please help me diaplay my tables (url included)

[Oleg Butuzov]

I did not saw your url, because i have bad connection...


may be

Code: Select all

<?
$user="user";
$pass="pass";
$host="localhost";

$link=mysq_connect($host, $user, $pass);

$mysql_select_db("dbname", $link);

$query=mysql_query("SELECT * FROM 'tablename' ", $link);

for ($i=0; $i < mysql_num_rows($query); $i++)
{
$data=mysql_fetch_array($query);

// now your data in array like $data[collum1]  $data[collum2] $data[collum3]

// do somthing whith this vars
}

?

[WiZARD]

<?php

$connect = mysql_connect("localhost","robbyjb","");
$mysql_select_db("robbyjb", $link);

$query=mysql_query("SELECT * FROM 'psytrance' ", $link);

for ($i=0; $i < mysql_num_rows($query); $i++)

while($row = mysql_fetch_array($result))
{
echo$row["record"];
}
?>
I viewed the psytrance.myi form & .frm .myd using zend at there ftp site & they appeared to be there intact.

Hi my dear little friend!
1 Check your connection for yor database:

Code: Select all

 $connect = mysql_connect("localhost","robbyjb","");
 if (!$connect)
 {echo "Sorry"}
 

2 Read carefuly message about your Warning
3 Read carefully http://www.nelie.org/robbyjb/sql1.php
4 What is value of your variable $resultid?
5 And once more check your database, whats field in, and how you call this.

I'm not tell about yours mistake you need see self.

P.S.
If you don't want what your site hack before you write code do not use original database name, user and his password. some peaple can use this

[Oleg Butuzov]

<?php

$connect = mysql_connect("localhost","robbyjb","");
$mysql_select_db("robbyjb", $link);

$query=mysql_query("SELECT * FROM 'psytrance' ", $link);

for ($i=0; $i < mysql_num_rows($query); $i++)

while($row = mysql_fetch_array($result))
{
echo$row["record"];
}
?>
I viewed the psytrance.myi form & .frm .myd using zend at there ftp site & they appeared to be there intact.

can oyu tell me where are you find in my code while

Code: Select all

for ($i=0; $i < mysql_num_rows($query); $i++)

while($row = mysql_fetch_array($result))
{
echo$row["record"];
}
?>

of course it is not work...

truly

Code: Select all

<?
for ($i=0; $i < mysql_num_rows($query); $i++)
{
$row = mysql_fetch_array($result);
echo$row["record"];
}
?>

[Jay]

<?php

$connect = mysql_connect("localhost","robbyjb","");
$mysql_select_db("robbyjb", $link);

$query=mysql_query("SELECT * FROM 'psytrance' ", $link);

for ($i=0; $i < mysql_num_rows($query); $i++)

while($row = mysql_fetch_array($result))
{
echo$row["record"];
}
?>
I viewed the psytrance.myi form & .frm .myd using zend at there ftp site & they appeared to be there intact.

Your code seems to be fine, if all you're trying to do is output the 'record' column. If you don't have a record column then nothing will be output. You might also try putting a space in your loop between the command and the variable, eg
echo $row["record"];