我是AJAx的新人。我的问题是我有ajax函数在页面加载时在php中传递变量ID错误是未定义变量:id但是当我查看firebug post id过去成功。这是我的ajax。
$('.btn_edit').click(function(e){
e.preventDefault();
var $this = $(this);
var id_id = $(this).attr('id');
alert(id_id);
$.ajax({
type: "POST",
url: "edit_query.php",
data:{id: id_id},
success: function() {
alert("Success Input");
这是我传递的php页面。
$id = $_POST['id'];
$sql = mysql_query("select * from user where uid = ".$id."");
$table = mysql_fetch_assoc($sql);
?>
答案 0 :(得分:1)
$sql = mysql_query("select * from user where uid = ".$id."");
应该是
$sql = mysql_query("select * from user where uid = $id ");
和
var id_id = $(this).attr('id');
alert(id_id);
$.ajax({
type: "POST",
url: "edit_query.php",
data:"id="+id_id,
success: function() {
alert("Success Input");
}
答案 1 :(得分:0)
试试这个
$.post( "edit_query.php", { id: id_id })
.done(function( data ) {
alert( data );
});
答案 2 :(得分:0)
试试这个
<?php
$id = $_POST['id'];
$sql = mysql_query('SELECT * FROM user WHERE uid = '.$id);
$row = mysql_fetch_assoc();
header('Content-Type: application/json');
echo json_encode($row);
exit;
$(function(){
var onClick, successHandler;
onClick = function (e) {
e.preventDefault();
$.post('edit_query.php',{id:$(this).attr('id')},successHandler,'json');
};
successHandler = function (json) {alert(json.uid);};
$('.btn_edit').click(onClick);
});