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                ??碼云GVP開源項目 12k star Uniapp+ElementUI 功能強大 支持多語言、二開方便! 廣告
                在某些應用場景之下,比如調用 API 的時候,我們可能需要判斷一個返回的字符串是不是一個合法的 JSON 字符串,可以考慮使用下面這種方式: ~~~ function isJson($string) { json_decode($string); return (json_last_error() == JSON_ERROR_NONE); } ~~~ 部分測試結果: ~~~ $jsonString = '{"foo": "bar"}'; dd(isJson($jsonString));// true $jsonString = "{ 'bar': 'baz' }"; dd(isJson($jsonString));// false $jsonString = '{ bar: "baz" }'; dd(isJson($jsonString));// false $jsonString = '{ bar: "baz", }'; dd(isJson($jsonString));// false ~~~
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