如何將datepicker
日期複製到PHP變量$adateStringValue
?使用jQuery datepicker和PHP時未定義的變量錯誤
當我使用「$adateStringValue = $_POST['datepicker'];
」時,我得到Undefined variable: adateStringValue error
。我該如何糾正這個問題?
以下是我的代碼。
要選擇日期:
<script>
$(document).ready(function() {
$("#datepicker").datepicker({
dateFormat: "dd-mm-yy"
});
$('form#dateform').submit(function(){
var aselectedDate = $('#datepicker').val();
localStorage.setItem('adate', aselectedDate);
});
});
</script>
abc.php:
<?php
echo "<form id=\"dateform\" name=\"dateform\" action=\"associate.php\" method=\"POST\"><br><br>
<table>
<tr><td>
<b>Select date<b></td><td>
<input id=\"datepicker\" name=\"datepicker\"value=\"$adateStringValue\"/>
</td></tr>
</table>
?>
我需要檢查這個選擇的日期排序,列在數據庫中(此列中的字符串格式日期) 。
if(isset($_POST['submit']))
{
$sql1="SELECT event_date from events_schedule";
$getDates_query= mysql_query($sql1,$con);
$adateStringValue = $_GET['datepicker'];
while($fetchdates = mysql_fetch_array($getDates_query)) {
if ($fetchdates['event_date'] == $adateStringValue) {
$message = "You have selected this date";
echo "<script>alert('$message');</script>";
}
else {
$message1 = "Select another date";
echo "<script>alert('$message1');</script>";
}
}
}
完整的代碼如下:
<link href="http://code.jquery.com/ui/1.10.4/themes/ui-lightness/jquery-ui.css" rel="stylesheet">
<script src="http://code.jquery.com/jquery-1.10.2.js"></script>
<script src="http://code.jquery.com/ui/1.10.4/jquery-ui.js"></script>
<script>
$(document).ready(function() {
$("#datepicker").datepicker({
dateFormat: "dd-mm-yy"
});
$('form#dateform').submit(function(){
var aselectedDate = $('#datepicker').val();
localStorage.setItem('adate', aselectedDate);
});
});
</script>
<?php
$con = mysql_connect("localhost","root","");
if(!$con){
die("Cannot connect:" .mysql_error());
}
mysql_select_db("trainingschedule",$con);
echo "<form id=\"dateform\" name=\"dateform\" action=\"associate.php\"method=\"POST\"><br><br>
<table>
<tr><td>
<b>Select a date <b></td><td><input id=\"datepicker\" name=\"datepicker\" size=\"15\" value=\"$adateStringValue \" /></td></tr>
</table>
<br>
<input type=\"submit\" name=\"submit\" value=\"Submit\"/>;
</form>";
if(isset($_POST['submit']))
{
$sql1="SELECT event_date from events_schedule";
$getDates_query= mysql_query($sql1,$con);
$adateStringValue = $_POST['datepicker'];
while($fetchdates = mysql_fetch_array($getDates_query)) {
if ($fetchdates['event_date'] == $adateStringValue) {
$message = "You have selected this date";
echo "<script>alert('$message');</script>";
}
else {
$message1 = "Select another date";
echo "<script>alert('$message1');</script>";
}
}
}
mysql_close($con);
?>
形式方法後,嘗試$ _ POST [「名」] – Rayon
@RayonDabre我嘗試使用POST。它並沒有解決問題。 – newbie
您在datepicker上的'name'和'value'屬性之間缺少一個空格。 –