如果这是一个重复的问题,请道歉,但我在这里找到的任何答案都没有给我带来帮助。我正在尝试为一个网站创建一个简单的登录功能,该网站使用对PHP的AJAX调用,它应返回JSON。我有以下PHP:
<?php
include("dbconnect.php");
header('Content-type: application/json');
$numrows=0;
$password=$_POST['password'];
$username=$_POST['username'];
$query="select fname, lname, memcat from members where (password='$password' && username='$username')";
$link = mysql_query($query);
if (!$link) {
echo 3;
die();
}
$numrows=mysql_num_rows($link);
if ($numrows>0){ // authentication is successfull
$rows = array();
while($r = mysql_fetch_assoc($link)) {
$json[] = $r;
}
echo json_encode($json);
} else {
echo 3; // authentication was unsuccessfull
}
?>
AJAX电话:
$( ".LogIn" ).live("click", function(){
console.log("LogIn button clicked.")
var username=$("#username").val();
var password=$("#password").val();
var dataString = 'username='+username+'&password='+password;
$.ajax({
type: "POST",
url: "scripts/sendLogDetails.php",
data: dataString,
dataType: "JSON",
success: function(data){
if (data == '3') {
alert("Invalid log in details - please try again.");
}
else {
sessionStorage['username']=$('#username').val();
sessionStorage['user'] = data.fname + " " + data.lname;
sessionStorage['memcat'] = data.memcat;
storage=sessionStorage.user;
alert(data.fname);
window.location="/awt-cw1/index.html";
}
}
});
}
正如我所说的,每当我运行这些来自&#34; data&#34;未定义。知道我哪里出错了吗?
非常感谢。