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javascript - 如何以 JSON 格式输出 pdo 结果以用于 javascript 函数？

问题描述

当我在 PHP 中回显结果时，它们似乎被正确格式化为 JSON，但是当我回显将它们保存在 JavaScript 中的变量时，我得到一个“未捕获的语法错误：意外的标识符”错误（当我删除回显时它消失了陈述）。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

$json如果我像这样直接在 JavaScript 中回显变量：

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

我收到“未捕获的语法错误：意外的标识符”错误消息，当我删除该行时该错误消息消失。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"}
];

...并在提供此硬编码时工作。所以我的问题是，如何正确输出 JSON 格式的 PDO 查询结果以用于 JavaScript 函数？

标签： javascriptjsonpdo

解决方案

我通过javascript而不是php中的json_encoding $results解决了这个问题并改变了

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

至

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

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