ajax php传递和接收变量实现思路及代码_PHP教程
- WBOY原創
- 2016-07-21 15:13:40856瀏覽
So, your jQuery might be something like.....
复制代码 代码如下:
$.ajax({
url: 'query.php',
data: {id:10},
datatype: json
success: function(results) {
if (results.msg == 'success') {
for (var i in data) {
$('#content').append(
'id = ' + results.data[i].id + ', description = ' + results.data[i].description + ', msrp = ' + results.data[i].msrp
);
}
} else {
$('#content').append(results.msg);
}
}
});
And your php....
复制代码 代码如下:
if (isset($_GET['id'])) {
$sql = "SELECT id, description, msrp FROM tbl WHERE id = '{$_GET['id']}'";
$return = array();
if ($result = mysql_query($sql)) {
if (mysql_num_rows($result)) {
$return['msg'] = 'success';
while ($row = mysql_fetch_assoc($result)) {
$return['data'][] = $row;
}
} else {
$return['msg'] = 'No results found';
} else {
$return['msg'] = 'Query failed';
}
header("Content-type: application/json");
echo json_encode($result);
}
陳述：
本文內容由網友自願投稿，版權歸原作者所有。本站不承擔相應的法律責任。如發現涉嫌抄襲或侵權的內容，請聯絡admin@php.cn