Question

您好我正在尝试使用phpmyadmin并连接到我的数据库。我曾经做过一次，但我从来没有在文本框中应用样式，所以我不知道它是否正确。出于某种原因，我收到一条错误消息，我告诉它说它是否无法连接。我无法弄清楚为什么它没有连接。我正在使用MAMP服务器端口：PHP 5.6.21

这是我的PHP代码：

<?php
if($_POST['formSubmit'] == "Submit") {
    $errorMessage = "";

    if(empty($_POST['formName'])) {
        $errorMessage .= "<li>You need to enter your name</li>";
    }
    if(empty($_POST['formEmail'])) {
        $errorMessage .= "<li>You need to enter your email</li>";
    }
    if(empty($_POST['formSubject'])) {
        $errorMessage .= "<li>Please enter a subject.</li>";
    }
    if(empty($_POST['formComment'])) {
        $errorMessage .= "<li>Please enter your question.</li>";
    }

    $varname = $_POST['formName'];
    $varemail = $_POST['formEmail'];
    $varsubject = $_POST['formSubject'];
    $varcomment = $_POST['formComment'];

    if(empty($errorMessage)) {
        $db = mysql_connect("localhost","root","root");
        if(!$db) die("Error connecting to MySQL database.");
        mysql_select_db("three_cats_database" ,$db);

        $sql = "INSERT INTO contact_form (name, email, subject, comment) VALUES (".
        PrepSQL($varname) . ", " .
        PrepSQL($varemail) . ", " .
        PrepSQL($varsubject) . "," .
        PrepSQL($varcomment) . ") ";
        mysql_query($sql);

        header("Location: thankyou.php");
        exit();
    }
}

// function: PrepSQL()
// use stripslashes and mysql_real_escape_string PHP functions
// to sanitize a string for use in an SQL query
//
// also puts single quotes around the string
//
function PrepSQL($value)
{
    // Stripslashes
    if(get_magic_quotes_gpc()) {
        $value = stripslashes($value);
    }

    // Quote
    $value = "'" . mysql_real_escape_string($value) . "'";

    return($value);
}
?>

这是我的HTML：

<p>Please fill out this form completely to contact us with any concerns or suggestions.</p><br>
    <div class="imgbg"><div class="img">
    <!-- FORM IS HERE -->
      <form action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post">
      <div class="contact-form margin-top">

        <label for='formName'><span>Name:</span>
        <input type="text" class="input_text" name="formName" id="name" maxlength="50" value="<?=$varname;?>"/>
        </label>

        <label for='formEmail'><span>Email:</span>
        <input type="text" class="input_text" name="formEmail" id="email" maxlength="50" value="<?=$varemail;?>"/>
        </label>

        <label for='formSubject'><span>Subject:</span>
        <input type="text" class="input_text" name="formSubject" id="subject" maxlength="50" value="<?=$varsubject;?>"/>
        </label>

        <label><span>Comment</span>
        <textarea class="message" name="formComment" id="feedback"><?php echo htmlspecialchars($varcomment);?></textarea>
        </label>

        <input type="submit" class="button" name="formSubmit" value="Submit" />
        </label>
      </div>
    </form>

最后一张我的数据库设置图片，用于证明姓名

Answer 1

以下是您遇到问题的代码。希望这可以帮助。我发表了评论，解释了整个代码中发生的事情，所以如果你想继续阅读它们，那会让你理解得更好一点。希望这会有所帮助。

if($_POST['formSubmit'] == "Submit") {
    $errorMessage = "";
    if(empty($_POST['formName'])) {
        $errorMessage .= "<li>You need to enter your name</li>";
    }
    if(empty($_POST['formEmail'])) {
        $errorMessage .= "<li>You need to enter your email</li>";
    } 
    if(empty($_POST['formSubject'])) {
        $errorMessage .= "<li>Please enter a subject.</li>";
    } 
    if(empty($_POST['formComment'])) {
        $errorMessage .= "<li>Please enter your question.</li>";
    }

    $varname = $_POST['formName'];
    $varemail = $_POST['formEmail'];
    $varsubject = $_POST['formSubject'];
    $varcomment = $_POST['formComment'];

    if(empty($errorMessage)) {
        try 
        { 
           // I used PDO because I prefer it over mysqli
            $db = new PDO('mysql:host=localhost:3306;dbname=three_cats_database, root, root');
        }
        catch(PDOException $e)
        {
            die('Unable to connect to database');
        }
        // Here is the sql statement that you provided.
        $sql = "INSERT INTO contact_form (name, email, subject, comment) 
            VALUES (?,?,?,?)";
        // Prepare helps with sql injections ect. It is still not completely hacker proof.
        $stmt = $db->prepare($sql);
        // Each question mark (?) means they increment from 1-4 in this case.
        $stmt->bindValue(1, $varname, PDO::PARAM_STR);
        $stmt->bindValue(2, $varemail, PDO::PARAM_STR);
        $stmt->bindValue(3, $varsubject, PDO::PARAM_STR);
        $stmt->bindValue(4, $varcomment, PDO::PARAM_STR);
        //If statement to check if the sql command went through.This is also good for error checking.  
        if($stmt->execute())
        {
            header("Location: thankyou.php");
        }
        else 
        {
            // There was an error in the statement
        }
    }
}

Answer 2

编辑此行以获取有关错误的更多信息：

if(!$db) die("Error connecting to MySQL database.");

可以成为：

if(!$db) die("Error connecting to MySQL database : ".mysql_error($db));

崩溃时您会看到更多信息：）

注意：为您的表创建一个用户，最好不要在您的php文件中写入root密码

连接到我的数据库时出错

2 个答案: