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我想用PHP中的ajax使用模态更新数据。我很开心,请告诉我我的错误在哪里。如何使用modal与ajax,php和mysql更新数据?
HTML代码
<button type="submit" class="btn btn-success" data-toggle="modal" data-cphone='.$row['country_phon'].' data-cname='.$row['country_name'].' >Update</button>';
<div id="myModal" class="modal fade" tabindex="-1" role="dialog">
<div class="modal-dialog">
<!-- Modal content-->
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal">×</button>
<h4 class="modal-title">Update Country</h4>
</div>
<form action="" method="post">
<div class="modal-body">
<input type="text" id="cn" name="pcountry">
</div>
<div class="modal-body">
<input type="text" id="cph" name="pphone">
</div>
<div class="modal-footer">
<button type="submit" class="btn btn-primary updt" name="updatecountry" >Save</button>
JS代码:在这里,当我点击按钮,模式会出现与使用jQuery的帮助数据库中获取的值。
<script>
$(document).ready(function(){
$(".btn").click(function(){
var cphone =$(this).data('cphone');
var cname = $(this).data('cname');
$("#cph").val(cphone);
$("#cn").val(cname);
$("#myModal").modal('show');
});
});</script>
JS代码:这是ajax代码。我想更新文本框的文本。
<script>
$(".updt").click(function(e){
e.preventDefault();
var cnt = $("#myModal").find("input[name='pcountry']").val();
var cp = $("#myModal").find("input[name='pphone']").val();
$.ajax({
dataType: 'json',
type:'POST',
url: 'test.php',
data:{pcountry:cnt, pphone:cp},
})
)};
</script></div>
PHP代码:
<?php
if(isset($_POST['updatecountry']))
{$country1 = $_POST['pcountry'];
$phone1 = $_POST['pphone'];
echo $country1;
echo $phone1;
echo "Updated Successfully";
mysqli_query($conn,"update country set country_name='$country1' , country_phon='$phone1' where country_id=18");
mysqli_close($conn);
}?>
此脚本正常工作。问题是在阿贾克斯部分 – kapildevsharma
@ kapildevsharma你可以显示错误的Ajax –