使用PHP解析表单JSON输出

How do I go about parsing the json request below in PHP to extract specific values (in value ) like the following: studentName, studentemail and courseName? There will be only one entry per request and I will not need the rest of the values. I need to store these as PHP variables to be written to MYSQL. The parsing is where I am having trouble

See json below

     {"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"}

You can use json_decode function, that parse json code to php objects and arrays. See documentation for more info.

One problem is your json is invalid. Check with:

http://jsonlint.com/ or
http://jsonformatter.curiousconcept.com/