How to use PHP to send data over to another page? [closed] - php

Question

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 8 years ago.
Improve this question
I have a website which is supposed to show some information about food courts from MySQL server and what I have done is that I made it such that the data would be shown in the website and if they click they would be redirected to another web page. My question is how would I be able to check what is the hyperlink they clicked so that I could display accurate information on the second webpage?
What I have done is that I tried to use get to send the information over but to no avail. Can someone guide me please? I am new to PHP and MySQL coding.
This is the PHP code for the website that I am using to display the food court information.
<?php
require 'dbFunction.php';
$con = getDbConnect();
if (!mysqli_connect_errno($con)) {
$result = mysqli_query($con,"SELECT * FROM foodcourt");
while($row = mysqli_fetch_array($result)) {
echo "Foodcourt " . $row['FCNo'] . "<br />";
echo $row['FCDescription'] . "<br />";
echo $row['FCImage'] . "<br />";
echo "<br>";
$data=$row['FCNo'];
}
} else {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_close($con);
?>
And this is the code I use to "receive" the information.
<?php
require 'dbFunction.php';
$con = getDbConnect();
$data = $_GET['$row[FCNo]'];
if (!mysqli_connect_errno($con)) {
$result = mysqli_query($con,
"SELECT * "
. "FROM StallDetails SD AND FoodDetails FD"
. "WHERE FD.StallID = SD.StallID "
. "AND FD.FCNo = ".$data);
while($row = mysqli_fetch_array($result)) {
echo "Foodcourt " . $row['FCNo'] . "<br />";
echo $row['Details']."<br />";
echo $row['Description'].",<br> ";
echo "Price: $".$row['Price']."<br> ";
} }
else {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_close($con);
This is my getDbConnect();
<?php
function getDbConnect() {
// get a database connect to studentacad database
$con = mysqli_connect("localhost:3306", "waduser01", "password123", "p1315774");
Return $con;
}
?>

Answer 1

You are doing it wrong. Your are using $row in get variable too and You are also not enclosing the $data in quotes ''.And one more thing you are not selecting the database. Use mysqli_select_db function to select your database.You are also using AND in selecting the table. use the code below
<?php
require 'dbFunction.php';
$con = getDbConnect();
mysqli_select_db($con,"db_name");
$data = $_GET['fc'];
if (!mysqli_connect_errno($con)) {
$result = mysqli_query($con,
"SELECT * "
. "FROM StallDetails SD , FoodDetails FD"
. "WHERE FD.StallID = SD.StallID "
. "AND FD.FCNo = '".$data."'");
while($row = mysqli_fetch_array($con,$result)) {
echo "Foodcourt " . $row['FCNo'] . "<br />";
echo $row['Details']."<br />";
echo $row['Description'].",<br> ";
echo "Price: $".$row['Price']."<br> ";
} }
else {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_close($con);
Hope this helps you

Answer 2

You don't need to send any data.
Usually, GET variables are used. Example:
Wayne
Would lead to a file called user.php, in which the GET variable id is checked:
$id = isset($_GET['id']) ? $_GET['id'] : false;
In this example, once you have the id, you can decide on what user details to display:
$userDetails = $user->get($id);
Another example:
Wayne
... translates into:
show=recipe
id=34

Answer 3

You are referencing your variable like this:
$_GET['$row[FCNo]'];
But it looks like the variable in the url is fc, so you would access it like this:
$_GET['fc'];