javascript - 在 Javascript 中读取 PHP 返回的 JSON

Question

我想从数据库中选择一些内容并将其返回给javascript。数据库返回了几行。我用 JSON 尝试了这个，如果我打印出来，也会得到一个结果。但是如果我想转换JSON字符串，总是出现下面的错误信息。 (在 JSON.parse 处)那么，我认为填充数组时可能出现错误？提前感谢大家!

Javascript:

$.ajax({
  url: "./select_firsttracks.php",
  type: "post",
  success: function(resultset) {
    $("#erroroutput").html(resultset);

    var arr = JSON.parse("{" + resultset + "}"); // --> "Uncaught SyntaxError: Unexpected token {"            
  },
  error: function(output) {
    $("#erroroutput").html("fatal error while fetching tracks from db: " + output);
  }
});

PHP:

$storage = array();
while($row = $result->fetch_array())
{
    $storage[] = 
        array
        (
            "id" => $row["id"],
            "trackname" => $row["trackname"],
            "artist" => $row["artist"],
            "genre" => $row["genre"],
            "url" => $row["url"],
            "musicovideo" => $row["musicovideo"]
        );

    echo json_encode($storage);
}

控制台输出:

[{"id":"1","trackname":"yes","artist":"Lady Gaga","genre":"Pop","url":"ftp:\/development","musicovideo":"1"}][{"id":"1","trackname":"yes","artist":"Lady Gaga","genre":"Pop","url":"ftp:\/development","musicovideo":"1"},{"id":"2","trackname":"no","artist":"Prinz Pi","genre":"Rap","url":"ftp:\/development","musicovideo":"1"}]

Answer 1

稍后回显 json

$storage = array();
while($row = $result->fetch_array())
{
    $storage[] = 
        array
        (
            "id" => $row["id"],
            "trackname" => $row["trackname"],
            "artist" => $row["artist"],
            "genre" => $row["genre"],
            "url" => $row["url"],
            "musicovideo" => $row["musicovideo"]
        );


}
echo json_encode($storage);

并更改:

 var arr = JSON.parse(resultset);