PHP中的json_编码未返回正确结果

我试图将数据输入到一个具有CRUD操作的网格中，该网格称为Jquery JTable，但从MYSQL表到Json的Json转换似乎存在某种错误，因为要在表上显示数据，它需要如下所示：

{
    "Result":"OK",
    "Records":[
      {"PersonId":1,"Name":"Benjamin Button","Age":17,"RecordDate":"\/Date(1320259705710)\/"},
      {"PersonId":2,"Name":"Douglas Adams","Age":42,"RecordDate":"\/Date(1320259705710)\/"},
      {"PersonId":3,"Name":"Isaac Asimov","Age":26,"RecordDate":"\/Date(1320259705710)\/"},
      {"PersonId":4,"Name":"Thomas More","Age":65,"RecordDate":"\/Date(1320259705710)\/"}
    ]
}

但每当我在php中实现

json_encode

函数时，它都会给我以下信息：

{
    "Result":"OK",
    "Records":[
        {"0":"1","PersonId":"1","1":"Benjamin Button","Name":"Benjamin Button","2":"17","Age":"17","3":"2011-12-27 00:00:00","RecordDate":"2011-12-27 00:00:00"},
        {"0":"2","PersonId":"2","1":"Douglas Adams","Name":"Douglas Adams","2":"42","Age":"42","3":"2011-12-26 00:00:00","RecordDate":"2011-12-26 00:00:00"},
        {"0":"3","PersonId":"3","1":"Isaac Asimov","Name":"Isaac Asimov","2":"26","Age":"26","3":"2011-12-28 00:00:00","RecordDate":"2011-12-28 00:00:00"},
        {"0":"4","PersonId":"4","1":"Thomas More","Name":"Thomas More","2":"61","Age":"61","3":"2011-12-27 00:00:00","RecordDate":"2011-12-27 00:00:00"},
        {"0":"5","PersonId":"5","1":"Ihsan Oktay Anar","Name":"Ihsan Oktay Anar","2":"44","Age":"44","3":"2012-01-03 20:55:02","RecordDate":"2012-01-03 20:55:02"}
    ]
}

以下是我的PHP代码：

<?php

//Get records from database
$result = mysqli_query($conn, "SELECT * FROM people;");

//Add all records to an array
$rows = array();
while($row = mysqli_fetch_array($result))
{
    $rows[] = $row;
}

//Return result to jTable
$jTableResult = array();
$jTableResult['Result'] = "OK";
$jTableResult['Records'] = $rows;
print json_encode($jTableResult);

?>

该表名为people，它有四个字段：

personId

、

Name

、

Age

和

RecordDate

---

如果您有任何想法，我们将不胜感激。

您可能希望尝试该功能

因为根据PHP文档，关于：

获取结果行作为关联数组、数字数组或两者兼有

这就是您目前在代码中遇到的情况，因为默认的抓取模式是

MYSQLI\u BOTH

：

... "1":"Benjamin Button","Name":"Benjamin Button" ...

您也可以使用

mysqli\u fetch\u array（）

，但您必须调整fetch模式，然后：

mysqli_fetch_array($result, MYSQLI_ASSOC);

---

您正在使用返回索引键（

0/1/2/etc.

）的

mysqli\u fetch\u array

），您想要使用

mysqli\u fetch\u assoc（）

或者简单地将其更改为：

mysqli\u fetch\u array（$result，mysqli\u assoc）

在结果中没有点循环作为数组，如果只是将它们单独存储为一个数组。您不妨只做

$rows=mysqli\u fetch\u array（$result）

并使用

$rows这看起来也像是一个评论，在他们开始投票给你之前，只需在上面展开更多的碎片。到目前为止，所有的内容都是这么说的。