使用 我有一个名为 但是当我设置我的$查询时：
它显示我想要的数据
当使用变量 some.php时出错？id = $ id
tb_syarat_layanan 的表。 当我尝试获取JSON时，数组显示没有数据。
＆lt;？php
//导入数据库
require_once（'koneksi.php'）;
$ query =“select * from tb_syarat_layanan WHERE id_layan = $ id”;
$ hasil = mysqli_query（$ con，$ query）;
if if（mysqli_num_rows（$ hasil）＆gt; 0）
{
$ response = array（）;
$ response [“data”] = array（）;
while（$ x = mysqli_fetch_array（$ hasil））
{
$ h ['id_layan'] = $ X [ “id_layan”];
$ h ['dokumen'] = $ x [“dokumen”];
array_push（$ response [“data”]，$ h）;
}
echo json_encode（$ response）;
} \ n else
{
$ response [“message”] =“No Data”;
echo json_encode（$ response）;
}
？＆gt;
\ n
$ query =“select * from tb_syarat_layanan WHERE id_layan = 1”;
$ id <设置$ query时，JSON显示无数据 / p>
Error when using some.php?id=$id
I have table named tb_syarat_layanan. When I try to get JSON, array it shows no Data.
<?php
//Importing database
require_once('koneksi.php');
$query = "select * from tb_syarat_layanan WHERE id_layan=$id";
$hasil = mysqli_query($con,$query);
if(mysqli_num_rows($hasil) > 0 )
{
$response = array();
$response["data"] = array();
while($x = mysqli_fetch_array($hasil))
{
$h['id_layan'] = $x["id_layan"];
$h['dokumen'] = $x["dokumen"];
array_push($response["data"], $h);
}
echo json_encode($response);
}
else
{
$response["message"] = "No Data";
echo json_encode($response);
}
?>
But when I set my $query like this :
$query = "select * from tb_syarat_layanan WHERE id_layan=1";
it show the data that I want
JSON shows no data when $query is set with variable $id
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