我想将用户输入的pdf文件上传到我的mysql数据库中。我提供的代码显示了带有pdf上载输入类型的用户注册表。我想通过AJAX将其发布到我的PHP脚本中，以输入到数据库中。我无法使此代码正常工作。任何建议，将不胜感激？
表单输入的摘要-
<form id='form1'>
<label>Are you a:</label>
<select name="type" id="type">
<option>Select...</option>
<option value="teacher">Teacher</option>
<option value="school">School</option>
</select><br>
<label>Teaching Council Number / School Roll
Number</label>
<input type="text" name="userNo" id="userNo"
value="" required><br>
<label>Full Name / School Name</label>
<input type="text" name="name" id="name" required>
<br>
<label>Level:</label>
<select name="level" id="level">
<option>Select...</option>
<option value="primary">Primary</option>
<option
value="secondary">Secondary</option>
</select><br>
<label>Phone Number</label>
<input type="number" name="phoneNo" id="phoneNo"
data-clear-btn="false" pattern="[0-9]*" value=""><br>
<label>Address</label>
<input type="text" name="location" id="location"
placeholder="" required><br>
<label>Email</label>
<input type="email" id="email" data-clear-
btn="false" required><br>
<label for="password">Password</label>
<input type="password" name="password"
id="password" value="" required><br>
<label for="password">Confirm Password</label>
<input name="password_confirm" required="required"
type="password" id="password_confirm" oninput="check(this)" /><br>
<label for="file">Curriculum Vitae <i>(PDF only)
</i></label>
<input type="file" name="file" id="cv" value="cv"
accept=".pdf"><br>
<label for="file">Garda Vetting <i>(PDF only)</i>
</label>
<input type="file" name="gardavetting"
id="gardavetting" value="gardavetting" accept=".pdf"><br>
<label>LinkedIn URL <i>(optional)</i></label>
<input type="text" id="linkedin" name="linkedin">
<br>
<button type="submit" id="submit1" name="submit1"
onclick="myFunction1();">Submit</button><br>
</form>
</div>
PHP-
<?php
// Selecting Database
include_once 'dbh.php';
//Here we fetch the data from the URL that was passed from our HTML
form
$type2 = $_POST['type'];
$userNo2 = $_POST['userNo'];
$name2 = $_POST['name'];
$level2 = $_POST['level'];
$phoneNo2 = $_POST['phoneNo'];
$location2 = $_POST['location'];
$email2 = $_POST['email'];
$password2 = $_POST['password'];
$cv2 = $_POST['cv'];
$gardavetting2 = $_POST['gardavetting'];
$linkedin2 = $_POST['linkedin'];
$sql = "INSERT INTO users (type, userNo, name, level, phoneNo,
location, email, password, cv, gardavetting, linkedin) VALUES
('$type2', '$userNo2', '$name2',
'$ level2'，'$ phoneNo2'，'$ location2'，'$ email2'，'$ password2'，'$ cv2'，'$ gardav etting2'，'$ linkedin2'）;“;
mysqli_query($conn, $sql);
?>
JS-
function myFunction1() {
var type = document.getElementById("type").value;
var userNo = document.getElementById("userNo").value;
var name = document.getElementById("name").value;
var level = document.getElementById("level").value;
var phoneNo = document.getElementById("phoneNo").value;
var location = document.getElementById("location").value;
var email = document.getElementById("email").value;
var password = document.getElementById("password").value;
var cv = document.getElementById("cv").value;
var gardavetting = document.getElementById("gardavetting").value;
var linkedin = document.getElementById("linkedin").value;
var dataString = '&type=' + type + '&userNo=' + userNo + '&name=' +
name + '&level=' + level + '&phoneNo=' + phoneNo + '&location=' +
location + '&email=' + email+ '&password=' + password + '&cv=' + cv +
'&gardavetting=' + gardavetting + '&linkedin=' + linkedin ;
if ( type== '' || userNo == '' || name == '' || level == '' ||
phoneNo == '' || location == '' || email == '' || password == '' || cv
== '' || gardavetting == '')
{
alert("Please Fill All Fields");
}
else
{
//AJAX code to submit form.
$.ajax({
type: "POST",
url: "http://localhost:8888/EduSubOct/signup.php",
data: dataString,
cache: false,
success: function(html) {
alert("Information Entered Successfully");
}
});
}
return false;
答案 0 :(得分：0)
尝试使用此代码进行文件存储和插入
<!DOCTYPE html>
<html>
<head>
<title>test</title>
</head>
<body>
<form id="form">
<label for="file">Curriculum Vitae <i>(PDF only)</i></label>
<input type="file" name="file" id="cv" value="cv" accept=".pdf"><br>
<label for="file">Garda Vetting <i>(PDF only)</i></label>
<input type="file" name="gardavetting" id="gardavetting" value="gardavetting" accept=".pdf"><br>
<label>LinkedIn URL <i>(optional)</i></label>
<input type="text" id="linkedin" name="linkedin"> <br>
<button type="submit" id="submit1" name="submit1">Submit</button><br>
</form>
</body>
<script
src="https://code.jquery.com/jquery-3.3.1.min.js"
integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8="
crossorigin="anonymous"></script>
<script>
$('#form').on('submit', function(e) {
e.preventDefault();
var form=document.getElementById('form');
var fdata=new FormData(form);
$.ajax({
type: "POST",
url: 'insert.php',
data: fdata,
contentType: false,
cache: false,
processData:false,
success: function(result)
{
if(result == 0)
{
alert('file stored');
}else{
alert('something went wrong');
}
}
});
});
</script>
</html>
insert.php
<?php
$servername = "host";
$username = "server database username";
$password = "server database password";
$dbname = "your db name";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$type2 = $_POST['type'];
$userNo2 = $_POST['userNo'];
$name2 = $_POST['name'];
$level2 = $_POST['level'];
$phoneNo2 = $_POST['phoneNo'];
$location2 = $_POST['location'];
$email2 = $_POST['email'];
$password2 = $_POST['password'];
$gardavetting2 = $_POST['gardavetting'];
$linkedin2 = $_POST['linkedin'];
// STORE PDF FILE IN FOLDER
if(isset($_FILES['file']['name']))
{
$cpath="resume/";
$file_parts = pathinfo($_FILES["file"]["name"]);
$file_path = 'resume'.time().'.'.$file_parts['extension'];
move_uploaded_file($_FILES["file"]["tmp_name"], $cpath.$file_path);
$cv2 = $file_path;
}
$sql = "INSERT INTO users (type, userNo, name, level, phoneNo,
location, email, password, cv, gardavetting, linkedin) VALUES('$type2', '$userNo2', '$name2', '$level2','$phoneNo2','$location2','$email2','$password2','$cv2','$gardav etting2','$linkedin2');";
if($sql){
echo '0';
}
mysqli_query($conn, $sql);
?>
如果存储任何文件（例如（图像，pdf，视频）），请使用表格序列化方法或表格数据方法来避免错误。在这里，我给出了一个代码，用于将pdf文件存储在本地文件夹中，并使用AJAX插入mysql数据库中，而无需刷新页面即可得到结果。