我有一个jQuery&AJAX脚本,该脚本正在抓取并将一些数据从HTML表传递到PHP脚本,然后将数据插入到我的表中...单击该按钮时,除检查数据库中的表是否为0。
startBalance1在HTML表中的值为1000。
HTML
<table class="forecast table table-striped" id="forecast" onload="myfunction()">
<tr><th scope="col">Month</th><th scope="col">Month Start</th><th scope="col">Investment</th><th scope="col">Return</th><th scope="col">Fee</th><th scope="col">Vat</th><th scope="col">Closing Balance</th><th scope="col">Month End</th></tr>
<tr><td>1</td><td class="date-forecast"></td>
<td><span>£</span><span id="startBalance1"></span></td>
<td><span>£</span><span id="simpleInt1"></span></td>
<td><span>£</span><span id="profitfee1"></span></td>
<td><span>£</span><span id="afterVAT1"></span></td>
<td><span>£</span><span id="amount1"></span></td>
<td class="date-forecast">Wed Jul 10 2019</td></tr>
<button type="button" name="submitfc" id="submitfc" class="btn btn-info mb-3" onclick="submitfc()">Save Forecast</button>
AJAX
function submitfc() {
var startBalance1 = $("startBalance1").val();
$.ajax({
url:"test-new.php",
type:"POST",
data:{startBalance_1:startBalance1},
success: function(data){
alert('success');
}
});
};
PHP
$connection = mysqli_connect("****", "
****", "****", "****");
if (!$connection) {
die("Database connection failed: " . mysqli_connect_error());
}
$startBalance_1 = $_POST['startBalance_1'];
$query = "insert into clientCashout(startBalance) values ('$startBalance_1')";
$result = mysqli_query($connection, $query);