从php中的mysql表中选择count（*）

我能够同时获取mysql查询结果的值和行。

但是我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。但是我没有得到结果。

我尝试了以下方法：

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

但是我没有成功显示（获取）实际值。

您需要使用as关键字为聚合设置别名，以便从中进行调用mysql_fetch_assoc

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

如果只需要该值：

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

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

试试这个代码。

请开始使用PDO。

从PHP 5.5.0开始不推荐使用mysql_ *，它将在7中完全删除。让我们现在更轻松地升级和开始使用它。

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

$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;

这是使用PHP在表格中不显示行的代码

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

对于mysqli用户，代码将如下所示：

$mysqli = new mysqli($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']);

要么：

$mysqli = new mysqli($db_host, $db_user, $db_pass, $db_name);

$result = $mysqli->query("SELECT COUNT(*) FROM Students")->fetch_array();
var_dump($result[0]);

 $howmanyuser_query=$conn->query('SELECT COUNT(uno)  FROM userentry;');
 $howmanyuser=$howmanyuser_query->fetch_array(MYSQLI_NUM); 
 echo $howmanyuser[0];

经过这么多小时的出色:)

$abc="SELECT count(*) as c FROM output WHERE question1=4";
$result=mysqli_query($conn,$abc);
if($result)
 {
    while($row=mysqli_fetch_assoc($result))
  {
        echo $row['c'];
  }     
 }

在此计数它在question1栏中出现的次数，其工作完全

您也可以使用它并升级到mysqli_（停止使用mysql_*扩展程序...）

$result = mysqli_query($conn, "SELECT COUNT(*) AS `count` FROM `Students`");
$row = mysqli_fetch_array($result);
$count = $row['count'];
echo'$count';

使用mysql v5.7.20，这就是我如何使用PHP v7.0.22从表中获取行数的方法：

$query = "select count(*) from bigtable";
$qresult = mysqli_query($this->conn, $query);
$row = mysqli_fetch_assoc($qresult);
$count = $row["count(*)"];
echo $count;

第三行将返回如下结构：

array(1) {
   ["count(*)"]=>string(4) "1570"
}

在这种情况下，结尾的echo语句将返回：

1570

我认为您的代码中有错别字，您应该删除倒数第二个分号：

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

您需要使用as关键字为聚合设置别名，以便从中进行调用mysqli_fetch_assoc

$result=mysqli_query($conn,"SELECT count(*) as total from Students");
$data=mysqli_fetch_assoc($result);
echo $data['total'];

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

结果：

我认为这是更好的答案。

$query = "SELECT count(*) AS total FROM table_name"; 
mysql_select_db('database_name');
$result = mysql_query($query); 
$values = mysql_fetch_assoc($result); 
$num_rows = $values['total']; 
echo $num_rows;

$qry_appr = "SELECT COUNT(*) FROM comments WHERE admin_panel_id ='$id' AND status = 'ON'";
$qry_data = mysqli_query($con, $qry_appr);
$approve_count = mysqli_fetch_array($qry_data);
$toatalCount = array_shift($approve_count);
echo $toatalCount;

这也可以，但是可以通过移动获取数组来返回0索引值。欢迎大家

在选项范围之外，没有提到这一选项。即使它不是最佳的或不必要的，也可以通过一个简单的循环来完成：

$result=mysqli_query($conn, "SELECT count(1) as total from Students");
foreach ($result as $rez) {$numb=$rez['total'];}
echo $numb;

如果只希望计数值，可以速记：

$cnt = mysql_num_rows(mysql_query('select * from students'));