关于php：JSON资料后出现非预期的非空白字元？

unexpected non-whitespace character after JSON data?

我想在输出中将此PHP代码通过jquery.each()回显name，star_type，service，但是我遇到了错误。如何解决？

错误：

An error has occured: [object Object] parsererror
SyntaxError: JSON.parse: unexpected non-whitespace character after
JSON data

我有这个PHP代码：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19

//$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);
}

这是在PHP代码上方输出的：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17

[{
"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"
}]

这是我的js代码出现错误：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19

$.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:\
" + x +"\
" + y +"\
" + z);
}
});

您没有在回显有效的json。试试这个：

1
2
3
4
5
6

$hotel_data = array();
foreach(...) {
// .. do stuff
$hotel_data[] = $data; // add $data to the end of the $hotel_data array
}
echo json_encode(array('data' => $hotel_data));

这会将所有$data数组包装为一个数组，并将其放入对象的data属性中。您可以按以下方式在js端访问此数据：

1
2
3

$.each(response.data, function(i, obj) {
alert(obj.name);
});

注意：我不确定我上面写的php语法，因为我写了php已经有一段时间了:)

您的JSOn完全无效。您不应在循环内回显json-ecnoded数组，而应在循环外回显：

1
2
3
4
5
6
7

$all_data = array();
foreach ($hotel_id as $val) {
//..what you have there now, but instead if echo json_encode($data); you do
$all_data[] = $data;
}
//and finally
echo json_encode('data'=>$all_data);

您的php输出不是有效的json，您在"address"之前错过了逗号。

您可以使用以下网址检查json：http：//json.parser.online.fr/