select count(*) from table of mysql in php

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:

  1. mysql_fetch_assoc()
  2. mysql_free_result()
  3. mysql_fetch_row()

But I didn’t succeed to display (get) the actual value.

Answers:

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Method 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'];

Method 2

If you only need the value:

$result = mysql_query("SELECT count(*) from Students;");
echo mysql_result($result, 0);

Method 3

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

Try this code.

Method 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());

Method 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'];

Method 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;

Method 7

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

Method 8

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]);

Method 9

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;

Method 10

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'];

Method 11

$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:

select count(*) from table of mysql in php

Method 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 🙂


All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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