PHP 将 mysql 结果传递给变量

Question

我想将 mysql 查询的结果传递给一个变量，该变量将我们数据库的电子邮件集成到 Mailchimp!但是我不知道如何将 mysql 结果传递给变量!

更精确的查询是:

$sql = "SELECT email, DATE_FORMAT(users.reg, '%Y-%m-%d') FROM users  
WHERE DATE(reg) = CURDATE()";
$result = $conn->query($sql);

if ($result->num_rows > 0) 
{
    while($row = $result->fetch_assoc())
    {
        echo " " . $row["email"]. "<br>";
    }
} else {
    echo "0 results";
}


$data = [
    'email'     => $row["email"],
    'status'    => 'subscribed'
];

syncMailchimp($data);

如何将查询结果传递给 $data 电子邮件？

Answer 1

尝试这样的事情:

<?php

$sql = "SELECT email, DATE_FORMAT(users.reg, '%Y-%m-%d') FROM users  
WHERE DATE(reg) = CURDATE()";
$result = $conn->query($sql);

$data = [];
if ($result->num_rows > 0) 
{
    while($row = $result->fetch_assoc())
    {
        $data = [
            'email'     => $row["email"],
            'status'    => 'subscribed'
        ];
        syncMailchimp($data);
    }

} 
else 
{
    echo "0 results";   
}

它会在循环期间将所有email 发送到mailchimp。 1 对 1。

或者您可以使用单独的循环将其添加到 syncMailchimp 中:

<?php

$sql = "SELECT email, DATE_FORMAT(users.reg, '%Y-%m-%d') FROM users  
WHERE DATE(reg) = CURDATE()";
$result = $conn->query($sql);

$emails = [];
if ($result->num_rows > 0) 
{
    while($row = $result->fetch_assoc())
    {
        $emails[] = [
            'email'     => $row["email"],
            'status'    => 'subscribed'
        ];
    }

} 
else 
{
    echo "0 results";   
}


foreach($emails as $email){
    syncMailchimp($email);
}