Greetings,
I have the following jquery mobile page:
<div data-role="content">
<center>
<img src="./images/logo320.png" />
</center>
<br><br>
<form method="POST" action="./ajax/login.php">
<label for="login_unameLabel">Username:</label><br>
<input type="text" name="login_uname" id="login_uname" /><br>
<label for="login_pwordLabel">Password:</label><br>
<input type="password" name="login_pword" id="login_pword" /><br>
<button id="login_submit" type="submit" data-theme="a">Submit</button>
</form>
</div>
./ajax/login.php returns either "OK" or "NOK". How can I capture this result in my page?
I keep getting the following error in Firebug:
k.data( [Break On This Error] false)a.mobile.activePage=k;h(k);y&&D&...dd(k).removeClass(" out in reverse "+
It's as if jquery mobile is performing some operation on the result? I don't want this to happen. Do I have to return a valid jquery mobile HTML page from my PHP?
Any insight greatly appreciated.
Many thanks in advance,
Solution: use <input type="button" id="login_submit" data-theme="a" value="Submit" />
Now I can capture click events via:
<script>
$(document).ready(function() {
$("#login_submit").click(function() {
alert('clicked');
});
});
</script>
using data-ajax="false"
in form.
then you can disable jquery mobile auto ajax call
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