[英]How to get rid from the same id in while loop when submitting form using ajax
我在while循环中使用ajax提交表单,但是由于循环,相同的表单id被使用了很多次,因此表单只提交了一次。 我认为我每次都要在表单循环中创建唯一的ID,但不知道如何。
到目前为止,这是我的代码,
<?php
$get_cmt ="SELECT * FROM comments WHERE post_id = $post_id ORDER BY id DESC";
$query_cmt = mysqli_query($db_conx,$get_cmt);
while($row_cmt=mysqli_fetch_array($query_cmt,MYSQLI_ASSOC)){
$comtr_id = $row_cmt['comtr_id'];
$comment_id = $row_cmt['id'];
?>
<form id="subcmt_smt" method="post">
<textarea name="subcmt"></textarea>
<input type="hidden" value="<?php echo $comment_id;?>" name="comment_id">
<input type="hidden" value="<?php echo $pager_id;?>" name="comtr_id">
</form>
<?php } ?>
<script src="jQuery v2.1.1"></script>
<script>
$("#subcmt_smt").submit(function(e) {
var form = $(this);
var url = form.attr('action');
e.preventDefault();
$.ajax({
type: "POST",
url: "submit_subcmt.php",
data: form.serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
}
});
});
</script>
submit_subcmt.php
<?php
$comtr_id =$_POST['comtr_id'];
$comment_id =$_POST['comment_id'];
echo $comtr_id;
echo $comment_id;
?>
为了说明我在上面所做的评论,您可以尝试与此类似的操作。
<?php
$get_cmt ="SELECT * FROM comments WHERE post_id = $post_id ORDER BY id DESC";
$query_cmt = mysqli_query($db_conx,$get_cmt);
while( $row_cmt=mysqli_fetch_array($query_cmt,MYSQLI_ASSOC) ){
$comtr_id = $row_cmt['comtr_id'];
$comment_id = $row_cmt['id'];
?> <!-- use a class attribute here -->
<form class="subcmt_smt" method="post">
<textarea name="subcmt"></textarea>
<input type="hidden" value="<?php echo $comment_id;?>" name="comment_id">
<input type="hidden" value="<?php echo $pager_id;?>" name="comtr_id">
</form>
<?php
}//end loop
?>
<script src="jQuery v2.1.1"></script>
<script>
/* and assign event handlers to form objects with this class as per above */
$("form.subcmt_smt").submit(function(e) {
var form = $(this);
var url = form.attr('action');
e.preventDefault();
$.ajax({
type: "POST",
url: "submit_subcmt.php",
data: form.serialize(),
success: function(data) {
alert(data);
}
});
});
</script>
尝试这个。
<?php
$get_cmt ="SELECT * FROM comments WHERE post_id = $post_id ORDER BY id DESC";
$query_cmt = mysqli_query($db_conx,$get_cmt);
while($row_cmt=mysqli_fetch_array($query_cmt,MYSQLI_ASSOC)){
$comtr_id = $row_cmt['comtr_id'];
$comment_id = $row_cmt['id'];
?>
<form class="subcmt_smt" method="post">
<textarea name="subcmt"></textarea>
<input type="hidden" value="<?php echo $comment_id;?>" name="comment_id">
<input type="hidden" value="<?php echo $pager_id;?>" name="comtr_id">
</form>
<?php } ?>
<script src="jQuery v2.1.1"></script>
<script>
$(".subcmt_smt").submit(function(e) {
var form = $(this);
var url = form.attr('action');
e.preventDefault();
$.ajax({
type: "POST",
url: "submit_subcmt.php",
data: form.serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
}
});
});
</script>
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