如何以JSON格式输出pdo结果以用于javascript函数？

Question

当我在PHP中回显结果时，它们似乎正确地格式化为JSON格式，但是当我回显将其保存在JavaScript中的变量时，我收到“未捕获的SyntaxError：意外的标识符”错误（当我删除回显时，该错误消失了）声明）。 这是代码：

$charset="utf8";
$dsn = "mysql:host=$host;dbname=$db;charset=$charset";
$options = [
    PDO::ATTR_ERRMODE            => PDO::ERRMODE_EXCEPTION,
    PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC,
    PDO::ATTR_EMULATE_PREPARES   => false,
];
try {
    $pdo = new PDO($dsn, $user, $pass, $options);
} catch (\PDOException $e) {
    throw new \PDOException($e->getMessage(), (int)$e->getCode());
}

$page = 'nike';
$stmt = $pdo->prepare('SELECT `url`, `alt`, `model`, `desc` FROM images WHERE page_id = (select id from pages where title = ?)');
$stmt->execute([$page]);
$results = $stmt->fetchAll(PDO::FETCH_ASSOC);
echo 'results are:' . $results; //array
$json = json_encode($results);
echo 'json is: ' . $json;// seemingly correct json formatted result set  

如果我直接在JavaScript内部回显$json变量，如下所示：

var imgs="<?php echo $json; ?>";

我收到“未捕获的SyntaxError：意外的标识符”错误消息，当我删除该行时，该消息消失。 我的JavaScript函数期望var imgs为：

var imgs = [
    {"url": "images/adidas_large/1.png", "model": "Kumacross", "desc": ""},
    {"url": "images/adidas_large/2.png", "model": "fig 2 model", "desc": "fig 2 desc"},
    {"url": "images/adidas_large/3.png", "model": "fig 3 model", "desc": "fig 3 desc"},
    {"url": "images/adidas_large/4.png", "model": "fig 4 model", "desc": "fig 4 desc"},
    {"url": "images/adidas_large/5.png", "model": "fig 5 model", "desc": "fig 5 desc"},
    {"url": "images/adidas_large/6.png", "model": "fig 6 model", "desc": "fig 6 desc"},
    {"url": "images/adidas_large/7.png", "model": "fig 7 model", "desc": "fig 7 desc"},
    {"url": "images/adidas_large/8.png", "model": "fig 8 model", "desc": "fig 8 desc"},
    {"url": "images/adidas_large/9.png", "model": "fig 9 model", "desc": "fig 9 desc"}
];

...，并在提供此硬编码后才能使用。 所以我的问题是，如何正确输出JSON格式的PDO查询结果以在JavaScript函数中使用？

Answer 1

我通过在javascript中而不是php中通过json_encoding $ results解决了这个问题，并进行了更改

var imgs = "<?php echo json_encode($results); ?>";

至

var imgs=<?php echo json_encode($results); ?>;