PHP - 使用表中的 MySQL 数据填充显示输入数据

Question

我正在尝试使用我的 MySQL 数据库表interest_type 中的描述填充结果

我的私人功能是

    private function Get_Interests_Types_From_DB()
    {
        $sql = "SELECT 
                    InterestID,
                    InterestDescription
                FROM
                    interest_type";

        $result = mysqli_query($this->Con, $sql);

        while($row = mysqli_fetch_array($result))
        {
            $arrayResult[] = $row;
        }

        return ($arrayResult);
    }

我尝试使用此函数的代码区域位于同一类的公共函数中。表单中的值是数字。我试图将 $interests 变量与表中的 InterestID 联系起来，然后打印 InterestDescription，而不是 $interests 的值。

function ProcessRegistrationForm()
    {   
        $fname = $_POST['firstname'];
        $lname = $_POST['lastname'];
        $email = $_POST['email'];
        $gender = $_POST['gender'];
        $interests = $_POST['interests'];

        if(!isset($_POST['firstname']) || !isset($_POST['lastname']) || !isset($_POST['email']) ||
            ($_POST['firstname']) == '' || ($_POST['lastname']) == '' || ($_POST['email']) == '')
        {
            echo("Please enter your first / last name and email.");
        }   
        else
        {
            echo("<h2>Results</h2>");
            echo("<div id='results'>");
            echo $fname; 
            echo("<br />");
            echo $lname; 
            echo("<br />");
            echo $email;
            echo("<br />");
            echo $gender;
            echo("<br />");

            $interestDescription = $this->Get_Interests_Types_From_DB();

            foreach($interests as $likes)
            {
                if($likes == $interestDescription['InterestID'])
                    echo $$interestDescription['InterestDescription'] . "<br />";
            }
            echo("<p style='font-weight: bold;'>Your data has been saved! We will contact you soon!</p>");
            echo("</div>");
        }

        $myClub = new Club("localhost","A340User","Pass123Word","info_club");

        $date = date("Y/m/d");

        $sql="INSERT INTO member
                    (`FirstName`,`LastName`,`Gender`,`Email`,`MemberSince`)
                VALUES
                    ('$fname','$lname','$gender','$email','$date');";

        $result = mysqli_query($this->Con,$sql);
        /*if($result == true) 
        {
            echo "Successful Insert<br />";
        }
        else
        {
            echo "Error Inserting class" . mysqli_error($this->Con) ." <br />";
        }*/

        for($i = 0; $i < sizeof($interests); $i++)
        {
            $interest = $interests[$i];

            $sql="INSERT INTO member_interests
                        (`Email`,`InterestID`)
                    VALUES
                        ('$email',$interest);";

            $result = mysqli_query($this->Con,$sql);
        }

我收到：Notice: Undefined index: InterestID in C:\xampp-portable\htdocs\A340\Assign5\Assign5_Club_Membership_Class.php on line 287

如何打印 InterestDescription 而不是 $interests 的值？

Answer 1

那里：

while($row = mysqli_fetch_array($result))
{
   $arrayResult[] = $row;
}

现在$arrayResult，实际上是您的$interestDescription，包含如下内容：

$interestDescription[0]["InterestID"] 
$interestDescription[1]["InterestID"]
...

而不是$interestDescription[InterestID"]

更新：

也许你需要这样的东西（假设 $interests 拥有一个 ID 数组）

        $interestDescription = $this->Get_Interests_Types_From_DB();
        foreach($interests as $likes) {
           foreach($interestDescription as $desc) {
              if($likes == $desc['InterestID']) {
                 echo $desc['InterestDescription'] . "<br />";
              }
           }
        }