Question

已经有一段时间了，因为我一直在搞乱PHP，但我现在需要。

我正在尝试从PHP MySQL中完成以下操作：

"holeScores": [
  {
    "type": "RoundHoleData",
    "xtraStrokes": 0,
    "strokes": 3
  },
  {
    "type": "RoundHoleData",
    "xtraStrokes": 1,
    "strokes": 5
  },
  {
    "type": "RoundHoleData",
    "xtraStrokes": 0,
    "strokes": 5
  }
]

以下是我尝试这样做的方法：

$userCard = array();

$sql = "SELECT * FROM fb_score WHERE userid=$player AND matchid=$matchid ORDER BY hole_no ASC";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    while($row = $result->fetch_assoc()) {

        $scoreInf = array(
            "type"=> "RoundHoleData",
            "xtraStrokes"=> $row["hXtra"],
            "strokes"=> $row["stroke"]
        );

        $userCard[] = $scoreInf;
    }
}



$matchResultArray = array("userid"=>$player
                      , "matchform"=>$matchformid
                      , "totScore"=>$totScore
                      , "par_score"=>$par_score
                      , "placement"=>$i
                      , "holeScores"=>$userCard
                      , "usercardInfo"=>$userCardInfo
                      );


echo json_encode($matchResultArray, 16);

这创造了这个结果：

"usercard":{  
    "0":{  
       "type":"RoundHoleData",
       "xtraStrokes":"0",
       "strokes":"4"
    },
    "1":{  
       "type":"RoundHoleData",
       "xtraStrokes":"1",
       "strokes":"5"
    },
    "2":{  
       "type":"RoundHoleData",
       "xtraStrokes":"1",
       "strokes":"5"
    }
}

无论如何我可以创建正确的json对象格式，如果是的话......怎么样？！？

任何帮助都表示赞赏，并提前感谢： - ）

Answer 1

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PHP创建没有键的数组对象

1 个答案: