我试图使用AJAX将数据从表单发送到PHP文件，以便存储在数据库中，但是我遇到了这个问题。我无法将数据发送到PHP文件。我不知道这是怎么回事
this is php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$db_name="menoon";
// Create connection
$conn = new mysqli($servername, $username, $password,$db_name);
// Check connection
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
$UserName=$_POST['username'];
$Email=$_POST['inputEmail3'];
$Password=md5($_POST['inputPassword3']);
$Address=$_POST['add'];
$fname=$_POST['name'];
$membership=$_POST['memberhhiip'];
if (filesize($_FILES['pic']['tmp_name']))
{
// cover_image is empty (and not an error)
$File=file_get_contents($_FILES['pic']['tmp_name']); //$_POST['file'];
$vaildMail="SELECT EMAIL From user_emails WHERE EMAIL='$Email'";
$r =mysqli_query($conn ,$vaildMail);
if(mysqli_num_rows($r)==0) //to check if there is an exist email
{
// echo "ID".$ID." ".$UserName." ".$Password."<br>";
$sql="INSERT INTO user (FULLNAME,HOMEADDRESS,UserName,PASSWORD,MEMBERSHIPTYPE,PROFILEPHOTO) values('$fname','$Address','$UserName','$Password','$membership',?)";
if($stmt = $conn->prepare($sql))
{ $stmt->bind_param("b",$_null);
$stmt->send_long_data(0,$File);
$stmt->execute();
$sql="SELECT ID FROM user WHERE UserName='$UserName'";
$s =mysqli_query($conn ,$sql);
$row = mysqli_fetch_row($s);
$sql=" INSERT INTO user_emails (USER_ID,EMAIL) values('$row[0]','$Email')";
$sr =mysqli_query($conn ,$sql);
{
ob_start();
header('Location:../index.php');
ob_end_flush();
die();
}
}
else {
echo "user name alredy exists";
}
}
else {echo "Email alredy exists";
}
}
else {$vaildMail="SELECT EMAIL From user_emails WHERE EMAIL='$Email'";
$r =mysqli_query($conn ,$vaildMail);
if(mysqli_num_rows($r)==0) //to check if there is an exist email
{
// echo "ID".$ID." ".$UserName." ".$Password."<br>";
$sql=" INSERT INTO user (FULLNAME,HOMEADDRESS,UserName,PASSWORD,MEMBERSHIPTYPE) values('$fname','$Address','$UserName','$Password','$membership')";
if ($s =mysqli_query($conn ,$sql))
{ $sql="SELECT ID FROM user WHERE UserName='$UserName'";
$s =mysqli_query($conn ,$sql);
$row = mysqli_fetch_row($s);
$sql=" INSERT INTO user_emails (USER_ID,EMAIL) values('$row[0]','$Email')";
$sr =mysqli_query($conn ,$sql);
{
ob_start();
header('Location:../index.php');
ob_end_flush();
die();
}
}
else
{ echo "user name alredy exists";
}
}
else {echo "Email alredy exists";}
}
?>
this is the html file including script
<div id="signup" class="modal fade" role="dialog">
<div class="modal-dialog">
<!-- Modal content-->
<div class="modal-content">
<div class="modal-header">
<h4 class="modal-title"SIGN UP</h4>
</div>
<div class="modal-body">
<form class="form-horizontal" role="form" id="#my_form" enctype="multipart/form-data">
<div class="form-group">
<div class="col-sm-10">
<input type="email" class="form-control"
name="inputEmail3" id="inputEmail3" placeholder="Email" required/>
</div>
</div>
<div class="form-group">
<div class="col-sm-10">
<input type="password" class="form-control"
name="inputPassword3" id="inputPassword3" placeholder="Password" required/>
</div>
</div>
<div class="form-group">
<div class="col-sm-10">
<input type="file" name="pic" class="" data-multiple-caption="{count} files selected" multiple />
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-2 col-sm-10">
<div class="checkbox">
<label>
<input type="checkbox"/> Remember me
</label>
</div>
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-2 col-sm-10">
<button type="submit" class="btn btn-primary " onclick="sendForm()">Sign Up</button>
</div>
</div>
</form>
</div>
<div class="modal-footer">
</div>
</div>
</div>
</div>
<script >
function sendForm(){
$.ajax({
method: "POST",
url: "signcontroller.php",
data: jQuery("#my_form").serialize(),
cache: false,
success: function(data){
alert(data);
}
});
}
</script>最佳答案
我看到的问题是，单击按钮而不是提交表单时，您正在运行脚本。由于您没有捕获要提交的表单，因此没有捕获任何值。尝试这样的方法：（取出格式名称为FIRST的主题标签，使其看起来像这样的id="my_form"）
$(document).on("submit", "#my_form", function(e){
// The e is the form submission event.
e.preventDefault(); //This prevents form from submitting normally
var formdata = $('#my_form').serialize(); //This will serialize the form data
$.ajax({
method: "POST",
url: "signcontroller.php",
data: formdata,
cache: false,
success: function(data){
alert(data);
}
});
});
关于javascript - Ajax没有将数据发送到php文件，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/42738624/