mysql_num_rows返回null每次

Question

if (isset($_GET["val"])) { 
    $val = $_GET["val"]; 
    $sql1 = "select count(*) as count from staff_log l where l.time_in is null and l.time_out is not null and l.date_today = curdate() and l.staff_id = ".$val.";"; 

    $result1 = mysql_num_rows($sql1); 
} 

Answer 1

加入这一行...

$result = mysql_query($sql1); 
$result1 = mysql_num_rows($result); 

必须在mysql_query()功能定义查询;

Answer 2

mysql_num_rows()返回，如在PHP manual所示：

行中一个结果集上或FALSE的成功失败的次数。

这意味着执行它时出错，这显然是因为您没有运行任何查询（mysql_query()）。

注意，PHP手册也有在页面的顶部以下警告：

这个扩展被废弃在PHP 5.5.0，它是PHP 7.0.0中删除。相反，应该使用MySQLi或PDO_MySQL扩展。

Answer 3

首先执行查询，然后NUM_ROWS

原来的MySQL扩展现在已经过时，并连接到数据库时，会产生E_DEPRECATED错误。而是使用MYSQLi或PDO_MySQL扩展。

试试这个

if (isset($_GET["val"])) 
{ 
$val = $_GET["val"]; 
$sql1 = "select count(*) as count from staff_log l where l.time_in is null and l.time_out is not null and l.date_today = curdate() and l.staff_id = ".$val.";"; 

$result = mysql_query($sql1); 
$count = mysql_num_rows($result); 

} 

Answer 4

你几乎权利，请查看下面的代码。您需要使用mysql_query（），mysql_query（）用于执行默认数据库的查询。

if (isset($_GET["val"])) 
{ 
    $val = $_GET["val"]; 
    $sql1 = "select count(*) as count from staff_log l where l.time_in is null and l.time_out is not null and l.date_today = curdate() and l.staff_id ='$val'"; 
    $sql2 = mysql_query($sql1); // Execute the query first 
    $result1 = mysql_num_rows($sql2); // Get the number of rows from executed query result. 
    echo "The count is $result1"; 
}