I have been trying to create a table on my database usin mysql but each time i dont jst get it right. I used dis comment: CREATE TABLE files_images ( number int, file_name varchar(225), file_link(20), file_size(20), screen_size(20), category(20) ); but it didnt work. Or am i suppose to run d comment on a php file? I also tried usin phpmyadmin but there were some many field i couldnt fill out like table comment & odas! Pls help me

CREATE TABLE files_images(
number int,
file_name varchar( 225 ) ,
file_link int( 20 ) ,
file_size int( 20 ) ,
screen_size int( 20 ) ,
category int( 20 )
)

I don't understand what you mean by - there were many fields I could not fill out using phpmyadmin.

Thanx but where do i run dat command. Should i run it on phpmyadmin cp or on a php file

in phpMyAdmin, after you create your database (you can also do this via phpMyAdmin), click your database name, click the "sql" tab to the left, and paste the above code in the window that ensues then click "go" and you should be good to go.

good luck!

hi, im simon, i really need ur help to brief me on how to upload my website, i swa ur reply on my post and really need someone to teach me this

Answer to post #5:

https://www.nairaland.com/nigeria/topic-68376.0.html
https://www.nairaland.com/nigeria/topic-11927.0.html
https://www.nairaland.com/nigeria/topic-213936.0.html

Just in case you aren't aware, there is a "search" button in the navigation area of this forum. Endeavor to use it every now and then before posting as chances are that your question has already been answered by others.

Also, please create another thread next time, as opposed to continuing on another, which isn't related to yours.

Thanks in advance for your cooperation

thanx it worked. My problem nw is connectin to my database. Is dis format correct <?php $con = mysql_connect("localhost","romeo",abc123"; if (!$con) die('could not connect: '. mysql_error()); //some code mysql_close($con); ?> & i used my subdomain 4 d localhost, but i dont knw what to use 4 d user & password. Pls help! Thanx 4 ur help i do appreciate

The format looks good to me.

When you run the code, what happens? Per the parameters, the host is what you will get from your webhost. Username/password is what you defined when you created the database.

good luck!

Pls help me correct dis code: $count = mysql_fetch_array(mysql_query("SELECT COUNT(*) FROM users WHERE hin >= '3'");

if($count[0]=="0"{
echo "<div class='leftline'>";
echo "<img src='images/ran.gif' alt=''/>No Site Active Now";
echo '</div>';
}else{



$sql ="SELECT * FROM users WHERE hin '3' ORDER BY

RAND() LIMIT 1";

$sites = mysql_query($sql);

while ($site = mysql_fetch_array($sites))

{

echo "<div class='leftline'>";
echo "<img src='images/ran.gif' alt=''/><a href='vsites.php?sid=$site[0]'>$site[4]</a><br/>";
echo '</div>';
} i ran it it generated errors pls help

and where is the error statement?

Also, it seems you are trying to do everything in one line. It isn't helping, at least me, in following your code. I don't even see a call to your database (with host, username and password). Looking at it without the error statement, I would say, knowing fully well that there are 1001 ways to skin a cat, that everything about your code is wrong.

Post the error statement let's see.

I only posted d code dat generated d error, dis d error comment: warning: mysql_fetch_array(): supplied argument is not valid mysql result resource in /www/zxq.net/m/w/a/mwap/htdocs/index.php on line 63. tanx 4 your help

much better

Chances are that your query is returning a single result set whereas, via your mysql_fetch_array() call, you are telling your code to expect an array. So you need to get rid of the mysql_fetch_array() call.

Alternatively, you could try this approach (note how I have broken down what you have condensed into one line, to a series of lines. This helps with readability and maintainability )

<?php
database calls.

$query = "SELECT name, subject, message FROM contact";
$result = mysql_query($query);

while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo "Name :{$row['name']} <br>" .
"Subject : {$row['subject']} <br>" .
"Message : {$row['message']} <br><br>";
}

include 'closedb.php';
?>

Thanx, get rid of d mysql_fetch_array()? Pls xplain i dnt understand.

Maybe you do, maybe you don't. It all depends. Again, are you intending to returning just one result set or many? If the latter, an array is warranted. If not, no. A good way to see how your query is working, and I do this all the time, is to copy it verbatim and paste/run it in phpMyAdmin. If you get an error, you note it and fix it and run it again till it works. If you don't, you note the results.

good luck!