Php JSON数据后出现意外的非空白字符？

我想通过

jquery.each（）

输出这个PHP代码echo

name

，

star\u type

，

service

，但我有错误。如何修复它

错误：

出现错误：
[对象对象]
解析器错误
SyntaxError:JSON.parse:后面出现意外的非空白字符 JSON数据

我有以下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);
}

[{
    "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PHP代码上方的输出：

//$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);
}

[{
    "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，它会出错：

$.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);
    }
});

---

您没有回显有效的json。试试这个：

$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));

这将把所有

$data

数组包装成一个数组，并将其放入对象的数据属性中。您可以在js端访问此数据，如下所示：

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

---

注意：我不确定我上面写的php语法，我写php已经有一段时间了：）

您没有响应有效的json。试试这个：

$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));

这将把所有

$data

数组包装成一个数组，并将其放入对象的数据属性中。您可以在js端访问此数据，如下所示：

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

---

注意：我不确定我上面写的php语法，我写php已经有一段时间了：）

您的php输出不是有效的json，您在

“address”

之前漏掉了一个逗号

---

您可以使用以下url检查您的json:

您的php输出不是有效的json，您在

“地址”

之前遗漏了一个逗号

---

您可以使用以下url检查json:

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

$all_data = array();
foreach ($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JSOn完全无效。您不应在循环内回显json ECNODE数组，而应在循环外回显：

$all_data = array();
foreach ($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);