PHP and MySQL problem

[dannysheps@hotmail.com]

Hello
i created a DB with 5 colums that is being updated with a form that i have in my site(created with ABVFP)

in the page where i put the PHP code ( that looks like that :
<?php
// Make a MySQL Connection
mysql_connect("localhost", "admin", "1admin") or die(mysql_error());
mysql_select_db("test") or die(mysql_error());

// Get all the data from the "example" table
$result = mysql_query("SELECT * FROM example")
or die(mysql_error());

echo "<table border='1'>";
echo "<tr> <th>Name</th> <th>Age</th> </tr>";
// 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['name'];
echo "</td><td>";
echo $row['age'];
echo "</td></tr>";
}

echo "</table>";
?>

but i dont want to show all the 5 colums only two or three of them .

im adding this code in HTML box...BUT NOTHING HAPPENS

thanks

[navaldesign]

Re: PHP and MySQL problem

Hi,

1. The echo command will output to the screen whatever you want. If, in example, you have 5 columns and you only want 2 of the values to appear, limit the echoing loop to the values that you want (as in example you posted, where only name and age will appear).


2. The code to use would be:

<?php
// Make a MySQL Connection

$db_host = "localhost";
$db_user = "cp_username_DB_username"; // Replace with actual value
$db_password = "your_password"; // Replace with actual value
$db_name = "cp_username_DB_name"; // Replace with actual DB name
$table = "_Form_nr_X"; // Replace with actual table that ABVFP has created

$db = mysql_connect($db_host, $db_user, $db_password);
if ($db == FALSE){
$error = "Could not connect to the Database Server. Please check user details !";
exit($error);
}
mysql_select_db($db_name, $db);
if (!mysql_select_db($db_name, $db)) {
$error = "Could not select Database. Please check user details !";
exit($error);
}
// Get all the data from the "example" table
$result = mysql_query("SELECT * FROM $table")
if (!result){
$error = "Could not get records from DB. Error:".mysql_error();
exit($error);
}
echo "<table border='1'>";
echo "<tr> <th>Name</th> <th>Age</th> </tr>";
// 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['name'];
echo "</td><td>";
echo $row['age'];
echo "</td></tr>";
}

echo "</table>";
?>



It needs to be placed in the Start of page .

Of course, it will NOT work until the details are correct (and not admin, admin1, test etc which i suppose are just virtual (example) values.

[dannysheps@hotmail.com]

Re: PHP and MySQL problem

hi
thanks , now i have another problem....if i just add the php code to an empty page (like test.php) it works fine
but when im trying to add it to an already made page that i have that i want to add the table to (in the middle) it doesnt do anything.

any suggetion ?

10x

[navaldesign]

Re: PHP and MySQL problem

Pste the entire code in a html box. Make it as large as you expect the table to be. Place it wherever you like in your page. Make sure that your page is published as php!

[dannysheps@hotmail.com]

Re: PHP and MySQL problem

10x again

[dannysheps@hotmail.com]

Re: PHP and MySQL problem

one more questionin my php page i have 3 more boxes one is Flash and 2 are js , and when everything is togther the php table doesnt work

when i remove them it is working

any one ?

10x

[navaldesign]

Re: PHP and MySQL problem

It is hard to help you at this point. There should be no effect on each other, unless you have somehow mixed the parts. One should see the code and debug

[dannysheps@hotmail.com]

Re: PHP and MySQL problem

thanks alot for your paiteint...
i just removed one of the part and it all works good

thanks