我正在尝试使用bootstrap的警报。特别是:
private final static long TIMEOUT = 10000;
private Handler mHandler;
final Lock lock = new ReentrantLock();
final Condition condition = lock.newCondition();
@Override
protected void onCreate(Bundle savedInstanceState) {
...
mHandler = new Handler();
...
}
private void timeOutSignal() {
mHandler.postDelayed(new Runnable() {
@Override
public void run() {
Log.d(">> ", "---> timeout notify");
lock.lock();
try {
condition.signal(); // releases lock and waits until doSomethingElse is called
} finally {
lock.unlock();
}
}
}, TIMEOUT);
}
public void buttonClick(View view) {
timeOutSignal();
Log.i("???", "... WAIT");
lock.lock();
try {
condition.await();
} catch (InterruptedException e) {
// todo
} finally {
lock.unlock();
}
Log.i("???", "... WAIT DONE !");
}
我想在成功点击提交按钮后弹出模态。我该怎么做?
以下是我的代码:
<div class="alert alert-success" role="alert">...</div>
这是我的PHP:
<!--____________________________ADD AGENT________________________-->
<div class="modal fade" id="addAgent" tabindex="-1" role="dialog" aria-labelledby="myModalLabel">
<div class="modal-dialog" role="document">
<form role="form" action="php/addAgent.php" method="POST">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">×</span></button>
<h4 class="modal-title" id="myModalLabel">Add Agent</h4>
</div>
<div class="modal-body">
<div class="row">
<div class="col-sm-12">
<div class="form-group">
<label for="fullname">Full Name</label>
<div class="row">
<div class="col-sm-4">
<input type="text" class="form-control" placeholder="First Name" name="fname">
</div>
<div class="col-sm-4">
<input type="text" class="form-control" placeholder="Middle Name" name="mname">
</div>
<div class="col-sm-4">
<input type="text" class="form-control" placeholder="Last Name" name="lname">
</div>
</div>
</div><!--___________FORM GROUP_____________-->
<div class="form-group">
<div class="row">
<div class="col-sm-4">
<label for="sel1">Type:</label>
<select class="form-control" name="agentType" id="sel1">
<option value="1">Broker</option>
<option value="2">Agent</option>
<option value="3">Sub-Agent</option>
</select>
</div>
<div class="col-sm-4">
<label for="sel1">Project:</label>
<select class="form-control" id="sel1">
<option>Mezza</option>
<option>Tivoli Gardens</option>
<option>Verawoods Residences</option>
</select>
</div>
</div>
</div>
<div class="form-group">
<div class="row">
<div class="col-sm-6">
<label for="email">Email Address</label>
<input type="email" class="form-control" name="email" id="email">
</div>
<div class="col-sm-4">
<label for="contact">Contact Number</label>
<input type="text" class="form-control" name="contact" id="contact">
</div>
</div>
</div>
<div class="form-group">
<div class="row">
<div class="col-sm-12">
<label for="homeAdd">Home Address</label>
<input type="text" class="form-control" name="homeAdd" id="homeAdd">
</div>
</div>
</div>
</form>
</div>
</div> </div>
<div class="modal-footer">
<input type="submit" class="btn btn-primary" value="submit"/>
<button type="button" class="btn btn-default" data-dismiss="modal"> Close</button>
</div>
</div>
</div>
</div><!--______________________ADD AGENTS MODAL_______________________-->
<!-- Button trigger modal -->
我在数据库中成功添加值后如何做到这一点?
答案 0 :(得分:0)
如果您没有使用AJAX来调用addAgent.php
并且您的PHP和HTML在一个文件中,那么一种方法是在PHP中设置变量,如下所示:
if ($insert) {
$alert_success = '<div class="alert alert-success" role="alert">...</div>';
}
然后在HTML中的任何位置添加一条回显该变量的行。
<?php
echo $alert_success;
?>
答案 1 :(得分:0)
您可以为表单使用提交处理程序:
// Add an ID to your form
$( "#formID" ).submit(function( event ) {
// Stop form from submitting normally
event.preventDefault();
// Get some values from elements on the page:
var $form = $( this ),
term = $form.find( "input[name='s']" ).val(),
url = $form.attr( "action" );
// Send the data using post
var posting = $.post( url, { s: term } );
// Put the results in a div
posting.done(function( data ) {
$(".modal-body, #orModalBodyID"). append('<div class="alert alert-success" role="alert">Post Success</div>')
});
});
中找到有关jQuery的post()方法的更多信息