make the debugger submit using ajax

This commit is contained in:
Alyssa Milburn 2014-07-29 00:51:00 +02:00
parent 79dc0cc376
commit 43faa22ede
1 changed files with 30 additions and 3 deletions

View File

@ -33,18 +33,45 @@
}
</style>
<script type="application/javascript" src="designer.js"></script>
<script src="//code.jquery.com/jquery-1.11.0.min.js"></script>
<script type="application/javascript">
window.onload = function() { design(); update(); }
window.onload = function() { design(); update(); };
function onSubmit() {
$('#submit-feedback').html("Submitting...");
// http://stackoverflow.com/questions/169506/obtain-form-input-fields-using-jquery
var values = {};
$.each($('#form').serializeArray(), function(i, field) {
values[field.name] = field.value;
});
var request = $.ajax({
accept: "text/plain",
url: "upload.cgi",
type: "POST",
dataType: "text",
data: values
});
request.done(function(msg) {
// TODO: red on error?
$('#submit-feedback').html(msg);
});
request.fail(function(jqXHR, status) {
// TODO: red on error?
$('#submit-feedback').html(jqXHR.responseText);
});
}
</script>
</head>
<body style="background: #222; color: #bbb;">
<h1>Tank Debugger</h1>
<div id="game_box"><canvas id="battlefield" style="border: 2px solid green;" width="450" height="450"></canvas></div>
<form action="designer.cgi" method="post">
<p>
<input type="button" onclick="resetTanks();" value="Run"> (Remember to submit (below) when you're done!)
</p>
<p id="debug"></p>
<form id="form">
<div id="stuff">
<div id="code">
<textarea id="program" name="program" rows="20" cols="80">get-turret 12 + set-turret! ( Rotate turret )
@ -53,7 +80,7 @@
1 sensor? { -50 50 set-speed! } if ( Turn if collision sensor triggered )
</textarea><br>
<fieldset id="submit">
<p><input type="submit" value="Submit"></p>
<p><input type="button" value="Submit" onclick="onSubmit();"> <span id="submit-feedback"> </span></p>
<p>
Before you can get going with a tank, you need a password. If you