Echo specific rows from a mysql database

Question

I have created a form that ask the user to add: first_name, last_name,location,status

After some time i have received 5 inputs. The mysql table is named users and the table data are like below:

id first_name location status  
== ========== ======== ========  
1   Chris       UK       Married  
2   Anton       Spain    Single  
3   Jacob       UK       Single  
4   Mike        Greece   Married  
5   George      UK       Married  

Also the way i am receiving the inputs its via POST method. So:

$firstname=$_POST['FIRST_NAME']; // First Name: <input type="text" name="FIRST_NAME">
$location=$_POST['LOCATION']; // Location: <input type="text" name="LOCATION">
$status=$_POST['STATUS']; // Status: <input type="text" name="STATUS">

I created a query to select all users from UK that are married:

$query = "SELECT * FROM users WHERE location='UK' AND status='Married'";
$result = mysqli_query($dbc,$query); //$dbc is the connection to my database

$row = mysqli_fetch_array($results, MYSQLI_BOTH);

In other words:

id first_name location status  
== ========== ======== ========  
1   Chris       UK       Married  
5   George      UK       Married

Questions:

1) Does the $row array look like below:

$row= array(
array(1 Chris UK Married),
array(5 George UK Married) 
);

2) How do i echo the content of the database after implementing the filtering WHERE location='UK' AND status='Married'?

i need it to be like this:

Hello Chris! You are from UK and you are married!

Hello George! You are from UK and you are married!

I know that i have to use foreach loop (echo arrays) but i have tried it and it doesnt work.One of the things i tried was something i found in php.net:

Unpacking nested arrays with list()

(PHP 5 >= 5.5.0)

PHP 5.5 added the ability to iterate over an array of arrays and unpack the nested array into loop variables by providing a list() as the value.

For example:

<?php
$array = [
[1, 2],
[3, 4],
];

foreach ($array as list($a, $b)) {
// $a contains the first element of the nested array,
// and $b contains the second element.
echo "A: $a; B: $b\n";
}
?> 

When i use the above i am receiving the error below:

Parse error: syntax error, unexpected T_LIST in C:\wamp\www....

Any suggestions?

As i understand, i somehow have to link the ID with the other data..

Thanks in advance.

Solution 1

You may use something like this:

$query = "SELECT * FROM users WHERE location='UK' AND status='Married'";
$result = mysqli_query($dbc,$query);

while($row = mysqli_fetch_array($result, MYSQLI_BOTH)){
    printf("Hello %s! You are from %s and you are %s!\n", $row['first_name'], $row['location'],$row['status']);
}

Solution 2

mysqli_fetch_array returns a single row. You'll need to loop to get all rows. You can get the value from the row and place it in a string like:

while($row = mysqli_fetch_array($results, MYSQLI_BOTH)){
    echo "Hi ".$row['first_name'] . " You are from ".$row['location']." and you are ".$row['status']."!";
}