Question

<form action="" method='POST'> 
<input  type='submit'  name='voted' value='Like'>
 <input  type='submit' name='votes' value='Dislike'>
</form>

<?php
$connect = mysqli_connect ("localhost", "root", "muarem1") or die ("Couldn not connect!");
mysqli_select_db ($connect, "likesystem") or die ("Couldn't find db");
$result = mysqli_query($connect,"SELECT * FROM fyci WHERE id=1");
while ($row = mysqli_fetch_assoc($result)) {
$voted = $row["voted"];
}
if (isset($_POST['voted'])){
$votes = $voted+1;
$update = mysqli_query($connect,"UPDATE fyci SET voted='$votes' WHERE id=1");
}else 
if (isset($_POST['votes'])){
$dislike = $voted-1;
$update = mysqli_query($connect,"UPDATE fyci SET voted='$dislike' WHERE id=1");
}
echo "Likes: $voted";
?>

提交表格并保持同一页面？谢谢！

如果点击提交，我想留在同一页面上。

应该怎么做？

Answer 1

如果您想留在页面上，请使用AJAX通话。有关示例，请参阅jQuery.post()。

Answer 2

您可以使用$ .post函数作为jquery

的选项提供程序之一

像

 $.post('name_of_your_file.php', $('form').serialize());

或

$.post('<?php echo $_SERVER["PHP_SELF"];?>', $('form').serialize());

Answer 3

使用AJAX你可以做到。我修改了一些HTML代码。添加了表单名称。

<强> HTML：

<form name="form1" action="vote.php" method='POST'>
    <input  type='submit'  name='voted' value='Like'>
    <input  type='submit' name='votes' value='Dislike'>
</form>

<强> JS：

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8/jquery.min.js"></script>
<script type="text/javascript">
var selector = $('form[name=form1]');
selector.submit(function(e){
    e.preventDefault();
    $.ajax({
        url : selector.attr('action'),
        data: selector.serialize(),
        type: selector.attr('method'),
        success: function(data){
            //alert(data);
        }
    });
});
</script>

<强> vote.php：

<?php
$connect = mysqli_connect ("localhost", "root", "muarem1") or die ("Couldn not connect!");
mysqli_select_db ($connect, "likesystem") or die ("Couldn't find db");
$result = mysqli_query($connect,"SELECT * FROM fyci WHERE id=1");
while ($row = mysqli_fetch_assoc($result)) {
    $voted = $row["voted"];
}
if (isset($_POST['voted'])){
    $votes = $voted+1;
    $update = mysqli_query($connect,"UPDATE fyci SET voted='$votes' WHERE id=1");
}else
    if (isset($_POST['votes'])){
        $dislike = $voted-1;
        $update = mysqli_query($connect,"UPDATE fyci SET voted='$dislike' WHERE id=1");
}
echo "Likes: $voted";

提交表单而不刷新

3 个答案: