Php

[murdocsvan]

Hey, all i need to do is show all the information in an MySQL table in an HTML table, regardless of the number of rows. How would i code this?

Thanks in advance.

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Query result table display class (metabase, split) - PHP Classes



Code Examples -> Create a dynamic HTML table from a query

[murdocsvan]

Okay here is my code:

 <?php

session_start();

//Check user is logged in in first place
if(!isset($_SESSION['user']))
    {
    header("location:******");
    }

//Stores mysql login details
$host="localhost"; // Host name 
$sql_username="*****"; // Mysql username 
$sql_password="*****"; // Mysql password 
$db_name="*****"; // Database name 
$tbl_name="admin"; // Table name 

//mysql Connect variable
$con = mysql_connect("$host","$sql_username","$sql_password");

//Query to retrieve information
$query = "SELECT * FROM admin";
$result = mysql_query($query,$con);

echo "<table border='1'>";
echo "<td><th>Username</th></tr>";
echo '<form action="user_delete.php" method="post">';
// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array($result))
    {
    // Print out the contents of each row into a table
    echo "<tr><td>";
    echo $row['id'];
    echo "</td></tr>";
    echo "<tr><td>"; 
    echo $row['username'];
    echo "</td><td>";
    echo '<input type="button" value="" />';
    echo "</tr></td>";
    } 

echo "</form></table>";
?>  

But for some reason it returns an error on the line:

 while($row = mysql_fetch_array($result)) 
  

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/vmrgjdq/public_html/test/users.php on line 29

Can anyone help?

Thanks.

[CrazeD]

 $result = mysql_query($query,$con); 
  

Change this line to

 $result = mysql_query($query,$con) or die (mysql_error()); 
  

And post back the message.

[murdocsvan]

It said "No Database selected", because i forgot to put in:

 //Database select
mysql_select_db($db_name, $con); 
  

So i put it in and it works. Thanks a lot!