Calculating Age Sql query - php

Question

good day guys i have my code here to get age from date of birth stored in my database.
$connect=mysql_connect("localhost","root","");
mysql_select_db("mydb",$connect);
$query = "select date_format(now(), '%Y') - date_format(dob, '%Y') - (date_format(now(), '00-%m-%d') < date_format(dob, '00-%m-%d'))
as age from data";
$result = mysql_query($query);
$values = mysql_fetch_assoc($result);
$age = $values['age'];
$query="select * from data";
$result=mysql_query($query);
if(mysql_num_rows($result)>0){
echo "<table align='center' border='1'>";
echo "<tr>";
echo "<th>id</th>";
echo "<th>Last Name</th>";
echo "<th>First Name</th>";
echo "<th>Age</th>";
echo "</tr>";
while($row=mysql_fetch_array($result)){
echo "<tr>";
echo "<td>".$row['id']."</td>";
echo "<td>".$row['lname']."</td>";
echo "<td>".$row['fname']."</td>";
echo "<td>".$age."</td>";
echo "</tr>";
}
echo "</table>";
}
all data is fetch but the only problem is the age of the first row is the same as the rest of my data. what wrong with my code? anybody help please thank you.

Answer 1

You are printing pre-fetched age from values array.
Instead read from current row and use to display.
Change your SQL query as below:
$query = "select id, lname, fname,
date_format(now(), '%Y') - date_format(dob, '%Y')
- ( date_format(now(), '00-%m-%d')
< date_format(dob, '00-%m-%d'))
as age
from data";
Change:
echo "<td>".$age."</td>";
To:
echo "<td>".$row['age']."</td>";

Answer 2

The problem is you are fetching the age for single time, get the dob in while loop and then calculate the age and print the age in while loop itself.

Answer 3

You are getting age once and using it for rest of your data.
Try something like this:
<?php
$connect = mysql_connect("localhost", "root", "");
mysql_select_db("mydb", $connect);
$query = "select * from data";
$result = mysql_query($query);
if (mysql_num_rows($result) > 0) {
echo "<table align='center' border='1'>";
echo "<tr>";
echo "<th>id</th>";
echo "<th>Last Name</th>";
echo "<th>First Name</th>";
echo "<th>Age</th>";
echo "</tr>";
while ($row = mysql_fetch_array($result)) {
$query1 = "select date_format(now(), '%Y') - date_format(dob, '%Y') - (date_format(now(), '00-%m-%d') < date_format(dob, '00-%m-%d'))
as age from data where id = $row[id]";
$result1 = mysql_query($query);
$values1 = mysql_fetch_assoc($result);
$age = $values1['age'];
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['lname'] . "</td>";
echo "<td>" . $row['fname'] . "</td>";
echo "<td>" . $age . "</td>";
echo "</tr>";
}
echo "</table>";
}

Answer 4

Considering your age formula is right, try this query.
// query will give all records and age for each record.
$query="SELECT a.*, (SELECT TIMESTAMPDIFF(YEAR,dob,NOW()) FROM data AS b WHERE a.id = b.id) AS age FROM data AS a";
// rather than getting two result sets, you have age in the same record now, access it as a field.
$result=mysql_query($query);
if(mysql_num_rows($result)>0){
echo "<table align='center' border='1'>";
echo "<tr>";
echo "<th>id</th>";
echo "<th>Last Name</th>";
echo "<th>First Name</th>";
echo "<th>Age</th>";
echo "</tr>";
while($row=mysql_fetch_array($result)){
echo "<tr>";
echo "<td>".$row['id']."</td>";
echo "<td>".$row['lname']."</td>";
echo "<td>".$row['fname']."</td>";
echo "<td>".$row['age']."</td>"; // get age from db results
echo "</tr>";
}
echo "</table>";
}