网学网PHP编辑为广大网友搜集整理了:ajax php传递和接收变量实现思路及代码绩等信息,祝愿广大网友取得需要的信息,参考学习。
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(
ajax php传递和接收变量实现思路及代码_网学
浏览:
GET[''id''])) {
$sql = "SELECT id, description, msrp FROM tbl WHERE id = ''{
ajax php传递和接收变量实现思路及代码_网学
浏览:
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);
}