Jump to content

Archived

This topic is now archived and is closed to further replies.

Ameslee

Integration into website

Recommended Posts

Hope someone can help me, im new at PHP and to tell you the truth this is my very first time at integration into a website. Can someone tell me how i display stuff from my database to my website pages? I just want records from my database to be displayed. HELP please!

Thanks

Share this post


Link to post
Share on other sites
there are several ways to display data from mysql database

here's the way i do it

[code]

<?php
     $hostname = localhost;
     $username = username;
     $password = password;

     $db = mysql_connect($hostname, $username, $password) or die(mysql_error());
     $connection  = mysql_select_db("mydb", $db);

     $sql = "SELECT fname, lname FROM table_name";
     $result = mysql_query($sql, $connection);

     while ($rows = mysql_fetch_array($result))
     {
       $fname = $rows['fname'];
       $lname = $rows['lname'];

       echo $fname;
       echo $lname;
     }
?>
[/code]


i'm havent tested it..... i probably missed a semi colon somewhere .. but it should be alrite .... and a simple google search will give u lots and lots of info on using php/mysql :)

Share this post


Link to post
Share on other sites
Can't you use a database include, like include("database.inc");?
or do you use the kind of way like below?

[!--quoteo(post=365806:date=Apr 18 2006, 11:21 AM:name=devofash)--][div class=\'quotetop\']QUOTE(devofash @ Apr 18 2006, 11:21 AM) [snapback]365806[/snapback][/div][div class=\'quotemain\'][!--quotec--]
there are several ways to display data from mysql database

here's the way i do it

[code]

<?php
     $hostname = localhost;
     $username = username;
     $password = password;

     $db = mysql_connect($hostname, $username, $password) or die(mysql_error());
     $connection  = mysql_select_db("mydb", $db);

     $sql = "SELECT fname, lname FROM table_name";
     $result = mysql_query($sql, $connection);

     while ($rows = mysql_fetch_array($result))
     {
       $fname = $rows['fname'];
       $lname = $rows['lname'];

       echo $fname;
       echo $lname;
     }
?>
[/code]
i'm havent tested it..... i probably missed a semi colon somewhere .. but it should be alrite .... and a simple google search will give u lots and lots of info on using php/mysql :)
[/quote]

Share this post


Link to post
Share on other sites
And also, when someone views the source are they going to be able to get the php code?

Share this post


Link to post
Share on other sites
include doesn't necessarily display the data, it just, like it's name implies, includes its info into the current document

and php is server code, it is parsed before it ever reaches the user, they can only see the output
[code]
<?php
echo "<h1>Hello<h1>";
?>

//displays in web source code
<h1>Hello<h1>
[/code]

Share this post


Link to post
Share on other sites
I can't get the above code to work, i have changed it to suit my purposes, here it is:

<?php
$hostname = localhost;
$username = *******_***;
$password = ********;

$conn = mysql_connect($hostname, $username, $password) or die(mysql_error());
$connection = mysql_select_db($database_conn, $conn);


$query = "SELECT name, age, gifts, about FROM Stories";
$mysql_result=mysql_query($query,$conn);

while ($rows = mysql_fetch_array($mysql_result))
{
$name = $rows['name'];
$age = $rows['age'];
$gifts = $rows['gifts'];
$about = $rows['about'];

<table>
<tr>
<td><b>Name</b><td>echo $name;
<tr><td><b>Age</b><td>echo $age;
<tr><td><b>Gifts and Talents</b><td>echo $gifts;
<tr><td><b>More about $name</b><td>echo $about;</tr>
</table>
}
?>

Share this post


Link to post
Share on other sites
ERROR MESSAGE:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/tafeamy/public_html/project/Klever Kids/stories.php on line 106

please help!

Thanks

Share this post


Link to post
Share on other sites
i have also changed the code to this:

?php

$connect = mysql_connect("localhost","*******_***","********");

$db = mysql_select_db("Stories",$connect);

$query = "SELECT * FROM Stories";

$mysql_result = mysql_query($query);

$display_array = mysql_fetch_array($mysql_result);

echo $display_array["name"];
echo $display_array["age"];
echo $display_array["gifts"];
echo $display_array["about"];

?>
and still get this error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/tafeamy/public_html/project/Klever Kids/stories.php on line 105

Share this post


Link to post
Share on other sites
Try This:

[code]<?php
$hostname = localhost;
$username = *******_***;
$password = ********;

$conn = mysql_connect($hostname, $username, $password) or die(mysql_error());
$connection = mysql_select_db($database_conn, $conn);


$query = "SELECT name, age, gifts, about FROM Stories";
$mysql_result=mysql_query($query,$conn);

echo("<table>");

while ($rows = mysql_fetch_array($mysql_result)){
echo("
<tr>
<td><b>Name</b><td>" . $rows['name'] . "
<tr><td><b>Age</b><td>" . $rows['age'] . "
<tr><td><b>Gifts and Talents</b><td>" . $rows['gifts'] . "
<tr><td><b>More about " . $rows['name'] . "</b><td>" . $rows['about'] . "</tr>");
}


echo("</table>");
?>[/code]

you cant output raw html within the <?php ?> tags, you need to echo them, also advanced warning: make sure to \ any " or ' within echos like this:

[code]<?php
echo("Foo \"Bar\"");
?>[/code]

to avoid any parse errors - sorry if the code is kinda messey, i was in a rush :)

Share this post


Link to post
Share on other sites
Thanks alot, this does work, fine.

[!--quoteo(post=365895:date=Apr 18 2006, 03:30 PM:name=Hooker)--][div class=\'quotetop\']QUOTE(Hooker @ Apr 18 2006, 03:30 PM) [snapback]365895[/snapback][/div][div class=\'quotemain\'][!--quotec--]
Try This:

[code]<?php
$hostname = localhost;
$username = *******_***;
$password = ********;

$conn = mysql_connect($hostname, $username, $password) or die(mysql_error());
$connection = mysql_select_db($database_conn, $conn);
$query = "SELECT name, age, gifts, about FROM Stories";
$mysql_result=mysql_query($query,$conn);

echo("<table>");

while ($rows = mysql_fetch_array($mysql_result)){
echo("
<tr>
<td><b>Name</b><td>" . $rows['name'] . "
<tr><td><b>Age</b><td>" . $rows['age'] . "
<tr><td><b>Gifts and Talents</b><td>" . $rows['gifts'] . "
<tr><td><b>More about " . $rows['name'] . "</b><td>" . $rows['about'] . "</tr>");
}
echo("</table>");
?>[/code]

you cant output raw html within the <?php ?> tags, you need to echo them, also advanced warning: make sure to \ any " or ' within echos like this:

[code]<?php
echo("Foo \"Bar\"");
?>[/code]

to avoid any parse errors - sorry if the code is kinda messey, i was in a rush :)
[/quote]

Share this post


Link to post
Share on other sites

×

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.