Question

我有以下PHP代码：

<?php
$email = 'email@email.com';
$phone = '+1 800 555 555';
//clean up phone number
$phone = preg_replace("/[^0-9]/", '', $phone);

$plan = 'europe';
$ip = '99.55.44.44';
$sendEmails = True;

function addInfoToDatabase($token) {
        //function that adds data to temporary table. token is passed through

        //put new user info in database
        $mysqli = new mysqli('localhost', 'php', '', 'php');

        $stmt = $mysqli->prepare("INSERT INTO `temp_users` VALUES (?, ?, ?, ?, ?, ?, '')");
        $stmt->bind_param('ssssss', $email, $phone, $plan, $sendEmails, $token, $ip);
        mysqli_stmt_execute($stmt);
        mysqli_stmt_close($stmt);
}
addInfoToDatabase('24_tok');
?>

并且运行时不会返回错误。但是，当我检查我的数据库时，尚未添加一行。我的代码看起来是正确的，为什么这个函数什么都不做？

编辑：D'哦，我想我的问题在于没有将变量传递给函数。这非常有效。

<?php
$email = 'email@email.com';
$phone = '+1 800 555 555';
//clean up phone number
$phone = preg_replace("/[^0-9]/", '', $phone);

$plan = 'europe';
$ip = '99.55.44.44';
$sendEmails = True;

function addInfoToDatabase($email, $phone, $plan, $sendEmails, $token, $ip) {
        //function that adds data to temporary table. token is passed through

        //put new user info in database
        $mysqli = new mysqli('localhost', 'php', '', 'php');

        $stmt = $mysqli->prepare("INSERT INTO `temp_users` VALUES (?, ?, ?, ?, ?, ?, '')");
        $stmt->bind_param('ssssss', $email, $phone, $plan, $sendEmails, $token, $ip);
        mysqli_stmt_execute($stmt);
        mysqli_stmt_close($stmt);
}
addInfoToDatabase($email, $phone, $plan, $sendEmails, '24_tok', $ip);
?>

Answer 1

您正在尝试在函数中使用变量$email，$phone，$plan，$sendEmails和$ip，但它们仅在外部设置功能。这意味着它们是全局变量，因此只有在将它们作为参数传递给函数时才能访问它们。

不这样做的结果是您将null值传递给数据库。如果使用NOT NULL（并且它们可能是）定义了任何相关列，则查询将失败。您可以通过运行此代码来验证失败的原因：

echo 'error '.$mysqli->errno().': '.$mysqli->error().'<br/>';

MySQLi语句的功能对数据库无效

1 个答案: