有没有办法根据 sql 查询返回的内容在 php 中创建变量？ [复制]

Question

我创建了一个查询，该查询从数据库中检索用户的街道地址、他们的州和城市。现在我试图将所有这些信息放入一个 $var 中，但我的代码似乎没有工作，我不知道为什么。我是 PHP 新手，所以对于我所犯的任何错误，我深表歉意。

require 'config.php';
$host ="myhost";
$user ="username";
$pass ="mypassword";
$db="mydb";
$mysqli = new mysqli($host, $user, $pass, $db);
if ( $mysqli->connect_errno ) {
    echo $mysqli->connect_error;
    exit();
}
$mysqli->set_charset('utf8');
$sql = "SELECT  addresses.street_address AS street_address, addresses.state AS state, addresses.City AS city
            FROM users
        LEFT JOIN addresses
            ON users.address_id=addresses.address_id
        WHERE username='".$_SESSION['username']."';";
$results = $mysqli->query($sql);
if(!$results) {
    echo $mysqli->error;
    exit();
}
$row = $results->fetch_assoc();

var_dump($row['state']);
var_dump($row['city']);
var_dump($row['street_address']);

var $address= $row['street_address']. $row['city']. $row['state'];

Answer 1

试试这个

$street = $row['street_address'];
 $city = $row['city'];
$state =  $row['state'];

$address = $street  .  '  '  .  $city .  '  '  .  $state;

echo $address;

Answer 2

您可以使用 list() 语言结构。 PHP 5.6 将从最右边的参数开始，但 7.* 将从最左边的参数开始，如手册所述https://www.php.net/manual/en/function.list.php。

PHP 5.6

list($city, $state, $streetAddress) = $rows;
$address = "$streetAddress | $city | $state";

PHP 7.*

list($streetAddress, $city, $state) = $rows;
$address = "$streetAddress | $city | $state";

// or

[$streetAddress, $city, $state] = $rows;
$address = "$streetAddress | $city | $state";