all the consequence of this failure.

You should probably use a different account to the root one to access your database, and you it's advisable to apply passowrds to all the MySQL accounts that you have.

geyids · #3 Mar 16th, 2007, 13:13

Im not root in the webserver im uploading my database to. if u dont mind pliz help me where im I supposed to put my password in this

Code: Select all

CREATE TABLE personnel
(
id int NOT NULL AUTO_INCREMENT,
firstname varchar(25),
lastname varchar(20),
nick varchar(12),
email varchar(35),
salary int,
PRIMARY KEY (id),
UNIQUE id (id)
);
INSERT INTO personnel VALUES ('1','John','Lever','John', 'john@everywhere.net','75000');
INSERT INTO personnel VALUES ('2','Camilla','Anderson','Rose', 'rose@flower.com','66000');

or im I supposed to put in the php one?

karinne · #4 Mar 16th, 2007, 13:29

In the php

Code: Select all

$db = mysql_connect("localhost", "root", "password goes here");

geyids · #5 Mar 16th, 2007, 14:27

OK i changed this line to

Code: Select all

$db = mysql_connect("localhost", "myusername", "mypassword");

and this is what I got...

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/(mylocation)/index.php on line 15
Full NameNick NameSalary

Did I make a mistake to replace "root" with my username? im not root in the server

karinne · #6 Mar 16th, 2007, 14:30

Ok ... could you repost updated php please ... and tell us which is line 15

geyids · #7 Mar 16th, 2007, 15:41

PHP: Select all



<?php
$db = mysql_connect("localhost", "myusername", "mypassword");
mysql_select_db("learndb",$db);
$result = mysql_query("SELECT * FROM personnel",$db);
echo "<TABLE>";
echo"<TR><TD><B>Full Name</B><TD><B>Nick Name</B><TD><B>Salary</B></TR>";
while ($myrow = mysql_fetch_array($result))
{
echo "<TR><TD>";
echo $myrow["firstname"];
echo " ";
echo $myrow["lastname"];
echo "<TD>";
echo $myrow["nick"];
echo "<TD>";
echo $myrow["salary"];
}
echo "</TABLE>";
?>

line 15 is...

PHP: Select all


while ($myrow = mysql_fetch_array($result))

karinne · #8 Mar 16th, 2007, 15:49

Remove ,$db from you mysql_query .... I'm pretty sure you don't need that.

And this

PHP: Select all


echo"<TR><TD><B>Full Name</B><TD><B>Nick Name</B><TD><B>Salary</B></TR>";

should be

PHP: Select all


echo"<TR><TD><B>Full Name</B></TD><TD><B>Nick Name</B></TD><TD><B>Salary</B></TD></TR>";

Don't forget to close your tags ... you need to fix it in the while ( ) section too.

geyids · #9 Mar 16th, 2007, 17:38

what do I have to "fix it in the while ( ) section too" sorry im a learner in both php and mysql. I've looked at the line and saw I dont have a ";" is that right?

karinne · #10 Mar 16th, 2007, 17:44

Quote:

Originally Posted by geyids

what do I have to "fix it in the while ( ) section too" sorry im a learner in both php and mysql.

closing the </td> properly.

Quote:

Originally Posted by geyids

I've looked at the line and saw I dont have a ";" is that right?

no ... a while loop doesn't have ; it's

Code: Select all

while (expression) {
 statement
}

http://ca3.php.net/manual/en/control...ures.while.php

Did you try what I suggested?!?

geyids · #11 Mar 16th, 2007, 17:46

This is the codes I have at present but they still aint working the error is the same...

PHP: Select all


<?php
$db = mysql_connect("localhost", "myusername", "mypassword");
mysql_select_db("learndb",$db);
$result = mysql_query("SELECT * FROM personnel",$db);
echo "<TABLE>";
echo"<TR><TD><B>Full Name</B></TD><TD><B>Nick Name</B></TD><TD><B>Salary</B></TD></TR>";
while ($myrow = mysql_fetch_array($result));
{
echo "<TR><TD>";
echo $myrow["firstname"];
echo " ";
echo $myrow["lastname"];
echo "<TD>";
echo $myrow["nick"];
echo "<TD>";
echo $myrow["salary"];
}
echo "</TABLE>";
?>

...this is the error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/(mylocation)/index.php on line 15
Full NameNick NameSalary

karinne · #12 Mar 16th, 2007, 17:48

Did you try this!

Quote:

Originally Posted by karinne

Remove ,$db from you mysql_query .... I'm pretty sure you don't need that.

geyids · #13 Mar 16th, 2007, 17:49

ok wait will check on it then tell you thnx hope u r not tired with this silly learner

karinne · #14 Mar 16th, 2007, 17:51

Quote:

Originally Posted by geyids

ok wait will check on it then tell you thnx hope u r not tired with this silly learner

I'm a fairly patient woman but I wish you would read the whole reply

geyids · #15 Mar 16th, 2007, 18:37

I removed the

PHP: Select all

$db

but doesn't help. What is wrong with this

PHP: Select all


mysql_fetch_array()

in...

PHP: Select all


while ($myrow = mysql_fetch_array($result))

is there no other alternative on this?

geyids · #16 Mar 17th, 2007, 18:20

I got the solution. I asked the guy who opened the account for me and he told me the error was the db created was not called learndb but...! So I changed it and its working now. Thanx anyway for ur help i've learned alot

karinne · #17 Mar 17th, 2007, 21:57

good stuff!

		Webforumz.com > Main Forums > Program Your Website > Databases
help with my first learning database