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使用php将表单数据发送到mysqli DTB失败(xampp环境)

mysqli forms php

thanos_zach · 技术社区 · 7 年前

这是我的html表单代码:问题是我不知道如何使用xampp成功地将表单提交到mysql DTB。(数据不会发送到DTB)。

    <!DOCTYPE html>
    <html>
    <head>
    <meta charset="utf-8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <title>My Form</title>
    <meta name="description" content="An interactive form">
    </head>

    <body>

    <form action="test.php" method="post" id="Personalinfo">

    <label for="fname">ÎÎ½Î¿Î¼Î±:</label>
    <input type="text" id="fname" name="firstname" placeholder="ÎÎ½Î¿Î¼Î± 
    Î ÎµÎ»Î¬ÏÎ·..">

    <input type="submit" value="Submit">

    </body>
    </html>

    <?php
    $servername = "localhost";
    $username = "username";
    $password = "";
    $dbname = "mydb";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
   // Check connection
   if ($conn->connect_error) {
   die("Connection failed: " . $conn->connect_error);
   } 

   $sql = "INSERT INTO Guests (firstname)
   VALUES ('?')";

   if ($conn->query($sql) === TRUE) {
   echo "New record created successfully";
   } else {
   echo "Error: " . $sql . "<br>" . $conn->error;
   }

   $conn->close();
   ?>

谢谢你,塔诺斯

2 回复 | 直到 7 年前

Nana Partykar 7 年前

$sql = "INSERT INTO Guests (firstname) VALUES ('?')";

'?' 替换为整数、字符串、double或blob值。

您放置了“?”,但是忘了用 bind_param $firstname 将值转换为 $stmt->bind_param("s", $firstname);

$firstname = $_POST['firstname'];
$sql = $conn->prepare("INSERT INTO Guests (firstname) VALUES (?)");
$sql->bind_param("s", $firstname);

if ($sql->execute() === TRUE) {

Lekens 7 年前

       <!DOCTYPE html>
            <html>
            <head>
            <meta charset="utf-8">
            <meta http-equiv="X-UA-Compatible" content="IE=edge">
            <title>My Form</title>
            <meta name="description" content="An interactive form">
            </head>

            <body>

            <form action="test.php" method="post" id="Personalinfo">

            <label for="fname">ÎÎ½Î¿Î¼Î±:</label>
            <input type="text" id="fname" name="firstname" placeholder="ÎÎ½Î¿Î¼Î± 
            Î ÎµÎ»Î¬ÏÎ·..">

            <input type="submit" name="submitForm" value="Submit">

            </body>
            </html> 
    **test.php file**
    <?php
        $servername = "localhost";
        $username = "root";
        $password = "";
        $dbname = "mydb";

        // Create connection
        $conn = new mysqli($servername, $username, $password, $dbname);
       // Check connection
       if ($conn->connect_error) {
       die("Connection failed: " . $conn->connect_error);
       } 

    if(isset($_POST['submitForm'])){
    $firstname = $_POST['firstname'];
       $sql = "INSERT INTO Guests (firstname)
       VALUES ('{$firstname}')";

       if ($conn->query($sql) === TRUE) {
       echo "New record created successfully";
       } else {
       echo "Error: " . $sql . "<br>" . $conn->error;
       }

    }else{
echo "Are you sure you enter a firstname and the name of your html submit is submitForm";
}

       $conn->close();
       ?>