从php中的mysql表中选择count(*)

Question

我能够获得mysql查询结果的值和行.

但我正在努力获得查询的单个输出.例如:

$result = mysql_query("SELECT COUNT(*) FROM Students;");

我需要显示结果.但我没有得到结果.

我尝试过以下方法:

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

但我没有成功显示(获得)实际值.

Answer 1

您需要使用as关键字对聚合进行别名以便从中调用它mysql_fetch_assoc

$result=mysql_query("SELECT count(*) as total from Students");
$data=mysql_fetch_assoc($result);
echo $data['total'];

Answer 2

如果您只需要值:

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

Answer 3

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

试试这个代码.

Answer 4

请开始使用PDO.

mysql_*自PHP 5.5.0起已弃用,将在7中完全删除.让我们更容易升级并立即开始使用它.

$dbh = new \PDO($dsn, $user, $password);
$sth = $dbh->prepare('SELECT count(*) as total from Students');
$sth->execute();
print_r($sth->fetchAll());

Answer 5

$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 6

这是使用 PHP 显示表中没有行的代码

$sql="select count(*) as total from student_table";
$result=mysqli_query($con,$sql);
$data=mysqli_fetch_assoc($result);
echo $data['total'];