filePHP.php
$query = $kon->prepare("SELECT * FROM t_kategori");
$query->execute();
while($row = $query->fetch(PDO::FETCH_ASSOC))
{
$json = array('id' => $row['id_kategori'], 'nama' => $row['nama_kategori']);
echo json_encode($json);
}
和index.php
$.post('filePHP.php', function(data){
console.log(data);
},'json');
但这不起作用,请解决???? thankz
答案 0 :(得分:2)
在 PHP
中尝试此操作$query = $kon->prepare("SELECT id_kategori,nama_kategori FROM t_kategori");
$query->execute();
$json=array();
while($row = $query->fetch(PDO::FETCH_ASSOC))
{
$arr=array('id'=>$row['id_kategori'],'nama'=>$row['nama_kategori']);
array_push($json,$arr);
}
echo json_encode($json);
阅读 array-push
答案 1 :(得分:1)
$query = $kon->prepare("SELECT * FROM t_kategori");
$query->execute();
$json_arr =array();
while($row = $query->fetch(PDO::FETCH_ASSOC))
{
$temp_arr = array();
$temp_arr['id'] => $row['id_kategori'];
$temp_arr['nama'] => $row['nama_kategori'];
array_push($json_arr,$temp_arr);
}
echo json_encode($json_arr);