[英]Need help inserting data to database using php,ajax,javascript
Recently, i take some template from the internet and trying to understand the code. 最近,我从互联网上获取了一些模板,并试图理解代码。 But i got stuck when trying to insert data to the database and the error message wont show.
但是,当我尝试向数据库中插入数据时,我被卡住了,错误消息不会显示。 Im sorry for my bad English
对不起我的英语不好
This is my piece of, master_menu.php 这是我的master_menu.php
<div class="form-group">
<label for="first_name">Nama Menu</label>
<input type="text" id="first_name" placeholder="Contoh : Ayam Goreng" class="form-control"/>
</div>
<div class="form-group">
<label for="last_name">Harga Pokok</label>
<input type="text" id="last_name" placeholder="Contoh : 15000" class="form-control"/>
</div>
<div class="form-group">
<label for="email">Harga Jual</label>
<input type="text" id="email" placeholder="Contoh : 15000" class="form-control"/>
</div>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Batal</button>
<button type="button" class="btn btn-primary" onclick="addRecord()">Tambahkan Menu</button>
</div>
And this is my piece of, function_script_master.js 这是我的function_script_master.js
function addRecord() {
// get values
var first_name = $("#first_name").val();
var last_name = $("#last_name").val();
var email = $("#email").val();
// Add record
$.post("ajax/addRecord.php", {
first_name: first_name,
last_name: last_name,
email: email
}, function (data, status) {
// close the popup
$("#add_new_record_modal").modal("hide");
//reload
readRecords();
// clear fields from the popup
$("#first_name").val("");
$("#last_name").val("");
$("#email").val("");
});
}
// READ records
function readRecords() {
$.get("ajax/readRecords.php", {}, function (data, status) {
$(".records_content").html(data);
});
}
And this is my piece of, addRecord.php 这是我的addRecord.php
<?php
if(isset($_POST['first_name']) && isset($_POST['last_name']) && isset($_POST['email'])){
// include Database connection file
include("function_connection.php");
alert('clcicked');
// get values
$first_name = $_POST['first_name'];
$last_name = $_POST['last_name'];
$email = $_POST['email'];
$query = "INSERT INTO MENUS(NAMA_MENU, HARGA_POKOK, HARGA_JUAL, STATUS) VALUES('$first_name', '$last_name', '$email', 'aktif')";
if ($conn->query($query) === TRUE) {
alert("Registrasi Sukses!");
} else {
alert("Username Yang Anda Inginkan Sudah Terpakai");
}
$conn->close();
echo "1 Record Added!";
}
?>
Try below code hope this helps, i just replaced alert with echo. 尝试下面的代码希望对您有帮助,我只是将echo替换为echo。
<?php
if(isset($_POST['first_name']) && isset($_POST['last_name']) && isset($_POST['email'])){
// include Database connection file
include("function_connection.php");
// get values
$first_name = $_POST['first_name'];
$last_name = $_POST['last_name'];
$email = $_POST['email'];
$query = "INSERT INTO MENUS(NAMA_MENU, HARGA_POKOK, HARGA_JUAL, STATUS) VALUES('$first_name', '$last_name', '$email', 'aktif')";
if ($conn->query($query) === TRUE) {
echo "Registrasi Sukses!";
} else {
echo "Username Yang Anda Inginkan Sudah Terpakai";
}
$conn->close();
echo "1 Record Added!";
}
?>
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.