[英]Ajax form submission inside modal box
好的,这是一个奇怪的小问题:
这是一个测试页面,用户单击该页面即可打开:
当用户单击查看结果时,模式框内有3个选择框。
box1 =>填充=> Box 2 =>填充Box 3
我的问题
当用户单击“提交”时,测试页将再次在模式框内打开,而不是根据选择框的选择显示查询结果,如下图所示
提交时
知道为什么在提交表单时当前页面会在modalbox中打开吗?
提交表格
<script type="text/javascript">
jQuery(document).click(function(e){
var self = jQuery(e.target);
if(self.is("#resultForm input[type=submit], #form-id input[type=button], #form-id button")){
e.preventDefault();
var form = self.closest('form'), formdata = form.serialize();
//add the clicked button to the form data
if(self.attr('name')){
formdata += (formdata!=='')? '&':'';
formdata += self.attr('name') + '=' + ((self.is('button'))? self.html(): self.val());
}
jQuery.ajax({
type: "POST",
url: form.attr("action"),
data: formdata,
success: function(data) { $('#resultForm').append(data); }
});
}
});
</script>
填充文本框
<script type="text/javascript">
$(document).ready(function()
{
$(".sport").change(function()
{
var id=$(this).val();
var dataString = 'id='+ id;
$.ajax
({
type: "POST",
url: "get_sport.php",
dataType : 'html',
data: dataString,
cache: false,
success: function(html)
{
$(".tournament").html(html);
}
});
});
$(".tournament").change(function()
{
var id=$(this).val();
var dataString = 'id='+ id;
$.ajax
({
type: "POST",
url: "get_round.php",
data: dataString,
cache: false,
success: function(html)
{
$(".round").html(html);
}
});
});
});
</script>
<label>Sport :</label>
<form method="post" id="resultForm" name="resultForm" action="result.php">
<select name="sport" class="sport">
<option selected="selected">--Select Sport--</option>
<?php
$sql="SELECT distinct sport_type FROM events";
$result=mysql_query($sql);
while($row=mysql_fetch_array($result))
{
?>
<option value="<?php echo $row['sport_type']; ?>"><?php echo $row['sport_type']; ?></option>
<?php
}
?>
</select>
<label>Tournamet :</label> <select name="tournament" class="tournament">
<option selected="selected">--Select Tournament--</option>
</select>
<label>Round :</label> <select name="round" class="round">
<option selected="selected">--Select Round--</option>
</select>
<input type="submit" value="View Picks" name="submit" />
</form>
<?php
显示结果
if(!empty($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest') {
echo $sport=$_POST['sport'];
echo $tour=$_POST['tournament'];
echo $round=$_POST['round'];
$sql="Select * FROM Multiple_Picks WHERE tournament ='$tour' AND round='$round' GROUP BY member_nr";
$result = mysql_query($sql);
?>
<?php
while($row=mysql_fetch_array($result)){
$memNr = $row['member_nr'];
$pick = $row['pick'];
$score = $row['score'];
?>
echo $memNr;
echo $pick;
echo $score;
}
}
?>
看来:
success: function(data) { $('#resultForm').append(data); }
success: function(data) { $('#resultForm').append(data); }
您告诉它将ajax响应放入resultForm中,该结果似乎在您的模式内部。 那不是正在发生的事情吗? 从您的问题和代码中很难说出应该发生什么与现在正在发生什么。
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