在本地变量中存储JSON数据

3
我从URL中获取YouTube视频的gdata。 它返回如下json代码。
{"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"}}}

但我只能在我的远程服务器上使用互联网连接,而不是本地系统。

现在我的问题是,

为了测试目的,在使用 json_decode() 之前,我想将上面的 json 代码 存储在本地变量中。但它会报语法错误。

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"}}};

如何将JSON数据存储在本地变量中


将JSON存储为字符串。 - Havelock
2个回答

4
将其存储在''中,就像字符串一样,如下所示。
$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"}}}';

如果我的 JSON 包含了 ' .. ',意味着它会结束一个事务。我该怎么办? - Ranjith
如果我要这样写 mysqli_real_escape_string({"apiVersion":......"),该如何转义?这也可能会导致错误。 - Ranjith
我使用 PHP 的 heredoc 功能来存储我的 JSON 数据。这对于任何 语法错误 都可以正常工作。但是,当我执行 json_decode($data); 后,它显示为空。 - Ranjith
@Ranjith 做一件事。首先使用 JSON_DECODE 解码 JSON,然后使用可用的 JSON 常量 http://www.php.net/manual/en/json.constants.php 再次对其进行编码,使用 JSON_ENCODE - Yogesh Suthar
我的 JSON 内容中也包含单引号 '',所以字符串会被打断并导致错误。 - Dashrath

0
$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"}}}';

应该可以工作,你需要用''或""将字符串括起来。无论如何,在字符串中转义"和'是一个好主意。


我已经检查了代码的方式。但是如果我的JSON值中有任何 ' '" ",它将会结束。 - Ranjith
这就是为什么我说你应该转义那些字符的原因;-) - Daniel Mensing
我如何在不将此JSON值分配给本地变量的情况下进行转义?这可能吗。 - Ranjith
这取决于您如何获取数据,但您可以使用str_replace或addslashes。请参阅php.net以了解如何使用它们。 - Daniel Mensing
@YogeshSuthar:是的。但是在这个例子中没有“引号”。 - Ranjith
在使用这种技术方法之前,我知道add slashesstr_replace,但如何传递json数据或将其存储在本地变量中呢? - Ranjith

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