Ajax Call back PHP interface return JSON data AJAX JSONP Ajax JSON instance AJAX get background JSON number

The PHP code is as follows:

  Header' Content-type:application/json '); Header' Content-type:text/html;charset=utf-8 ');$email= $_get[' Email '];$user= [];$conn= @Mysql_connect ("localhost","Test","123456")or die("Failed in connecting database"); mysql_select_db ("Test",$conn); mysql_query ("Set names ' UTF-8 '");$query= the Select*From userinformation where email = ' ".$email."'";$result= mysql_query ($query);if(NULL== ($row= Mysql_fetch_array ($result))){Echo$_get[' Callback ']."(no such user)";} Else{$user [' Email '] = $email;$user [' nickname '] = $row [' nickname '];$user [' Portrait '] = $row [' Portrait '];Echo$_get[' Callback ']."(".Json_encode ($user).")";}?>

The JS code is as follows:

Two questions were encountered:

1. First question:

Uncaught syntaxerror:unexpected Token:

The solution is as follows:

This have just happened to me, and the reason is none of the reasons above. I was using the JQuery command Getjson and adding callback=? to use JSONP (as I needed to go Cross-domain), and returning the J SON code and {"foo":"bar"} getting the error.

This is because I should has included the callback data, something likejQuery17209314005577471107_1335958194322({"foo":"bar"})

Here are the PHP code I used to achieve this, and which degrades if JSON (without a callback) is used:

   $ret   [  ' foo '  ]   =   "Bar"  ;   Finish   ();   function   Finish   ()   {  header   (  "Content-type:application/json"  );   if   (  $_get   [  ' callback '  ]   {  print   $_get   [  ' callback '  ].   "("  ;  }   print   json_encode   (  $GLOBALS   [  ' ret '  ]);   if   (  $_get   [  ' callback '  ]   {  print   ")"  ;  }   exit  ;  }   

Hopefully that'll help someone in the future.

2. The second question:

Parse the JSON data. As you can see from the JavaScript above, I'm not using Jquery.parsejson () to get started with these methods, but I always report

Vm219:1 uncaught syntaxerror:unexpected token o in JSON at position 1

Error, and then not Jquery.parsejson () This method, but everything is normal. I don't know why.

The above describes the Ajax call back to the PHP interface to return JSON data, including the ajax,json aspects of the content, I hope to be interested in PHP tutorial friends helpful.

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