Question

嗨，我要从sql返回一些数据，并想将其格式化为Json，我知道我可以使用json.encode（），但是数据有点嵌套，我不确定如何实现。

这就是我希望json看起来的样子。

[
{"coords":
            {"lat":53.745,"lng":-0.338},
                "iconImage":"https://developers.google.com/maps/documentation/javascript/examples/full/images/beachflag.png",
                    "content":"<h1>Tony G</h1>"},
{"coords":{"lat":53.747,"lng":-0.340},
    "iconImage":"https://maps.gstatic.com/mapfiles/ms2/micons/blue.png",
        "content":"<h1>fred</h1>"}
]

到目前为止，这是我的代码。

require("../PHP/phpsqlajax_dbinfo.php");
$connection=odbc_connect($database, $username, $password);

//Select Test statement 
$query="select 53.745 as lat,-0.338 as lng,'https://developers.google.com/maps/documentation/javascript/examples/full/images/beachflag.png' as iconImage, '<h1>Tony G</h1>' as content union all
select 53.745 as lat,-0.310 as lng,'https://maps.gstatic.com/mapfiles/ms2/micons/blue.png' as iconImage, '<h1>fred</h1>' as content ";

$result=odbc_exec($connection,$query);
//work through result and create JSON
while (odbc_fetch_row($result)){

//what do I do here?

} 

echo json_encode(//theData I would like to return) ;

感谢您的帮助

Answer 1

只需添加到数组然后编码：

$json = [];

while ($row = odbc_fetch_row($result)){

    $json[] = [
        'coords' => ['lat' => $row['lat'],'lng' => $row['lng']],
        'iconImage' => $row['iconImage'],
        'content' => $row['content'],
    ];
}

echo json_encode($json);

从SQL结果集中创建一个更复杂的json对象

1 个答案: