小编Art*_*Art的帖子

我遇到了问题.我有一个json代码如下.我想解析它们但得到一个错误:SyntaxError:JSON.parse:意外的字符我不知道错误在哪里.有人会帮忙吗？

我的js代码:

  function retreive() {
  var userInfo = new Array();
  userInfo[0] = $("#contactId").val();
  userInfo[1] = $("#pw").val();
  var cId = $.ajax({ 
    url: "server.php", 
    type: "POST", 
data: {phpData : userInfo}, 
    datatype: "json",
success:function(msg) {
    responseJson = JSON.parse(msg.responseText);
var outputHtml = "";
    for (var i=0; i<responseJSON.user.mary.length; i++) {
outputHtml += responseJSON.user.mary[i].sender[i].sendDate + 
", " + responseJSON.user.mary[i].sender[i].time + 
", " + responseJSON.user.mary[i].sender[i].timezone + 
", " + responseJSON.user.mary[i].sender[i].message + "<br/>"}
divMessage = document.getElementById("message");
    divMessage.innerHTML = outputHtml;
}
});   

}

我的PHP代码:

    $data = '{
  "user" : …

我运行以下代码时遇到问题:

function newUser($email,$pwd,$pwd2,$firstname,$surname,$isAdmin=0){
  $email = $this->verify('Email',$email,10,40);
  $pwd = $this->verify('Password',$pwd,6,20);
  $pwd2 = $this->verify('Password',$pwd2,6,20);
  $firstname = $this->strToTitle($this->verify('Name',$firstname,2,40));
  $surname = $this->strToTitle($this->verify('Title',$surname,2,40));
  if ($pwd != $pwd2)
    return -1;
    $key=md5("secure")
  $result = $this->query("INSERT INTO user (email, pw, firstname, surname, isAdmin) VALUES (".$email.", AES_ENCRYPT(".$pwd.",".$key."), ".$firstname.", ".$surname.", ".$isAdmin.")");
  if (mysql_affected_rows()>0)
    return mysql_insert_id();
  else
    return 0;
}

它始终在第76行的F:\ xampp\htdocs\sql.php中提示"解析错误:语法错误,意外'$结果'(T_VARIABLE)"

任何人都可以给我一些建议吗？非常感谢!!