ajax php传递和接收变量实现思路及代码
author:一佰互联 2019-04-30   click:156
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);
}