How to display data from a database - php

Question

I have successfully created a PHP script which connects to my database. When the user types in their email address I want to display some information on the same page from the database relating to that email address. The database fields that I want to display are: name, tracking, status.
My question is how can I display this information. I know I have code at the bottom which redirects to a page if login successful which iI no longer want.
PHP Script:
<?php
$host="######"; // Host name
$username="######"; // Mysql username
$password="######"; // Mysql password
$db_name="######"; // Database name
$tbl_name="orders"; // Table name
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
// username and password sent from form
$myusername=$_POST['myusername'];
$myusername = stripslashes($myusername);
$myusername = mysql_real_escape_string($myusername);
$sql="SELECT * FROM $tbl_name WHERE username='$myusername'";
$result=mysql_query($sql);
$count=mysql_num_rows($result);
if($count==1){
// Register $myusername, $mypassword and redirect to file "login_success.php"
session_register("myusername");
header("location:trackyourorder.html"); //DO NOT WANT TO REDIRECT
}
else {
echo "Wrong Username";
}
?>

Answer 1

You can use the fetch_row function in combination with a while loop. This is the easiest way. Read more about the function here: http://www.php.net/manual/en/mysqli-result.fetch-row.php
For example:
if ($result = mysqli_query($link, $query)) {
/* fetch associative array and print result*/
while ($row = mysqli_fetch_row($result)) {
printf ("%s (%s)\n", $row[0], $row[1]);
}

Answer 2

If I understand your question correctly, you and asking how to output data from the database directly to the screen. Without offering any formatting, this might be what you are looking for.
$sql="SELECT * FROM $tbl_name WHERE username='$myusername'";
$result=mysql_query($sql);
$results_array=mysql_fetch_array($result);
Once you have the $results_array, you can access any fetched columns from the database using normal keyed array syntax, such as:
echo $results_array["username"];
Hopefully that helps.

Answer 3

As an example:
$result = mysql_query($sql);
if(! $result) die("Error executing query");
while($row = mysql_fetch_assoc($result) {
echo $row["name"];
echo $row["tracking"];
echo $row["status"];
}
Or, you can do it the object oriented way:
$result = mysql_query($sql);
if(! $result) die("Error executing query");
while($row = mysql_fetch_object($result) {
echo $row->name;
echo $row->tracking;
echo $row->status;
}
On a related note, if the only fields you want are name, tracking, and status, you should only select those fields in your query. It will make the query faster, as well as prevent data from being fetched that you don't want (which can be more secure)
"SELECT name, tracking, status FROM $tbl_name WHERE username='$myusername'";

Answer 4

A traditional way from start to finish could be as follows.
<?php
$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Persons");
while($row = mysqli_fetch_array($result))
{
echo $row['FirstName'] . " " . $row['LastName']; //**Put all your formatting in here!**
echo "<br />";
}
mysqli_close($con);
?>