我想在输出这个PHP代码回声name
,star_type
,service
由jquery.each()
,但我有错误。如何解决它?JSON数据后意外的非空白字符?
错误:
An error has occured:
[object Object]
parsererror
SyntaxError: JSON.parse: unexpected non-whitespace character after JSON data
我有这样的PHP代码:
//$hotel_id = $this->input->post('hotel_id');
$hotel_id = array('1','2','3');
//print_r($hotel_id);
foreach ($hotel_id as $val) {
$query_r = $this->db->query("SELECT * FROM hotel_submits WHERE id LIKE '$val' ORDER BY id desc");
$data = array();
foreach ($query_r->result() as $row) {
$data_s = json_decode($row->service, true);
$data_rp = json_decode($row->address, true);
$data[] = array(
'name' => $row->name,
'star_type' => $row->star . '-' . $row->type,
'site' => $row->site,
'service' => $data_s,
'address' => $row->address
);
}
echo json_encode($data);
}
这是输出上面的PHP代码:
[{
"name": "how",
"star_type": "5-hotel",
"site": "www.sasaas.assa",
"service": ["shalo", "jikh", "gjhd", "saed", "saff", "fcds"]"address": "chara bia paeen"
}][{
"name": "hello",
"star_type": "4-motel",
"site": "www.sasasa.asas",
"service": ["koko", "sili", "solo", "lilo"]"address": "haminja kilo nab"
}][{
"name": "hi",
"star_type": "3-apparteman",
"site": "www.saassaas.aas",
"service": ["tv", "wan", "hamam", "kolas"],
"address": "ok"
}]
而且这是我的js代码,得到错误:
$.ajax({
type: "POST",
dataType: "json",
url: 'get_residence',
data: dataString_h,
cache: false,
success: function (respond) {
//alert(respond);
$.each(respond[0].name, function (index, value) {
alert(value);
});
},
"error": function (x, y, z) {
alert("An error has occured:\n" + x + "\n" + y + "\n" + z);
}
});