PHP中的json_decode返回null

Question

我正在尝试编码json字符串，但它一直返回null。 我在这里尝试了一些建议

$url = "https://www.google.com/finance?output=json&start=0&num=200&noIL=1&q=[currency%20%3D%3D%20%22USD%22%20%26%20%28%28exchange%20%3D%3D%20%22NYSEMKT%22%29%20%7C%20%28exchange%20%3D%3D%20%22NYSEARCA%22%29%20%7C%20%28exchange%20%3D%3D%20%22NYSE%22%29%20%7C%20%28exchange%20%3D%3D%20%22NASDAQ%22%29%29%20%26%20%28change_today_percent%20%3E%3D%20-101%29%20%26%20%28change_today_percent%20%3C%3D%20-3%29%20%26%20%28volume%20%3E%3D%20150001%29%20%26%20%28volume%20%3C%3D%20313940000%29%20%26%20%28last_price%20%3E%3D%2010%29%20%26%20%28last_price%20%3C%3D%20229301%29]&restype=company&ei=T5mTVKG5IYT1igKEoYHQCQ";

$obj = file_get_contents($url);
$obj = json_decode($obj);
var_dump($obj);

Answer 1

JSON不支持\\ x转义，因此按定义它是无效的JSON，因此为什么json_decode（）返回null。 解码有效的JSON \\ u0022可以正常工作。

$url = "https://www.google.com/finance?output=json&start=0&num=200&noIL=1&q=[currency%20%3D%3D%20%22USD%22%20%26%20%28%28exchange%20%3D%3D%20%22NYSEMKT%22%29%20%7C%20%28exchange%20%3D%3D%20%22NYSEARCA%22%29%20%7C%20%28exchange%20%3D%3D%20%22NYSE%22%29%20%7C%20%28exchange%20%3D%3D%20%22NASDAQ%22%29%29%20%26%20%28change_today_percent%20%3E%3D%20-101%29%20%26%20%28change_today_percent%20%3C%3D%20-3%29%20%26%20%28volume%20%3E%3D%20150001%29%20%26%20%28volume%20%3C%3D%20313940000%29%20%26%20%28last_price%20%3E%3D%2010%29%20%26%20%28last_price%20%3C%3D%20229301%29]&restype=company&ei=T5mTVKG5IYT1igKEoYHQCQ";

$obj = file_get_contents($url);
$obj = trim($obj);
$obj = str_replace('\x', '\u00', $obj);
$obj = json_decode($obj,true);
var_dump($obj);

Answer 2

试试这个代码

<?php
function convert($string) {
    return preg_replace_callback('#\\\\x([[:xdigit:]]{2})#ism', function($matches) {
        return htmlentities(chr(hexdec($matches[1])));
    }, $string);
}
$url = "https://www.google.com/finance?output=json&start=0&num=200&noIL=1&q=[currency%20%3D%3D%20%22USD%22%20%26%20%28%28exchange%20%3D%3D%20%22NYSEMKT%22%29%20%7C%20%28exchange%20%3D%3D%20%22NYSEARCA%22%29%20%7C%20%28exchange%20%3D%3D%20%22NYSE%22%29%20%7C%20%28exchange%20%3D%3D%20%22NASDAQ%22%29%29%20%26%20%28change_today_percent%20%3E%3D%20-101%29%20%26%20%28change_today_percent%20%3C%3D%20-3%29%20%26%20%28volume%20%3E%3D%20150001%29%20%26%20%28volume%20%3C%3D%20313940000%29%20%26%20%28last_price%20%3E%3D%2010%29%20%26%20%28last_price%20%3C%3D%20229301%29]&restype=company&ei=T5mTVKG5IYT1igKEoYHQCQ";

$obj = file_get_contents($url);
$obj = convert($obj);
$obj = json_decode($obj);
echo '<pre>'; print_r($obj); echo '</pre>';
?>

Answer 3

作为推荐这里 （！和@showdev已指出，谢谢），你应该用自己的RSS API和解析XML作为替代，因为谷歌并没有它认为必要的荣誉Web标准（在这里，他们不能在JSON;在其他地方，他们失败，例如SMTP，尤其是IMAP）。

JSON有非常详细的规范，在生成器中很容易遵循。 出于安全考虑，建议严格进行JSON解析-尤其是对于外部输入。 因此，请使用Google的RSS变体。