</div>
</div>
</div>
<hr class="my12 outline-none baw0 bb bc-powder-2">
<div class="grid fw-nowrap fc-black-600">
<div class="grid--cell mr8">
<svg aria-hidden="true" class="svg-icon iconLightbulb" width="18" height="18" viewbox="0 0 18 18"><path d="M9.5.5a.5.5 0 0 0-1 0v.25a.5.5 0 0 0 1 0V.5zm5.6 2.1a.5.5 0 0 0-.7-.7l-.25.25a.5.5 0 0 0 .7.7l.25-.25zM1 7.5c0-.28.22-.5.5-.5H2a.5.5 0 0 1 0 1h-.5a.5.5 0 0 1-.5-.5zm14.5 0c0-.28.22-.5.5-.5h.5a.5.5 0 0 1 0 1H16a.5.5 0 0 1-.5-.5zM2.9 1.9c.2-.2.5-.2.7 0l.25.25a.5.5 0 1 1-.7.7L2.9 2.6a.5.5 0 0 1 0-.7z" fill-opacity=".4"></path><path opacity=".4" d="M7 16h4v1a1 1 0 0 1-1 1H8a1 1 0 0 1-1-1v-1z" fill="#3F3F3F"></path><path d="M15 8a6 6 0 0 1-3.5 5.46V14a1 1 0 0 1-1 1h-3a1 1 0 0 1-1-1v-.54A6 6 0 1 1 15 8zm-4.15-3.85a.5.5 0 0 0-.7.7l2 2a.5.5 0 0 0 .7-.7l-2-2z" fill="#FFC166"></path></svg>
</div>
<div class="grid--cell lh-md">
<p class="mb0">
<b>Want to improve this question?</b> <a href="/posts/32894241/edit">Update the question</a> so it's <a href="/help/on-topic">on-topic</a> for Stack Overflow.
</p>
<p class="mb0 mt6">Closed <span title="2015-10-02 01:08:47Z" class="relativetime">4 years ago</span>.</p>
</div>
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</aside>
i try to insert from html to mysql (Phonegap). But he don´t make that... What is my mistake?. Sry for my bad englisch and thx for helping... I have ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ HTML:
<div id="landmark-1" data-landmark-id="1">
<form>
<div id="block1">Datum:<input type="date" id="date" name="date"></div>
<div id="block2">Baustelle: <input name="bau" id="bau" class="tb1" type="text" size="18" /></div>
<div id="block4">
<input type="submit" value="Speichern">
</div>
</form>
</div>
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Post.js
$(document).bind('deviceready', function(){
$(function(){
$('form').submit(function(){
var landmarkID = $(this).parent().attr('data-landmark-id');
var postData = $(this).serialize();
alert(postData+'&lid='+landmarkID);
$.ajax({
type: 'POST',
data: postData+'&lid='+landmarkID,
//change the url for your project
url: "http://voss-hm.de/zeitserver/save.php",
success: function(data){
console.log(data);
alert('Your comment was successfully added');
},
error: function(){
console.log(data);
alert('There was an error adding your comment');
}
});
return false;
});
});
});
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ PHP:
<?php
$server = "--------------";
$username = "--------------";
$password = "--------------";
$database = "--------------";
$con = mysql_connect($server, $username, $password) or die ("Could not connect: " . mysql_error());
mysql_select_db($database, $con);
$id = $_Post["id"];
$date = ($_Post["date"]);
$baustelle = ($_Post["bau"]);
$sql = "INSERT INTO timesave.$id (datum, baustelle, anfang, ende, pause) VALUES ('$date' , '$baustelle', '', '', '')";
if (!mysql_query($sql, $con)) {
die('Error: ' . mysql_error());
} else {
echo "Comment added";
}
mysql_close($con);
?>
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
</div>
$_POST
not $_Post
mysql_
-functions! Use mysqli_
or PDO instead.There are a few mistakes here.
First, use $_POST instead of $_Post (case matters).
Second, you should not use mysql_* function anymore. Please read this tutorial : https://www.binpress.com/tutorial/using-php-with-mysql-the-right-way/17
Your script is vulnerable to a lot of attacks right now, you really need to learn to escape POST values. Your script will also not work if you have single quote in your posted values. The link below will teach you how to handle this.
Returning JSON-encoded objects from your PHP script would also be a good idea.