Question

我想保留目标字段的原始值并使用json_decode将以下字符串用作对象：

{
    "translatorID": "f4a5876a-3e53-40e2-9032-d99a30d7a6fc",
    "label": "ACL",
    "creator": "Nathan Schneider",
    "target": "^https?://(www[.])?aclweb\\.org/anthology-new/[^#]+",
    "minVersion": "1.0.7",
    "maxVersion": "",
    "priority": 100,
    "browserSupport": "gcs",
    "inRepository": true,
    "translatorType": 4,
    "lastUpdated": "2012-01-01 01:42:16"
}

Answer 1

使用json_decode解析之前，您可以执行的操作是：

$string = str_replace('\\', '\\\\\\\\', $string);
var_dump(json_decode($string, true));

这必须是json解析器中的错误该方法不是很干净，但至少你得到的结果。

Answer 2

你试过剥掉斜线吗？

这对我有用：

$string = '{
    "translatorID": "f4a5876a-3e53-40e2-9032-d99a30d7a6fc",
    "label": "ACL",
    "creator": "Nathan Schneider",
    "target": "^https?://(www[.])?aclweb\.org/anthology-new/[^#]+",
    "minVersion": "1.0.7",
    "maxVersion": "",
    "priority": 100,
    "browserSupport": "gcs",
    "inRepository": true,
    "translatorType": 4,
    "lastUpdated": "2012-01-01 01:42:16"
}';

var_dump( json_decode(stripslashes ($string)));

保留字符串的原始值并将其解码为json

2 个答案: