Question

嗨大家我在将表单数据添加到数据库时遇到了问题。由于某种原因，未插入数据。这是我的代码：

<?php include_once 'secure/connect.php'; ?>
<?php 
$name = "Your Name";
$email = "Your Best Email";
$msg_to_user = "";
if ($_POST['name'] != ""){
    //Be sure to filter this data to deter SQL injection
    $name =  $_POST['name'];
    $name = stripslashes($name);
    $name = strip_tags($name);

    $email = $_POST['email'];
    $email = stripslashes($email);
    $email = strip_tags($email);

    $sql = mysql_query("SELECT * FROM newsletter WHERE email='$email'");
    $numRows = mysql_num_rows($sql);
    if(!$email){
        $msg_to_user = '<h4><font color="FF0000">Please Type an email address ' . $name . '</font></h4>';
    } else if ($numRows > 0) {
        $msg_to_user = '<h4><font color="FF0000">' . $email . ' is already in our system</font></h4>';
    } else {
        $sql_insert = mysql_query("INSERT INTO newsletter (name, email, dateTime) VALUES ('$name', '$email', now())") or die (mysql_error());
        $msg_to_user = '<h4><font color="0066FF">Thanks' . $name . ', You have been added successfully</font></h4>';
        $name = "";
        $email = "";
    }
}
?>

我的html表单如下所示：

<div class="topForm">
<H3 style="text-align:center">SIGN UP FOR OUR NEWSLETTER</H3>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="text" name="name" value="<?php echo $name; ?>"/>
<input type="text" name="email" value="<?php echo $email; ?>"/><br/>
<input name="mySubmitBtn" type="submit"  value="SUBMIT">
<?php echo $msg_to_user; ?>
</form>
</div>

非常感谢所有人菲利普

这就是我现在所拥有的，而且还没有任何工作......

<?php 
$name = "Your Name";
$email = "Your Best Email";
$msg_to_user = "";
if ($_POST['name'] != ""){

    include_once 'secure/connect.php';

    //Be sure to filter this data to deter SQL injection
    $name =  $_POST['name'];
    $name = stripslashes($name);
    $name = strip_tags($name);

    $email = $_POST['email'];
    $email = stripslashes($email);
    $email = strip_tags($email);

    $sql = mysql_query("SELECT * FROM newsletter WHERE email='$email'");
    $numRows = mysql_num_rows($sql);
    if(!$email){
        $msg_to_user = '<h4><font color="FF0000">Please Type an email address ' . $name . '</font></h4>';
    } else if ($numRows > 0) {
        $msg_to_user = '<h4><font color="FF0000">' . $email . ' is already in our system</font></h4>';
    } else {
        $sql_insert = mysql_query("INSERT INTO newsletter (name, email) VALUES ('".$name."', '".$email."')") or die (mysql_error());
        $msg_to_user = '<h4><font color="0066FF">Thanks' . $name . ', You have been added successfully</font></h4>';
        $name = "";
        $email = "";
    }
}
?>

Answer 1

不考虑其他错误或不一致。另请注意，您应该使用mysqli或pdo。但php使用time()

$sql_insert = mysql_query("
                           INSERT INTO newsletter 
                           (name, email, dateTime) 
                           VALUES 
                           ('$name', '$email', ".time().")
                         ");

或者如果您想要日期时间而不是时间戳，则可以使用date()功能。

Answer 2

您必须从代码中更改now()。并使用以下代码。

$time = time()  ;
$sql_insert = mysql_query("INSERT INTO newsletter (name, email, dateTime) VALUES ('".$name."', '".$email."', '".$time."' )") or die (mysql_error());

Answer 3

确保您已连接到数据库！看看echo mysql_error();说什么
如果提交了表单，请捕获值，然后清理
插入查询

ps：看看以下内容：

if(isset($_POST['name']) ...

echo mysql_insert_id();

time() not now()

查看插入的新数据的id

您的代码应该有效，如果您按照这些步骤操作，并且您已连接到数据库

将表单数据插入MySQL数据库

3 个答案: