将 json 数据存储在局部变量中答案

【问题标题】：Store json data in a local variable将 json 数据存储在局部变量中
【发布时间】：2013-04-26 10:26:35
【问题描述】：

我从URL 得到gdata 的youtube video。它像这样返回json code。

{"apiVersion":"2.1","data":{"id":"4TSJhIZmL0A","uploaded":"2008-07-15T18:11:59.000Z","updated":"2013-05-01T21:01:49.000Z","uploader":"burloandbardsey","category":"News","title":"bbc news start up theme","description":"bbc","thumbnail":{"sqDefault":"http://i.ytimg.com/vi/4TSJhIZmL0A/default.jpg","hqDefault":"http://i.ytimg.com/vi/4TSJhIZmL0A/hqdefault.jpg"},"player":{"default":"http://www.youtube.com/watch?v=4TSJhIZmL0A&feature=youtube_gdata_player","mobile":"http://m.youtube.com/details?v=4TSJhIZmL0A"},"content":{"5":"http://www.youtube.com/v/4TSJhIZmL0A?version=3&f=videos&app=youtube_gdata","1":"rtsp://v5.cache7.c.youtube.com/CiILENy73wIaGQlAL2aGhIk04RMYDSANFEgGUgZ2aWRlb3MM/0/0/0/video.3gp","6":"rtsp://v5.cache7.c.youtube.com/CiILENy73wIaGQlAL2aGhIk04RMYESARFEgGUgZ2aWRlb3MM/0/0/0/video.3gp"},"duration":15,"aspectRatio":"widescreen","rating":4.6683936,"likeCount":"354","ratingCount":386,"viewCount":341066,"favoriteCount":0,"commentCount":155,"accessControl":{"comment":"allowed","commentVote":"allowed","videoRespond":"allowed","rate":"allowed","embed":"allowed","list":"allowed","autoPlay":"allowed","syndicate":"allowed"}}}

但我只在remote server 上工作internet connection 而不是local system。

现在我的问题是，

为了我的测试目的，在使用json_decode()之前，我想将上面的json code存储在一个局部变量中。但它给了syntax error。

例如，

$myJson = {"apiVersion":"2.1","data":{"id":"4TSJhIZmL0A","uploaded":"2008-07-15T18:11:59.000Z","updated":"2013-05-01T21:01:49.000Z","uploader":"burloandbardsey","category":"News","title":"bbc news start up theme","description":"bbc","thumbnail":{"sqDefault":"http://i.ytimg.com/vi/4TSJhIZmL0A/default.jpg","hqDefault":"http://i.ytimg.com/vi/4TSJhIZmL0A/hqdefault.jpg"},"player":{"default":"http://www.youtube.com/watch?v=4TSJhIZmL0A&feature=youtube_gdata_player","mobile":"http://m.youtube.com/details?v=4TSJhIZmL0A"},"content":{"5":"http://www.youtube.com/v/4TSJhIZmL0A?version=3&f=videos&app=youtube_gdata","1":"rtsp://v5.cache7.c.youtube.com/CiILENy73wIaGQlAL2aGhIk04RMYDSANFEgGUgZ2aWRlb3MM/0/0/0/video.3gp","6":"rtsp://v5.cache7.c.youtube.com/CiILENy73wIaGQlAL2aGhIk04RMYESARFEgGUgZ2aWRlb3MM/0/0/0/video.3gp"},"duration":15,"aspectRatio":"widescreen","rating":4.6683936,"likeCount":"354","ratingCount":386,"viewCount":341066,"favoriteCount":0,"commentCount":155,"accessControl":{"comment":"allowed","commentVote":"allowed","videoRespond":"allowed","rate":"allowed","embed":"allowed","list":"allowed","autoPlay":"allowed","syndicate":"allowed"}}};

