@J Salaria我了解你与jquery AJAX和PHP代码有问题，因为你没有得到你想要的结果你的问题。有不同的方式通过jquery ajax发送数据，我会详细解释。 ？
$_POST["lExpensesId"]，您在一个HTML <form>得到这个ID。因为在这里，我会向您展示3点不同的熟练的方式通过Ajax来发送数据..
注意：你的代码很容易受到SQL INJECION。 I'LL也向您展示的方法来OVERCOME.IF你想了解更多有关SQL注入点击这个LINK SQL INJECTION LINK
HTML表单代码：
<form action="" id="send_lExpensesId_form" method="post">
<input type="text" name="lExpensesId" id="lExpensesId" >
<input type="submit" name="submit" >
</form>
第一种方式用于发送数据THIS IS虽然HTML <FORM>
<script>
$(document).ready(function(){
$("#send_lExpensesId_form").submit(function(e){
e.preventDefault();
var form_serialize = $(this).serialize();
$.ajax({
type:'POST',
url:'script_admin-add-category.php',
data:form_serialize,
success:function(data){
if(data == "YES"){
alert("EMAIL");
}else{
alert(data);
}
}
});
});
});
</script>
第二种方法用于发送数据不过这是在HTML <FORM>
<script>
$(document).ready(function(){
$("#send_lExpensesId_form").submit(function(e){
e.preventDefault();
var form_serialize = new FormData($(this)[0]);
$.ajax({
type:'POST',
url:'script_admin-add-category.php',
data:form_serialize,
contentType: false,
processData: false,
success:function(data){
if(data == "YES"){
alert("EMAIL");
}else{
alert(data);
}
}
});
});
});
</script>
用于发送数据的第三方式使用WHEN A LINK CLICKED OR TO DELETED THROUGH ID OR CLASS
<script>
$(document).ready(function(){
$("#send_lExpensesId_form").submit(function(e){
e.preventDefault();
var lcl_ExpensesId = $("#lExpensesId").val();
$.ajax({
type:'POST',
url:'script_admin-add-category.php',
data:{lExpensesId:lcl_ExpensesId},
success:function(data){
if(data == "YES"){
alert("EMAIL");
}else{
alert(data);
}
}
});
});
});
</script>
这与mysqli_real_escape_string(); PHP代码针对SQL INJECTION
<?php
$servername = "localhost";
$username = "root";
$password = "admin";
$dbname = "demo";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
if(isset($_POST["lExpensesId"])){
$lExpensesId = mysqli_real_escape_string($conn, $_POST["lExpensesId"]);
$Lquery = "SELECT ExpensesId FROM tblexpensestype WHERE ExpensesId = '$lExpensesId'";
if($query_result = mysqli_query($conn, $Lquery)){
if(mysqli_num_rows($query_result)){
echo 'YES';
}else{
echo "Proceed";
}
}else{
echo "Error".mysqli_connect_error();
}
}
?>
这其他PHP代码MYSQLI->PREPARED这是更好地与SQL INJECTION
<?php
// WITH MYSQLI PREPARED STATEMENT AGAINST SQL INJECTION
$sql = $conn->stmt_init();
$Lquery = "SELECT ExpensesId FROM tblexpensestype WHERE ExpensesId =?";
if($sql->prepare($Lquery)){
$sql->bind_param('i',$lExpensesId);
$sql->execute();
$sql->store_result();
if($sql->num_rows > 0){
echo 'YES';
}else{
echo "Proceed";
}
}
else
{
echo "Error".mysqli_connect_error();
}
?>
我希望你得到答案为你问题，如果你有其他疑问感到自由和意见如下
什么响应代码是你在你的网络控制台的故事吗？这应该始终是您第一站调试ajax，以验证请求是否发生以及它得到的响应。 – Taplar
我只收到“是”。其实，它应该显示YES和EMAIL –
我敢打赌，它只是肯定的。 – epascarello