Question

我不太确定我在这里做错了什么，我正在尝试检查行数以查看用户名是否存在。

我目前收到此错误：

Notice: Undefined property: mysqli::$num_rows in C:\xampp\htdocs\xampp\oop\ServerSide-Project-Message-Ninja\register.php on line 24

感谢任何帮助：）

register.php

<?php 

require_once("config/config.php");
require_once("classes/database_class.php");
require_once("classes/html_class.php");
require_once("classes/security_class.php");


$objHTML = new makeHTML($register_form_array);

// $objHTML->viewArray();


echo $objHTML->makeHeader('Register for an account on Message Ninja', $keywords="Register, Signup", $description="");


if ($objHTML->checkData()) {
    $data = $objHTML->getFilteredData();

    $db = new database($dsn);
    echo $data['username'];
    $sql = "SELECT * FROM members WHERE username = '$data[username]'";
    $res = $db->checkIfRecordExists($sql);
    $row_count = $res->num_rows;

    // if ( $row_count > 0) {
    //  echo "User already exists";
    // } else {
    //  echo "This would normally insert data";
    //  // $sql = "INSERT INTO Members VALUES (NULL,'$data[username]', '$data[password]', 
 //  //          '$data[first_name]', '$data[last_name]', '$data[email]' )";
    //  // if ($db->insert($sql)) {
    //  //  header("location: thankyou.php");
    //  // }
    // }
}


 ?>
<div id="form_wrapper">
<?php



echo $objHTML->openForm();
echo $objHTML->makeRegisterFormRows();
echo $objHTML->makeSubmit('Register');

echo $objHTML->closeForm();

?>
</div>
<?php
echo $objHTML->makeFooter();
?>

database_class.php

<?php   

class database{
    private $objDbConnection;

    function __construct($dsn) {

            $this->objDbConnection = new mysqli($dsn['host'], $dsn['username'],
                                    $dsn['password'], $dsn['database']);

         if (mysqli_connect_errno()) {
                    die("Database connection failed". mysqli_connect_error());
            } 
    }

    function insert($sql) {
         mysqli_query($this->objDbConnection, $sql);

        if(mysqli_errno($this->objDbConnection)) {
            die("Failed query: $sql". $this->objDbConnection->error);
        }
        $intInsertID = $this->objDbConnection->insert_id;
        return $intInsertID;
    }


    function checkIfRecordExists($sql){
        mysqli_query($this->objDbConnection, $sql);

        if(mysqli_errno($this->objDbConnection)) {
            die("Failed query: $sql". $this->objDbConnection->error);
        }
        $intInsertID = $this->objDbConnection;
        return $intInsertID;
    }

 }


 ?>

Answer 1

你班上没有这样的方法。即使你有一个方法num_rows 你应该在$ db object

上调用它

 $db->num_rows($sql) //you will need a query also as a parameter

行

   $res = $db->checkIfRecordExists($sql);

只返回objDbConnection对象

Answer 2

使用

$sql = "SELECT * FROM tbl_Freelancers WHERE User_Username = '".$username."'";
    $result = mysqli_query($sql);

然后你需要使用另一个内置的PHP函数 - $row_count = mysqli_num_rows($result);

Answer 3

<?php
$con = mysql_connect("localhost", "peter", "abc123");
if (!$con)
{
  die('Could not connect: ' . mysql_error());
}

$db_selected = mysql_select_db("test_db",$con);

$sql = "SELECT * FROM person";
$result = mysql_query($sql,$con);
echo mysql_num_rows($result);

mysql_close($con);
?>

Answer 4

您需要在where子句

中更改您缺少squte ' '的查询

$sql = "SELECT * FROM members WHERE username = '".$data['username']."''";//in here

如何让num_rows工作？

4 个答案: