Retrieving information

QuickStilz

How do I get the information of a certain item from a database returned as PHP array structure form? ThankS!

For example, if you sign on PHPBuilder, I would be able to retrieve your information based on your username.

Roger_Ramjet

Here's how:

<?php // db.php 

$dbhost = "localhost"; 
$dbuser = "username"; 
$dbpass = "password";

function dbConnect($db="") { 
   global $dbhost, $dbuser, $dbpass; 

   $dbcnx = @mysql_connect($dbhost, $dbuser, $dbpass) 
       or die("The site database appears to be down."); 

   if ($db!="" and !@mysql_select_db($db)) 
       die("The site database is unavailable."); 

   return $dbcnx; 
} 
?>

<?php		//index.php
//disable all browser caching MUST BE FIRST LINES WITH NO PRECEEDING SPACES ETC
// Date in the past
header("Expires: Mon, 26 Jul 1997 05:00:00 GMT");

// always modified
header("Last-Modified: " . gmdate("D, d M Y H:i:s") . " GMT");

// HTTP/1.1
header("Cache-Control: no-store, no-cache, must-revalidate");
header("Cache-Control: post-check=0, pre-check=0", false);

// HTTP/1.0
header("Pragma: no-cache");

// include neccessary files
	require_once './db.php';
	require_once './common.php';
	require_once './accesscontrol.php';
?>
<!DOCTYPE html PUBLIC "-//W3C/DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
//add anything you want in the head of the page
</head>
<body>

<?  

	dbConnect("database");
//LOG OUT button clicked	
	if (isset($_REQUEST['log_out'])) {  

  		$_SESSION = array();
  		session_destroy;
  		echo '<script language="javascript">window.location = "http://www.yourdomain.co.uk/goodbye.php/"</script>';
  	}

//--- SET QUERY TO RUN ---
if (isset($_REQUEST['go_search'])) {
	$field = $_REQUEST['field'];
	$value = $_REQUEST['value']
	$sql = "SELECT * FROM tbl WHERE " . $field . "= '" . $value . "' ORDER BY " . $_REQUEST['order'];
}
//--- END SET QUERY ---

//--- RUN QUERY ---
	$result = mysql_query($sql) or die($sql . '<br />' . mysql_error());
// query failed 
	if (!$result) { 
		error('query failed my friend');
		exit;
		} 	
//--- END RUN QUERY ---	

//--- COUNT ROWS AND BRANCH ---
  $count = mysql_num_rows($result);

//--- NO RESULT RETURNED ---
	if ($count==0) {
		echo $field . ' = ' . $value . ' not found.';
	}
//--- END NO RESULT ---

//--- SINGLE ROW RETURNED ---
	if ($count==1) {
		$row = mysql_fetch_array($result);
// display account  		
  		$addr = addr_line($row['addr1'],$row['addr2'],$row['addr3'],$row['addr4'],$row['addr5'],$row['post']);
  		echo '<p>track no: ' . $track . '<br />';
  		echo 'account: ' . $row['ref'] . '<br />';
  		echo 'name: ' . $row['name'] . '<br />';
  		echo 'amount: ' . $row['curr'] . $row['amt'] . '<br />';
  		echo 'balance due: ' . $row['curr'] . $row['amt_due'] . '<br />';
  		echo 'status: ' . $row['code'] . ' ' . $row['status'] . '<br />';
  		echo 'address: ' . $addr . '</p>';
//etc
	}
//--- END SINGLE ROW ---

