我有两个数据库表，guestlist和考勤
在一个HTML页面上，我有一个window.onload脚本，我想通过AJAX检查访客列表。如果url查询中的firstname和lastname出现在guestlist表中，则加载页面。如果没有，请加载一条错误消息。
当页面正确加载时，firstname和lastname会预先填充在两个输入字段中。用户完成表单的其余部分，然后单击提交，将他们的名字和姓氏插入考勤表中。
如果名字和姓氏已经出现在考勤表中，则加载一条错误消息。如果firstname AND lastname没有出现在考勤表中，请将表单信息提交到考勤表中。
当谈到Ajax时，我不是团队中的亮点。这是我目前拥有的代码：
HTML
<body>
<div id="formDiv">
<form id="partyForm" name="party" action="party_insert" method="post">
<h1>Welcome to The Party</h1>
<input name="first_name" id="firstname" class="input" type="text" maxlength="99" placeholder="First Name"><br/>
<input name="last_name" id="lastname" class="input" type="text" maxlength="99" placeholder="Last Name"><br/>
<input name="costume" id="costume" class="input" type="text" maxlength="999" placeholder="What are you supposed to be?"><br/>
<div id="buttonDiv">
<a class="button" id="submit" style="cursor:pointer;">SUBMIT</a>
</div>
</form>
</div>
<script>
window.onload = function () {
var fname_init = decodeURIComponent(getUrlVars()["fname"]);
var lname_init = decodeURIComponent(getUrlVars()["lname"]);
if(fname_init !== "undefined" && lname_init !== "undefined"){
var newString = 'fname='+encodeURIComponent(fname_init)+'&lname='+encodeURIComponent(lname_init);
$.ajax({
type: "GET",
url: "guestList.php",
data: newString,
success: function(){
alert("ON THE LIST");
$('#firstname').val(fname_init);
$('#lastname').val(lname_init);
},
error: function(){
alert("NOT ON THE LIST");
window.location = 'error1.html?fname='+encodeURIComponent(fname_init)+'lname='+encodeURIComponent(lname_init);
}
})
}
}
$("#submit").click(function() {
validate();
});
function submit(){
var fname = $("#firstname").val();
var lname = $("#lastname").val();
var cost = $("#costume").val();
var dataString = 'fname='+encodeURIComponent(fname)+'&lname='+encodeURIComponent(lname)+'&cost='+encodeURIComponent(cost);
$.ajax({
type: "POST",
url: "partyEntry.php",
data: dataString,
success: function() {
alert("ENJOY THE PARTY");
clearForms();
}
});
}
function validate(){
if ($("#firstname").val() == ""){
alert("Please Enter your First Name");
} else {
if ($("#lastname").val() == ""){
alert("Please Enter your Last Name");
}else{
if ($("#costume").val() == ""){
alert("You have to have a costume to be eligible for this raffle");
}else{
submit();
}
}
}
}
function clearForms() {
$('#partyForm')[0].reset();
}
function getUrlVars()
{
var vars = [], hash;
var hashes = window.location.href.slice(window.location.href.indexOf('?') + 1).split('&');
for(var i = 0; i < hashes.length; i++)
{
hash = hashes[i].split('=');
vars.push(hash[0]);
vars[hash[0]] = hash[1];
}
return vars;
}
</script>
</body>
guestList.php
<?php
$host = "localhost";
$user = "root";
$password = "";
$database = "party";
$link = mysql_connect($host, $user, $password);
mysql_select_db($database);
//SURVEY INFORMATION
$fname = mysql_real_escape_string($_REQUEST['fname']);
$lname = mysql_real_escape_string($_REQUEST['lname']);
$checkClient = "SELECT * FROM guestlist WHERE first_name = ".$fname." AND last_name = ".$lname;
mysql_query($checkClient) or die(mysql_error());
mysql_close($link);
?>
partyEntry.php
<?php
$host = "localhost";
$user = "root";
$password = "";
$database = "party";
$link = mysql_connect($host, $user, $password);
mysql_select_db($database);
//SURVEY INFORMATION
$fname = mysql_real_escape_string($_REQUEST['fname']);
$lname = mysql_real_escape_string($_REQUEST['lname']);
$cost = mysql_real_escape_string($_REQUEST['cost']);
$addClient = "INSERT INTO attendance (first_name, last_name, costume) VALUES ('$fname','$lname', '$cost')";
mysql_query($addClient) or die(mysql_error());
mysql_close($link);
?>
我得到的错误是，即使一个名字不在客人名单上，它仍然会显示他们在名单上。所以我在Ajax调用guestlist.php时一定做错了什么，但我不知道是什么。我在编写ajax调用脚本以检查客人是否已经被安排到考勤表中时也遇到了问题。
正如我在评论中所说，您必须从guestList.php返回一个值，这样的操作应该有效：
$checkClient = "SELECT * FROM guestlist
WHERE first_name = ".$fname." AND
last_name = ".$lname;
$result = mysql_query($checkClient);
$count = mysql_num_rows($result);
mysql_close($link);
// output 1 or 0 stating if the user is on the list or not
echo ($count ? 1 : 0);
exit();
然后在您的ajax回调中，您将执行如下检查：
success:function(e) {
alert((e == 1 ? "User is on list" : "User isn't on list"));
根据REST原则，使用HTTP200响应POST请求意味着成功创建了资源。您可以使用HTTP400进行响应，还可以以text/html/json/xml格式提供有关错误的详细信息。
试着这样做吧，
添加以下代码
$query = mysql_query($addClient) or die(mysql_error());
if(mysql_num_rows($query) > 0)
{
header('HTTP/1.1 500 Internal Server Error');
echo 'this is an error message';
}
php脚本永远不会抛出错误，至少当您尝试执行无效查询时是这样。执行查询时没有任何错误，因为它的格式很好，所以不会抛出错误，因为您没有从数据库中获取行。