我有一个页面有一个表格。它显示选择选项时,选择一个选项用户单击按钮,它运行updatephp.php查询更新。我需要动态更新选择并在屏幕上显示“更新”或“无结果”等成功/错误消息,我该如何实现这一点。我不是很擅长阿贾克斯,请有人引导我。表格动态刷新php和ajax
displaytable.php
<form method="POST" action="choosecake.php">
<select id="bakeryid" name="bakeryid">
<option value="">Select</option>
<?php
$sql = "SELECT bakeryid, datefrom FROM cakes";
$sqlresult = $link->query($sql);
$sqllist = array();
if(mysqli_num_rows($sqlresult) > 0) {
while($row = mysqli_fetch_array($sqlresult))
{
echo "<option value=".$row['bakeryid'].">".$row['datefrom']."</option>";
}
$sqlencode = json_encode($sqllist);
echo $sqlencode;
} else {
echo 'No Results were found';
}
?>
</select>
<input type="hidden" value="<?php echo $bakeryid;?>" name="bakeryid"/>
<input type="submit" value="Submit" name="submit"/>
</form>
我在这里看不到ajax – cameronjonesweb 2015-02-08 23:45:07