我想上传一个图像与其他一些变量和表单提交,做我的php代码保存图像的用户profile_picture
表。AJAX上传图像提交
我想以与保存数据更改相同的形式上传图像。
这是什么样子的图片,所以你可以有一个更好的理解:
我做到了,我用的POST方法之前,但对于AJAX我不知道怎么样去做吧。
我的JavaScript代码(注意,这是不行的,我只是想有一个去 - 它在执行console.log返回Illegal invocation
):
<script>
function updateMyAccount() {
var fd = new FormData($("#fileinfo"));
var password = document.getElementById("myAccountNewPassword").value;
var profilePicture = document.getElementById("myAccountNewProfilePic").value;
$.ajax({
type: "POST",
url: "includes/form_submit.php",
data: {
SaveAccountChanges: true,
securePassword_Val: password,
fd
},
success: function(msg){
if(msg == 1) {
update_myAccount_success();
} else {
general_error_forms();
}
},
});
return false;
}
</script>
我的PHP代码:
//My account AJAX POST
if(($_POST['SaveAccountChanges']) == true & isset($_POST['securePassword_Val']))
{
$member_config->doUpdateAccountInfo($con);
}
然后我的功能,上传的图片,并将其保存到数据库:
function doUpdateAccountInfo($con)
{
//Upload users image to our /uploads directory
$uploaddir = 'uploads/';
$uploadfile = $uploaddir . basename($_FILES['fileToUpload']['name']);
$save_to_database = ("uploads/" . $_FILES["fileToUpload"]["name"]);
$normalPassword = mysqli_real_escape_string($con, $_POST["securePassword_Val"]);
$pwd = password_hash($normalPassword, PASSWORD_DEFAULT);
$username = $_SESSION["username"];
if(!empty($_FILES['fileToUpload']) & !empty($_POST['securePassword_Val']))
{
if(move_uploaded_file($_FILES['fileToUpload']['tmp_name'], $uploadfile)) {} else { die('Could not upload file.<br>Contact the Administrator of the website for more infomration.'); }
$query = "UPDATE users SET password = '$pwd', profile_picture = '$save_to_database' WHERE username='$username'";
$result = mysqli_query($con, $query) or die('error');
echo '<div class="panel -success"><div class="panel-body"><p>You have successfully updated your <b><i>password and profile picture</i></b>!</p></div>';
//echo '1';
}
else if (empty($_FILES['fileToUpload']) & empty($_POST['securePassword_Val']))
{
$query = "UPDATE users SET password = '$pwd' WHERE username='$username'";
$result = mysqli_query($con, $query) or die('error');
echo '<div class="panel -success"><div class="panel-body"><p>You have successfully updated your <b><i>profile picture</i></b>!</p></div>';
//echo '1';
}
else if (empty($_POST['securePassword_Val']) & !(empty($_FILES['fileToUpload'])))
{
if(move_uploaded_file($_FILES['fileToUpload']['tmp_name'], $uploadfile)) { echo 'Successfully uploaded image'; } else { die('Could not upload file.<br>Contact the Administrator of the website for more infomration.'); }
$query = "UPDATE users SET profile_picture = '$save_to_database' WHERE username='$username'";
$result = mysqli_query($con, $query) or die('error');
echo '<div class="panel -success"><div class="panel-body"><p>You have successfully updated your <b><i>password</i></b>!</p></div>';
//echo '1';
}
else if (empty($_POST['securePassword_Val']) & empty($_FILES['fileToUpload']))
{
$result = mysqli_query($con, $query) or die('error');
//echo '<div class="panel -danger"><div class="panel-body"><p>You have failed to update your <b><i>password and profile picture</i></b>!</p></div>';
echo '0';
}
else
{
//echo '<div class="panel -danger"><div class="panel-body"><p>An error occured!</p></div>';
echo '0';
}
}
我看过一个被张贴的链接,现在有这样的代码:
<script>
function updateMyAccount() {
var fdata = new FormData($("#data"));
fdata.append("securePassword_Val",$("#myAccountNewPassword").val());
fdata.append("SaveAccountChanges",true);
$.ajax({
type: "POST",
url: "includes/form_submit.php",
data:
//SaveAccountChanges: true,
//securePassword_Val: password,
fdata
,
async: false,
success: function(msg){
if(msg == 1) {
update_myAccount_success();
} else {
general_error_forms();
}
},
cache: false,
contentType: false,
processData: false
});
return false;
}
</script>
我怎么会去通过这种方法制作的图片上传?
很多关于如何使用Ajax上传文件的问题。 – epascarello
['FormData'构造函数](https://developer.mozilla.org/en/docs/Web/API/FormData/FormData)不接受一个jQuery对象,而是一个本地表单元素。在你的情况下,你将不得不使用[FormData.append()](https://developer.mozilla.org/en-US/docs/Web/API/FormData/append)来添加文件。 – Mikey
@Mikey好的,谢谢。你有一个例子吗? – Benza