select count(*) from table of mysql in php - php

Question

I am able to get both the value and row of the mysql query result.
But I am struggling to get the single output of a query. e.g.:
$result = mysql_query("SELECT COUNT(*) FROM Students;");
I need the result to display. But I am not getting the result.
I have tried with the following methods:
mysql_fetch_assoc()
mysql_free_result()
mysql_fetch_row()
But I didn't succeed to display (get) the actual value.

Answer 1

You need to alias the aggregate using the as keyword in order to call it from mysql_fetch_assoc
$result=mysql_query("SELECT count(*) as total from Students");
$data=mysql_fetch_assoc($result);
echo $data['total'];

Answer 2

If you only need the value:
$result = mysql_query("SELECT count(*) from Students;");
echo mysql_result($result, 0);

Answer 3

$result = mysql_query("SELECT COUNT(*) AS `count` FROM `Students`");
$row = mysql_fetch_assoc($result);
$count = $row['count'];
Try this code.

Answer 4

Please start using PDO.
mysql_* is deprecated as of PHP 5.5.0 and will be removed entirely in 7. Let's make it easier to upgrade and start using it now.
$dbh = new \PDO($dsn, $user, $password);
$sth = $dbh->prepare('SELECT count(*) as total from Students');
$sth->execute();
print_r($sth->fetchColumn());

Answer 5

here is the code for showing no of rows in the table with PHP
$sql="select count(*) as total from student_table";
$result=mysqli_query($con,$sql);
$data=mysqli_fetch_assoc($result);
echo $data['total'];

Answer 6

$num_result = mysql_query("SELECT count(*) as total_count from Students ") or exit(mysql_error());
$row = mysql_fetch_object($num_result);
echo $row->total_count;

Answer 7

You can as well use this and upgrade to mysqli_ (stop using mysql_* extension...)
$result = mysqli_query($conn, "SELECT COUNT(*) AS `count` FROM `Students`");
$row = mysqli_fetch_array($result);
$count = $row['count'];
echo $count;

Answer 8

With mysql v5.7.20, here is how I was able to get the row count from a table using PHP v7.0.22:
$query = "select count(*) from bigtable";
$qresult = mysqli_query($this->conn, $query);
$row = mysqli_fetch_assoc($qresult);
$count = $row["count(*)"];
echo $count;
The third line will return a structure that looks like this:
array(1) {
["count(*)"]=>string(4) "1570"
}
In which case the ending echo statement will return:
1570

Answer 9

For mysqli users, the code will look like this:
$mysqli = new mysqli($db_host, $db_user, $db_pass, $db_name);
$result = $mysqli->query("SELECT COUNT(*) AS Students_count FROM Students")->fetch_array();
var_dump($result['Students_count']);
or:
$mysqli = new mysqli($db_host, $db_user, $db_pass, $db_name);
$result = $mysqli->query("SELECT COUNT(*) FROM Students")->fetch_array();
var_dump($result[0]);

Answer 10

$db = new PDO('mysql:host=localhost;dbname=java_db', 'root', 'pass');
$Sql = "SELECT count(*) as `total` FROM users";
$stmt = $db->query($Sql);
$stmt->execute();
$total = $stmt->fetch(PDO::FETCH_ASSOC);
print '<pre>';
print_r($total);
print '</pre>';
Result:

Answer 11

You need to alias the aggregate using the as keyword in order to call it from mysqli_fetch_assoc
$result=mysqli_query($conn,"SELECT count(*) as total from Students");
$data=mysqli_fetch_assoc($result);
echo $data['total'];

Answer 12

$howmanyuser_query=$conn->query('SELECT COUNT(uno) FROM userentry;');
$howmanyuser=$howmanyuser_query->fetch_array(MYSQLI_NUM);
echo $howmanyuser[0];
after the so many hours excellent :)