如何在 PHP 中解析下面的 json 请求以提取特定值（在 value 中），如下所示：studentName、studentemail 和 courseName？每个请求只有一个条目，我不需要其余的值。我需要将这些存储为要写入 MYSQL 的 PHP 变量。解析是我遇到麻烦的地方
见下面的 json
{"count":7,"data":[{"name":"Timestamp","ordinal":0,"value":"3/5/2013 2:45:31 PM"},{"name":"studentName","ordinal":1,"value":"mikejones"},{"name":"studentEmail","ordinal":2,"value":"mikejones%40gmail.com"},{"name":"courseName","ordinal":3,"value":"Grade%20Test"},{"name":"cpQuizInfoPassFail","ordinal":4,"value":"Fail"},{"name":"cpInfoPercentage","ordinal":5,"value":"75"},{"name":"institutionCode","ordinal":6,"value":"1212"}]}{"count":0}{"subject":"Course Transcript","footer":"Thank you for your participation!","footerURL":"https://sites.google.com/site/test393223837/googleformsdemotranscript/footerLogo.jpg","correctURL":"http//abc.com,"enabled":true,"correctResponseText":"Correct Response","showCorrectAnswers":true,"description":"Please keep this for your records as proof of course completion and certification.","incorrectURL":"https://sites.google.com/site/test3232393837/_/rsrc/1342229575450/googleformsdemotranscript/redX.png","fromName":"The Training Team","replyTo":"","headerURL":"https://sites.google.com/site/test3938232337/googleformsdemotranscript/CaptivateDev.jpg","isHTML":true,"title":"Course Transcript","to":"joe@gmail.com","userResponseText":"Your Response"}