CODEWITHSUNDEEP
phpsql
Adarsh
Adarsh

Reputation: 33

Output mysql query result in a php variable

I have this variable ($sql) in PHP:

$sql = "SELECT DATE_FORMAT(FROM_DAYS(DATEDIFF(CURRENT_DATE, dob)),'%y Years %m Months %d Days') AS age WHERE id=$row[id]";

How to output the produced data as a PHP variable?

The query is taking the date of birth from my table and calculating the age from today's date.

How to echo the age in PHP?

Upvotes: 2

Views: 4276

Answers (3)

Kingston
Kingston

Reputation: 619

You can try as below: 

$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

$sql = "SELECT DATE_FORMAT(FROM_DAYS(DATEDIFF(CURRENT_DATE, dob)),'%y Years %m Months %d Days') AS age WHERE id=$row[id]";

$result = $conn->query($sql);

if ($result->num_rows > 0) {
    while($row = $result->fetch_assoc()) {
    echo "age: " . $row["age"]. "<br>";}
} 
else {
echo "No result!";
}

You can reference from http://www.w3schools.com

Upvotes: 0

Adhan Timothy Younes
Adhan Timothy Younes

Reputation: 570

Just try this:

$con = mysqli_connect("localhost","user","pass", "database_name"); //your connection
$sql = "SELECT DATE_FORMAT(FROM_DAYS(DATEDIFF(CURRENT_DATE, dob)),'%y Years %m Months %d Days') AS age FROM table_name WHERE id=".$row['id'];    
$query = mysqli_query($con, $sql);
$result = mysqli_fetch_assoc($query);
echo $result['age'];

Don't forget to replace table_name.

Upvotes: 4

Ehsan Ilahi
Ehsan Ilahi

Reputation: 298

try this code

i see your code you miss table name 

$sql = "SELECT DATE_FORMAT(FROM_DAYS(DATEDIFF(CURRENT_DATE, dob)),'%y Years %m Months %d Days') AS age WHERE id=$row[id]";

 $result = mysql_query($sql);
    while($query_data = mysql_fetch_row($result))
    $age= $query_data[0];
    print $age;

Upvotes: 0

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