我的PHP网页正在返回一个JSON字符串。我写了以下函数来获取这些数据并在jQuery移动列表视图中显示它们。从php页面获取JSON数据
function LoadJsonDataFunction()
{
$.getJSON("my_web_page.php", function(obj) {
$.each(obj, function(key, value){
$("ul").append("<li>"+value.fname+"</li>");
});
});
}
这里是我的列表视图代码:
<ul data-role=listview> </ul>
我呼吁人们在身体标记功能
,但是当我在执行程序显示“取消定义”,并没有数据。
然后我改变$ .getJSON()请求像这样,然后它的工作完美。
$.getJSON("some_page_returning_same_json_string.json",function(obj) { .....
让我知道我该如何解决这个问题。
PS。这是我的PHP页面输出..
{
"employees":[
{
"fname": "sdsdsd",
"lname": "sdsd",
"phone": "sdsd",
"gender": "female",
"dob": "1990-03-11",
"address": "03",
"nic": "erer",
"email": "erererer",
"empid": "ererere",
"designation": "sdsds",
"qualifications": "dsds"
}
]
}
这里是我的PHP代码
<?php
header('Content-Type: application/json');
/*
Following code will list all the employees
*/
// array for JSON response
$response = array();
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
// get all employees from employees table
$result = mysql_query("SELECT * FROM emp_master") or die(mysql_error());
// check for empty result
if (mysql_num_rows($result) > 0) {
// looping through all results
// employees node
$response["employees"] = array();
while ($row = mysql_fetch_array($result)) {
// temp user array
$employee = array();
$employee["fname"] = $row["fname"];
$employee["lname"] = $row["lname"];
$employee["phone"] = $row["phone"];
$employee["gender"] = $row["gender"];
$employee["dob"] = $row["dob"];
$employee["address"] = $row["address"];
$employee["nic"] = $row["nic"];
$employee["email"] = $row["email"];
$employee["empid"] = $row["empid"];
$employee["designation"] = $row["designation"];
$employee["qualifications"] = $row["qualifications"];
//push single employee into final response array
array_push($response["employees"], $employee);
}
// success
// $response["success"] = 1;
// echoing JSON response
echo json_encode($response);
} else {
// no employees found
$response["success"] = 0;
$response["message"] = "No employees found";
// echo no users JSON
echo json_encode($response);
}
?>
你是如何返回来自PHP端的json数据...? –
你是否从php返回json字符串和'json_encode/json_decode'? – Jai
您是否试图在JavaScript中添加另一个foreach? –