我如何将json data 存储在局部变量中

【问题讨论】：

将 JSON 存储为字符串

标签： php json gdata

【解决方案1】：

$myJson = '{"apiVersion":"2.1","data":{"id":"4TSJhIZmL0A","uploaded":"2008-07-15T18:11:59.000Z","updated":"2013-05-01T21:01:49.000Z","uploader":"burloandbardsey","category":"News","title":"bbc news start up theme","description":"bbc","thumbnail":{"sqDefault":"http://i.ytimg.com/vi/4TSJhIZmL0A/default.jpg","hqDefault":"http://i.ytimg.com/vi/4TSJhIZmL0A/hqdefault.jpg"},"player":{"default":"http://www.youtube.com/watch?v=4TSJhIZmL0A&feature=youtube_gdata_player","mobile":"http://m.youtube.com/details?v=4TSJhIZmL0A"},"content":{"5":"http://www.youtube.com/v/4TSJhIZmL0A?version=3&f=videos&app=youtube_gdata","1":"rtsp://v5.cache7.c.youtube.com/CiILENy73wIaGQlAL2aGhIk04RMYDSANFEgGUgZ2aWRlb3MM/0/0/0/video.3gp","6":"rtsp://v5.cache7.c.youtube.com/CiILENy73wIaGQlAL2aGhIk04RMYESARFEgGUgZ2aWRlb3MM/0/0/0/video.3gp"},"duration":15,"aspectRatio":"widescreen","rating":4.6683936,"likeCount":"354","ratingCount":386,"viewCount":341066,"favoriteCount":0,"commentCount":155,"accessControl":{"comment":"allowed","commentVote":"allowed","videoRespond":"allowed","rate":"allowed","embed":"allowed","list":"allowed","autoPlay":"allowed","syndicate":"allowed"}}}';

应该可以，您需要用 '' 或 "" 将字符串括起来。在任何情况下，最好在字符串中转义 ", 和 '。

【讨论】：

我已经用代码的方式检查过了。但是我的json值有' '或者" "都表示会结束。
这就是为什么我说你应该逃避那些字符 ;-)
如果不在本地变量中分配这个 json 值，我该如何转义？有没有可能。
这取决于您如何获取数据，但您可以使用 str_replace 或 addlashes。有关如何使用它们的参考，请参见 php.net
@YogeshSuthar：是的。但在这个例子中没有quotes。

【解决方案2】：

将其存储在'' 中，就像字符串一样

$myJson = '{"apiVersion":"2.1","data":{"id":"4TSJhIZmL0A","uploaded":"2008-07-15T18:11:59.000Z","updated":"2013-05-01T21:01:49.000Z","uploader":"burloandbardsey","category":"News","title":"bbc news start up theme","description":"bbc","thumbnail":{"sqDefault":"http://i.ytimg.com/vi/4TSJhIZmL0A/default.jpg","hqDefault":"http://i.ytimg.com/vi/4TSJhIZmL0A/hqdefault.jpg"},"player":{"default":"http://www.youtube.com/watch?v=4TSJhIZmL0A&feature=youtube_gdata_player","mobile":"http://m.youtube.com/details?v=4TSJhIZmL0A"},"content":{"5":"http://www.youtube.com/v/4TSJhIZmL0A?version=3&f=videos&app=youtube_gdata","1":"rtsp://v5.cache7.c.youtube.com/CiILENy73wIaGQlAL2aGhIk04RMYDSANFEgGUgZ2aWRlb3MM/0/0/0/video.3gp","6":"rtsp://v5.cache7.c.youtube.com/CiILENy73wIaGQlAL2aGhIk04RMYESARFEgGUgZ2aWRlb3MM/0/0/0/video.3gp"},"duration":15,"aspectRatio":"widescreen","rating":4.6683936,"likeCount":"354","ratingCount":386,"viewCount":341066,"favoriteCount":0,"commentCount":155,"accessControl":{"comment":"allowed","commentVote":"allowed","videoRespond":"allowed","rate":"allowed","embed":"allowed","list":"allowed","autoPlay":"allowed","syndicate":"allowed"}}}';

【讨论】：

如果我的 json 包含 ' .. ' 意味着它将结束一个。我该怎么办？
我该如何逃跑，如果我要这样走mysqli_real_escape_string({"apiVersion":......");这也是一个错误。
我使用 php heredoc 功能来存储我的 json 数据。 syntax error 中的任何一个都可以。但是在我完成json_decode($data); 之后，它什么也没有显示。
@Ranjith 做一件事。首先使用 JSON_DECODE 解码 JSON，然后使用 JSON_ENCODE 使用可用的 JSON 常量 php.net/manual/en/json.constants.php 再次对其进行编码
我的 json 在其内容中也包含 '' （单引号）。所以字符串中断并给出错误。