ajax php傳遞和接收變量實現思路及代碼

So, your jQuery might be something like.....
復制代碼 代碼如下:
$.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....
復制代碼 代碼如下:
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);
}