Very easy PHP / Mysql Problem..

[Spherian]

Hello..

Firstly can i say sorry for asking for help on my first post. ;)

Okay, I need a few examples of how to display information from a mysql database, if any one has the time.. Here is my query in more detail.

I know how to connect to the database (Thats about it) So i am now connected, There are 7 fields, and one of the fields is "Reports" I want to display on screen the all the details of the row IF Reports has a value of 1 or more.


Hmm.. Confused...

Any comments would be appreciated.

[^BuGs^]

$array = mysql_query("SELECT * FROM TABLENAMEHERE WHERE SOMETHING ARGUMENT VALUE");
while ($i=mysql_fetch_array($array)) {

/* do what in here
vars returned are: $i[nameoffield]; */

}

[Spherian]

Would this be valid


$array = mysql_query("SELECT * FROM xlinks WHERE reports > 0 ");

?? Basically, show all that are more than 0.

[Halide]

I would use $result instead of $array

:)

and it looks ok to me...

[Halide]

im going to qbasic class :D later

[Spherian]

Hey people, I am getting this error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /usr/local/psa/home/vhosts/httpdocs/new/reported2.php on line 24

This is the bottm piece of script:


$conn = @MYSQL_CONNECT($server, $user, $pass);
//Select DB
@MYSQL_SELECT_DB($mydb) or die ("<h1>Database '$mydb' does not exist or mySQL is not connected!

Please check your settings ...</h1>");
//Check
if(!$conn) die("<h1>mySQL-Connection-Error
Check your settings ...</</h1>");

$result = mysql_query("SELECT * FROM l_xlinks WHERE reports >0");
while ($i=mysql_fetch_array($result)) {

echo("$i[name]");


}
?>


Hmm, i dont mean to be a pest, but any info would be great, also i am very new to Mysql. So b gental.. :)

[Halide]

If you want a good reference, check out the HUGE documentation that they have.

Also, what line is 24 in your script? What editor are you using?

[Halide]

I gotta go.. bbl

[Spherian]

line 24 is this:
while ($i=mysql_fetch_array($result)) {

I will have a read up now. :)

[^BuGs^]

$result = mysql_query("SELECT * FROM l_xlinks WHERE reports > 0");

[roninblade]

... and try printing out mysql_error()