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我在這裏閱讀了許多類似的線索,看看我做錯了什麼,但我的AJAX調用似乎是正確的。我在這裏錯過了什麼?沒有警報彈出,所以我認爲這是JS方面。AJAX:我能正確插入數據嗎?
$("#SignupSubmit").click(function()
{
var fName = $("#txtSignFName").val();
var lName = $("#txtSignLName").val();
var email = $("#txtSignEmail").val();
var pw = $("#txtPW").val();
if(fName == "" || lName == "" || email == "" || pw == "")
{
alert();
}
else
{
$.ajax({
type: "POST",
url: "actionPages/signUp.php",
dataType: 'json',
data: {
fName:fName,
lName:lName,
email:email,
pw:pw
},
success: function(response) {
alert(response);
}
});
}
});
而PHP(這是正確的?):
<?php
require "../connectionPages/localConnect.php";
$fName = $_POST["fName"];
$lName = $_POST["lName"];
$email = $_POST["email"];
$pw = $_POST["pw"];
if($fName == null || $lName == null || $email == null || $pw == null)
$message = "missing required data";
else
{
$SQL = "INSERT INTO `customer/User` (custFName,
custLName,
custEmail,
custPassword)
VALUES ('$fName', '$lName','$email', '$pw')";
$mysqli->query($SQL);
if($mysqli->affected_rows > 0)
{
$message = "Record successfully inserted <br><a href='..'>Back to Main Page</a>";
$SQL = $mysqli->insert_id; /* $SQL = "SELECT max(custID) as ID FROM `customer/User`"; */
$res = $mysqli->query($SQL) or trigger_error($mysqli->error."[$SQL]");
json_encode($res);
}
else {
$message = "Unable to insert record: " . $mysqli->error;
}
$mysqli->close();
}
哦。它仍然不起作用。我沒有注意到。是的,我試圖得到最後的ID – Calisto
檢查您的瀏覽器控制檯,並在發送數據或發送數據之後發現問題。在這裏發佈控制檯消息以進一步幫助... – Naga
add'error:function(e){console.log(e); '''成功後':function(response){alert(response); }' – Naga