PHP:json_encode格式

Question

我有一张桌子上有一份商家名单.每行都有详细信息,例如商家名称,ID,区域,数字等等.

我正在查询mysql数据库,目前我只检索商家名称和ID.然后我使用json_encode函数获得如下输出:{"BusinessA":"business_A_ID","BusinessB":"business_B_ID"}

我希望能够检索所有业务细节并使用json编码函数来获得如下输出:

[
    {
        "businessName": "BusinessA",
        "businessID": "business_A_ID"
    },
    {
        "businessName": "BusinessB",
        "businessID": "businees_B_ID"
    }
]

这是我正在使用的当前代码:

$businessRow = array();
        $businessResult = mysql_query("SELECT * FROM businessTable");
        while($row = mysql_fetch_assoc($businessResult)) {
            $businessRow[$row['businessName']] = $row['businessID'];
            }
            $result = json_encode($businessRow);
            echo $result;

所以我的问题是我需要将此行更改为:

$businessRow[$row['businessName']] = $row['businessID'];

为了获得上面的JSON输出？

谢谢

Answer 1

像这样更改你的脚本:

$businessRow[] = array(
    'businessName' => $row['businessName'],
    'businessID'   => $row['businessID']
);