Question

我遇到此代码的电子邮件部分的问题。当我提交记录时，它说插入成功插入，但对于电子邮件，我得到通知：数组到字符串转换，它不想发送订购产品的电子邮件到输入的电子邮件。我无法弄清楚问题是什么。我将继续尝试不同的方法。

process_insert.php

 <html>
    <head>
    <title></title>
    </head>
    <body>
    <?php
        ini_set('display_errors', 1);
        error_reporting(~0);

        $serverName = "localhost";
        $userName = "root";
        $userPassword = "";
        $dbName = "blog_samples";

        $conn = mysqli_connect($serverName,$userName,$userPassword,$dbName);

        $rows_count = count($_POST["name"]);

        for($i=0;$i<$rows_count;$i++){

            // PREVENTING SQL INJECTION !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

            $employee_name = mysqli_real_escape_string($conn,$_POST["employee_name"][$i]);
            $name = mysqli_real_escape_string($conn,$_POST["name"][$i]);
            $code = mysqli_real_escape_string($conn,$_POST["code"][$i]);
            $quantity = intval($_POST["quantity"][$i]);
            $price = mysqli_real_escape_string($conn,$_POST["price"][$i]);


            $sql = "INSERT INTO order_table ( employee_name, name, code, quantity, price) 
                VALUES ('$employee_name', '$name', '$code', '$quantity', '$price')";

            $query = mysqli_query($conn,$sql);
        }

        if(mysqli_affected_rows($conn)>0) {
            echo "Record add successfully";
        }



    $to = "test123@gmail.com";

    $subject = "Supplies";
    $headers = "From: user@gmail.com";  

    $message =

    "employee_name: " . $_POST['employee_name'] . " 

    " ."name: ".  $_POST['name'] ." 

    ". "code: " . $_POST['code'] . " 

    " ."quantity: ".  $_POST['quantity'] . " 

    ". "price: " . $_POST['price'] . "";


    mail($to,$subject,$message,$headers); 


    ?>
    </body>
    </html>

Answer 1

您必须将您的电子邮件代码放在循环中，如下所示： -

 <html>
    <head>
    <title></title>
    </head>
    <body>
    <?php
        ini_set('display_errors', 1);
        error_reporting(~0);

        $serverName = "localhost";
        $userName = "root";
        $userPassword = "";
        $dbName = "blog_samples";

        $conn = mysqli_connect($serverName,$userName,$userPassword,$dbName);

        $rows_count = count($_POST["name"]);

        for($i=0;$i<$rows_count;$i++){

            // PREVENTING SQL INJECTION !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

            $employee_name = mysqli_real_escape_string($conn,$_POST["employee_name"][$i]);
            $name = mysqli_real_escape_string($conn,$_POST["name"][$i]);
            $code = mysqli_real_escape_string($conn,$_POST["code"][$i]);
            $quantity = intval($_POST["quantity"][$i]);
            $price = mysqli_real_escape_string($conn,$_POST["price"][$i]);


            $sql = "INSERT INTO order_table ( employee_name, name, code, quantity, price) 
                VALUES ('$employee_name', '$name', '$code', '$quantity', '$price')";

            $query = mysqli_query($conn,$sql);
            if(mysqli_affected_rows($conn)>0) {
                echo "Record add successfully";
                $to = "test123@gmail.com";

                $subject = "Supplies";
                $headers = "From: user@gmail.com";  

                $message =

                "employee_name: " . $employee_name . " 

                " ."name: ".  $name ." 

                ". "code: " . $code . " 

                " ."quantity: ".  $quantity . " 

                ". "price: " . $price . "";


                mail($to,$subject,$message,$headers); 
            }

        }
    ?>
    </body>
    </html>

注意： - 您的代码容易受到SQL INJECTION的攻击。使用prepared statements作为评论

中建议的@RiggsFolly

收到要发送的电子邮件时遇到问题。继续将Array转换为字符串转换

1 个答案: