@J Salaria我瞭解你與jquery AJAX和PHP代碼有問題，因爲你沒有得到你想要的結果你的問題。有不同的方式通過jquery ajax發送數據，我會詳細解釋。 ？
$_POST["lExpensesId"]，您在一個HTML <form>得到這個ID。因爲在這裏，我會向您展示3點不同的熟練的方式通過Ajax來發送數據..
注意：你的代碼很容易受到SQL INJECION。 I'LL也向您展示的方法來OVERCOME.IF你想了解更多有關SQL注入點擊這個LINK SQL INJECTION LINK
HTML表單代碼：
<form action="" id="send_lExpensesId_form" method="post">
<input type="text" name="lExpensesId" id="lExpensesId" >
<input type="submit" name="submit" >
</form>
第一種方式用於發送數據THIS IS雖然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>
第二種方法用於發送數據不過這是在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>
用於發送數據的第三方式使用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>
這與mysqli_real_escape_string(); PHP代碼針對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();
}
}
?>
這其他PHP代碼MYSQLI->PREPARED這是更好地與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();
}
?>
我希望你得到答案爲你問題，如果你有其他疑問感到自由和意見如下
什麼響應代碼是你在你的網絡控制檯的故事嗎？這應該始終是您第一站調試ajax，以驗證請求是否發生以及它得到的響應。 – Taplar
我只收到「是」。其實，它應該顯示YES和EMAIL –
我敢打賭，它只是肯定的。 – epascarello