当我将记录保存在表中时,它会成功地工作。
当更改预订号码并集中精力时,它也会起作用。
但当点击票号时,它只选择第一条记录,不适用于其他记录。在哪里解决问题
<script>
$(function() {
$("#BookingNo").on("focusout", function() {
$.ajax({
url: "ticket-booking-status.php",
type: "POST",
data: {
BookingNo: $("#BookingNo").val()
},
success: function(data) {
$("#txtShowData").html(data);
}
});
});
$("#SaveTicketBooking").on("click", function() {
$.ajax({
url: "ticketbooking-insert.php",
type: "POST",
data: {
SaveTicketBooking: "SaveTicketBooking",
BookingNo: $('#BookingNo').val(),
TicketNo: $('#TicketNo').val()
},
success: function(data) {
$("#txtSaveTicketBooking").html(data);
}
});
});
});
</script>
<div class="row">
<form onsubmit="return false" method="POST" class="form-horizontal" enctype="multipart/form-data">
<div class="col-lg-3 col-md-3 col-sm-3">
<label class="control-label" for="BookingNo">Booking No</label>
<input id="BookingNo" name="BookingNo" class="form-control" type="text" placeholder="" style="width: 100%;" value="<?php echo $BookingNo; ?>" autofocus required />
</div>
<div class="col-lg-3 col-md-3 col-sm-3">
<label class="control-label" for="TicketNo">Ticket No</label>
<input type="text" id="TicketNo" name="TicketNo" class="form-control input-md" maxlength="16" pattern=".{11,}" required />
</div>
<div class="col-lg-3 col-md-3 col-sm-3">
<label class="control-label"> </label>
<input type="submit" id="SaveTicketBooking" name="SaveTicketBooking" value="Save Booking" class="btn btn-save btn-block">
</div>
</form>
</div>
<div id="txtSaveTicketBooking"></div>
<div id="txtShowData"></div>
<script>
$(function() {
$("#EditTicketBookingShow").on("click", function() {
$.ajax({
url: "ticket-booking-edit.php",
type: "POST",
data: {
EditTicketBookingShow: "EditTicketBookingShow",
SrNoShow: $("#SrNoShow").val(),
TicketNoShow: $("#TicketNoShow").val(),
BookingNoShow: $("#BookingNoShow").val()
},
success: function(data) {
$("#txtEditData").html(data);
}
});
});
});
</script>
<?php if (isset($_POST['BookingNo'])) { $BookingNo = $_POST['BookingNo']; } ?>
<?php
$sql = "Select * from tblTicketBooking Where BookingNo = '$BookingNo' Order By SrNo";
$stmt=sqlsrv_query($conn, $sql);
?>
<div style="overflow-x: auto;">
<table id="viewreport" class="display" style="font-size:small">
<thead>
<tr>
<th></th>
<th>Ticket</th>
</tr>
</thead>
<tbody>
<?php
while($r=sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC))
{
$BookingNo = $r['BookingNo'];
$TicketNo = $r['TicketNo']; ?>
<tr>
<td>
<form onsubmit="return false" method="post" class="form-horizontal" enctype="multipart/form-data">
<input id="SrNoShow" name="SrNoShow" type="hidden" value="<?php echo $r['SrNo']; ?>">
<input id="BookingNoShow" name="BookingNoShow" type="hidden" value="<?php echo $BookingNo; ?>">
<input id="TicketNoShow" name="TicketNoShow" type="hidden" value="<?php echo $TicketNo; ?>">
<input type="submit" id="EditTicketBookingShow" name="EditTicketBookingShow" value="EDIT" class="btn btn-warning btn-xs">
</form>
</td>
<td>
<?php echo $TicketNo; ?>
</td>
</tr>
<?php } ?>
</tbody>
</table>
</div>
<script type="text/javascript">
$('#viewreport').removeClass('display').addClass('table table-bordered');
</script>
if (isset($_POST['EditTicketBookingShow']))
{
$BookingNo = $_POST['BookingNoShow'];
$SrNo = $_POST['SrNoShow'];
$TicketNo = $_POST['TicketNoShow'];
$getTicket = "Select * From tblTicketBooking Where TicketNo = '$TicketNo'"; $stmtTicket = sqlsrv_query($conn, $getTicket);
$rowTicket = sqlsrv_fetch_array($stmtTicket, SQLSRV_FETCH_ASSOC); ?>
<script>
document.getElementById("BookingNo").value = '<?php echo $rowTicket['
BookingNo '] ?>';
document.getElementById("TicketNo").value = '<?php echo $rowTicket['
TicketNo '] ?>';
</script>
<?php } ?>
1条答案
按热度按时间knsnq2tg1#
我认为您必须创建一个简单的编辑页面,例如,在编辑页面上使用get方法获取值,并像where booking no=“123”那样获取/更新
表代码