Question

我是PHP的初学者。当我尝试做mysql_num_rows时，它给出了一个错误：

警告：mysql_num_rows（）期望参数1是资源，对象在第22行的D：\ xampp \ htdocs \ kcp \ police \ viewcase.php中给出

这是代码：

<?php
session_start();
if(empty($_SESSION['pusername'])||empty($_SESSION['pid']))
{
    header('location://localhost/kcp/police_login.html');
    exit();

}
else
{
    $station=$_SESSION['station'];
    $con = mysqli_connect('localhost','root','','kcp');

    if(!$con)
    {
        echo 'NOT CONNECTED TO DB';
    }
    $sql="SELECT * FROM complaint WHERE station='$station'";
    mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
    $res=mysqli_query($con,$sql) or die("llllllll");

    $numrow=mysql_num_rows($res);
    echo $numrow;
}

?>

Answer 1

你不能混淆mysql和mysqli。

$numrow = mysqli_num_rows($res);
echo $numrow;

mysqli_num_rows() Reference

Answer 2

使用mysqli_num_rows而不是mysql_num_rows

mysql_num_rows不起作用

2 个答案: