所以我正在构建一个网页,以及其中的所有内容,但有一件事情很好。 所以在我的“MDT”页面上,我选择一个状态(例如状态0,它发送“恐慌按钮”),并将其发送到数据库。当control.html检查数据库(每秒),并选择“状态”列说“紧急按钮”时,它会发出警报。当我的状态发生变化时,该警报就会结束。
我想要的是一个警报,一个弹出基本细节的模态表格(例如活动状态0,呼号,姓名等)
到目前为止:
<center><h2>Available Units</h2>
<div id="units">
<ul class="units">
<?php
$sql = "SELECT * FROM mdt WHERE CAD = 'NULL';";
$result = mysqli_query($link, $sql);
while ($row = mysqli_fetch_array($result)) {
echo '<li class="' . statusToClass(displayData($row['status'])) . '">' . displayData($row['callsign']) . ' - ' . displayData($row['name']) . ' <span class="glyphicon glyphicon-cog editUnitBtn" aria-hidden="true"></span><br/>Status: ' . displayData($row['status']) . '</li>';
}
?>
<?php
$sql = "SELECT * FROM mdt WHERE status = 'Panic Button';";
$result = mysqli_query($link, $sql);
if ($row = mysqli_fetch_array($result)) {
echo '<audio autoplay="false" src="panic.mp3">';
sleep(2);
}
?>
<?php
$sql = "SELECT * FROM mdt WHERE status = 'Panic Button';";
$result = mysqli_query($link, $sql);
if ($row = mysqli_fetch_array($result)) {
echo "<script>$('#panicModal').modal('show')</script>";
}
?>
然后靠近页面底部:
<div class="modal fade" id="panicModal" tabindex="-1" role="dialog">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
<h4 class="modal-title" id="myModalLabel">ATTENTION - ACTIVE STATE 0</h4>
</div>
<div class="modal-body">
<?php
$sql = "SELECT * FROM mdt WHERE status = 'Panic Button';";
$result = mysqli_query($link, $sql);
$row = mysqli_fetch_array($result);
?>
<label>Callsign</label>
<?php echo displayData($row['callsign']); ?>
<Label>Collar Number</label>
<?php echo displayData($row['name']); ?>
<Label>Last Known Location</label>
<?php
if (displayData($row['CAD']) == "NULL") {
echo ' <i>Currently unassigned</i>';
} else {
$cadid = displayData($row['CAD']);
$sql = "SELECT * FROM cads WHERE id = '$cadid';";
$result = mysqli_query($link, $sql2);
$row = mysqli_fetch_array($result2);
echo displayData($row['location']); } ?>
<Label>Call Type</label>
<?php
if (displayData($row['CAD']) == "NULL") {
echo ' <i>Currently unassigned</i>';
} else {
$cadid = displayData($row['CAD']);
$sql2 = "SELECT * FROM cads WHERE id = '$cadid';";
$result2 = mysqli_query($link, $sql2);
$row2 = mysqli_fetch_array($result2);
echo displayData($row['type']); } ?>
</div>
</div>
</div>
任何想法?在互联网的某个地方,它提到将jquery放在页面的顶部,但是当我这样做时它没有任何区别,取决于我把它放在哪里,停止所有js功能工作。然而,底部的jquery脚本,这一切都很好。
control.html:https://pastebin.com/9u6Su1uD control.js:https://pastebin.com/1FeXUN6b
非常感谢任何帮助。 TIA 泰勒