Display mysql query results on a new page from links - php

Question

I create a table and display entries from MySQL in a brief table report format. Each of the entries in the table have further properties that I would like to display on a separate page when I click on the link in their names.
<?php
require_once 'includes/db.php';
connect();
$query = "SELECT * FROM persons order by date_entered DESC LIMIT 5";
$response = mysqli_query($con, $query);
if($response){
echo '<div id="frm"> <h2> Last 5 candidates entered </h2> <table align="left">
<tr>
<td align="left"><b>First Name</b></td>
<td align="left"><b>Email</b></td>
<td align="left"><b>Sex</b></td>
<td align="left"><b>City</b></td>
<td align="left"><b>Phone Number</b></td>
<td align="left"><b>Education</b></td>
<td align="left"><b>Salary</b></td>
</tr>';
while($row = mysqli_fetch_array($response)){
echo '<tr>
<td align="left">' . '' . $row["first_name"] . " " .$row['last_name'] . '' . '</td>
<td align="left">' . $row['email'] . '</td>
<td align="left">' . $row['sex'] . '</td>
<td align="left">' . $row['city'] . '</td>
<td align="left">' . $row['phone'] . '</td>
<td align="left">' . $row['education'] . '</td>
<td align="left">' . $row['salary'] . '</td>';
echo '</tr>';
}
echo '</table>';
} else {
echo "Couldn't issue database query<br />";
echo mysqli_error($con);
}
mysqli_close($con);
?>
This gives me a table report with links in the names:
brief report with link in names
My question is: if I click Mary Swanson in the table, what variables do I use in person.php so I can retrieve the MySQL data for that record?
Thanks.

Answer 1

You could pass this id trough the link as a GET variable and use it on the other side to query the database for that specific record like this:
<?php
require_once 'includes/db.php';
connect();
$query = "SELECT * FROM persons order by date_entered DESC LIMIT 5";
$response = mysqli_query($con, $query);
if($response){
echo '<div id="frm"> <h2> Last 5 candidates entered </h2> <table align="left">
<tr>
<td align="left"><b>First Name</b></td>
<td align="left"><b>Email</b></td>
<td align="left"><b>Sex</b></td>
<td align="left"><b>City</b></td>
<td align="left"><b>Phone Number</b></td>
<td align="left"><b>Education</b></td>
<td align="left"><b>Salary</b></td>
</tr>';
while($row = mysqli_fetch_array($response)){
echo '<tr>
<td align="left">' . '' . $row["first_name"] . " " .$row['last_name'] . '' . '</td>
<td align="left">' . $row['email'] . '</td>
<td align="left">' . $row['sex'] . '</td>
<td align="left">' . $row['city'] . '</td>
<td align="left">' . $row['phone'] . '</td>
<td align="left">' . $row['education'] . '</td>
<td align="left">' . $row['salary'] . '</td>';
echo '</tr>';
}
echo '</table>';
} else {
echo "Couldn't issue database query<br />";
echo mysqli_error($con);
}
mysqli_close($con);
?>
Then you can access the variable on the other side like this $_GET["id"])
<?php
require_once 'includes/db.php';
connect();
$query = "SELECT * FROM persons where id = " . $_GET["id"]);
$response = mysqli_query($con, $query)
After this you can print the values returned by the query as if where a multiple row select, but it will only come back with one row, you still need to do the while since what it returns is an array.