//--- MULTIPLE ROWS RETURNED ---
// query returned more than 1 debt so display them for user to select one 	
	else {
		echo '<div id="multi">';
		echo '<h5>' . $debt_count . ' debts found </h5><h6>[click on a debt to view details]</h6>';
  	echo '<table border="1" cellspacing="0" cellpadding="3">';
  	echo '<tr><th>track</th><th>' . $client . ' ref</th><th>name</th><th>amount</th><th>balance</th><th>status</th></tr>';
		// Iterate through the results
		while ($row = mysql_fetch_array($result)) {
			$link = '<a href="index.php?go_search=1&field=track&value=' . $row['track'] . '">';
  			echo '<tr><td>' . $link . $row['track'] . '</a></td>';
  			echo '<td>' . $link . $row['ref'] . '</a></td>';
  			echo '<td>' . $link . $row['name'] . '</a></td>';
  			echo '<td align="right">' . $link . $row['curr'] . $row['amt'] . '</a></td>';
  			echo '<td align="right">' . $link . $row['curr'] . $row['amt_due'] . '</a></td>';
  			echo '<td title="' . $row['status'] . '">' . $link . $row['code'] . '</a></td></tr>';
 		}
  	echo '</table>';
  	echo '</div>';	
	}
//--- END MULTIPLE ROWS

?>

intenz

Roger Ramjet, too complex for him. Remember he's posting on "Newbies"

QuickStilz, here's how...

<?PHP
$query = mysql_query( "SELECT * FROM phpbuilder_users WHERE username = 'intenz'" );
$array = mysql_fetch_array( $query );
print_r( $array );
?>

QuickStilz

First off, I would like to thank you for the quick reply.

In this line:
$query = mysql_query( "SELECT * FROM phpbuilder_users WHERE username = 'intenz'" );

How would I replace the username 'intenz' with something that the user input (and how do I set up so that the user can input)? Moreover, how do I know if the username is 'intenz'? What if it's some other username? Is there a way for the system to identify whoever is logging in? ThankS!

Roger_Ramjet

You first need to set up an html Form with a text input where the user submits the name

See this for Form syntax: http://www.htmlhelp.com/reference/html40/forms/form.html

You then substitute the value submitted as a variable in the query string

$user_in = $_POST['usr'];

$query = mysql_query( "SELECT * FROM phpbuilder_users WHERE username = ' " . $user_in . " ' " );

Start with the input form, post here for help with it.

Here is the input form I use

<p><form method="post" action="<?=$_SERVER['PHP_SELF']?>">
    User: <input type="text" name="usr" size="10" /><br />
    Password: <input type="password" name="pwd" SIZE="10" /><br />
    <input type="submit" value="Log in" />
  	</form></p>

If the user has not yet logged in, my first page offers this and also accepts the input back: that's the <?=$_SERVER['PHP_SELF']?> bit. For the moment, you could try just posting to a seperate page that can then echo the input so you know the form works. Substitute the correct address in the action= part.

<p><form method="post" action="http://your.domain/yoursite/check_login.php">
    User: <input type="text" name="usr" size="10" /><br />
    Password: <input type="password" name="pwd" size="10" /><br />
    <input type="submit" value="Log in" />
  	</form></p>

QuickStilz

How do I loop through an array and insert the data into a table? ThankS!

Roger_Ramjet

Like this

//--- MULTIPLE ROWS RETURNED --- 
// query returned more than 1 debt so display them for user to select one      

    else { 
        echo '<div id="multi">'; 
        echo '<h5>' . $debt_count . ' debts found </h5><h6>[click on a debt to view details]</h6>'; 
      echo '<table border="1" cellspacing="0" cellpadding="3">'; 
      echo '<tr><th>track</th><th>' . $client . ' ref</th><th>name</th><th>amount</th><th>balance</th><th>status</th></tr>'; 
        // Iterate through the results 
        while ($row = mysql_fetch_array($result)) { 
            $link = '<a href="index.php?go_search=1&field=track&value=' . $row['track'] . '">'; 
              echo '<tr><td>' . $link . $row['track'] . '</a></td>'; 
              echo '<td>' . $link . $row['ref'] . '</a></td>'; 
              echo '<td>' . $link . $row['name'] . '</a></td>'; 
              echo '<td align="right">' . $link . $row['curr'] . $row['amt'] . '</a></td>'; 
              echo '<td align="right">' . $link . $row['curr'] . $row['amt_due'] . '</a></td>'; 
              echo '<td title="' . $row['status'] . '">' . $link . $row['code'] . '</a></td></tr>'; 
        } 
      echo '</table>'; 
      echo '</div>';     

    } 
//--- END MULTIPLE ROWS

The while loop creates a new table row for each record and inserts the individual fields into <td> cells.