我试图创建一个带有循环的JSON数组,我将语言名称存储在数据库中并从查询中检索,之后我需要将它转换成这种格式
$arrayName = array('lang-1' => null , 'lang-2' => null, ..... ,'lang-n' => null );我怎样才能做到这一点
PHP代码如下所示
include_once "inc/init.php";
header('Content-Type: application/json');
$arrayName = array();
$query = $db->query("SELECT * FROM `medium`");
while ($data = mysqli_fetch_assoc($query)) {
array_push($arrayName, $data['medium']);
}
echo json_encode($arrayName);我要用这种格式
[
"Kannada",
"Telugu",
"Tamil",
"Urdu",
"Spanish",
"Arabian"
]我正在努力推动价值,但我没有得到那种格式,请帮助我
谢谢
发布于 2017-03-18 18:36:19
基于您的评论,我认为$data['language']代表了您语言的名称。然后,您可以通过以下代码实现您的目标:
include_once "inc/init.php";
header('Content-Type: application/json');
$arrayName = array();
$query = $db->query("SELECT * FROM `medium`");
while ($data = mysqli_fetch_assoc($query)) {
$arrayName[$data['language']] = null; // $array[$key] = $value
}
echo json_encode($arrayName);发布于 2017-03-18 18:50:44
使用array_push您将得到相同的结果
include_once "inc/init.php";
header('Content-Type: application/json');
$arrayName = array();
$query = $conn->query("SELECT * FROM `medium`");
while ($data = mysqli_fetch_assoc($query)) {
array_push($arrayName,$data['language ']);
}
$arrayName=array_fill_keys($arrayName, 'null');
echo json_encode($arrayName);https://stackoverflow.com/questions/42878122
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