嘿我正在将文件上传到选定的文件夹,现在我可以选择并上传一个文件。我知道如何在php中处理多个文件,但我不知道如何通过AJAX发送所有文件。感谢您提供的任何帮助
AJAX
function submitForm() {
console.log("submit event");
var fd = new FormData(document.getElementById("fileinfo"));
fd.append("label", "sound");
fd.append('label', document.getElementById('selected_folder').value);
$.ajax({
url: "upload.php",
type: "POST",
data: fd,
enctype: 'multipart/form-data',
processData: false, // tell jQuery not to process the data
contentType: false // tell jQuery not to set contentType
}).done(function( data ) {
console.log("PHP Output:");
console.log( data );
alert("upload success!")
});
return false;
}
PHP
<?php
if ($_POST["label"]) {
$subfolder = $_POST["label"];
}
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = end($temp);
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"] < (10000*1024))
&& in_array($extension, $allowedExts)) {
if ($_FILES["file"]["error"] > 0) {
// echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
} else {
$filename = $_FILES["file"]["name"];
echo "Upload: " . $_FILES["file"]["name"] . "<br>";
echo "Type: " . $_FILES["file"]["type"] . "<br>";
echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br>";
if (file_exists("uploaded/".$subfolder .'/'. $filename)) {
//already exists
} else {
move_uploaded_file($_FILES["file"]["tmp_name"],
"uploaded/".$subfolder .'/'. $filename);
// "Stored in: " . "uploaded/".$subfolder .'/'. $filename;
}
}
} else {
echo "Invalid file";
}
?>
答案 0 :(得分:2)
您可以使用表单数据传递多个文件,如下所示
<强> HTML 强>
<input id="fuDocument" type="file" accept="image/*" multiple="multiple" />
<强> JS 强>
var fd = new FormData();
var files = $("#fuDocument").get(0).files; // this is my file input in which We can select multiple files.
fd.append("label", "sound");
for (var i = 0; i < files.length; i++) {
fd.append("UploadedImage" + i, files[i]);
}
$.ajax({
type: "POST",
url: 'Url',
contentType: false,
processData: false,
data: fd,
success: function (e) {
alert("success");
}
})
现在将fd
对象传递给你ajax调用它正在使用我的代码
答案 1 :(得分:2)
首先,您必须在输入标记中使用“multiple”属性。 像
<input id="fileUpload" type="file" accept="image/*" name="my_file[]" multiple />
然后在Javascript onChange函数 -
var data = new FormData();
var imgData = document.getElementById('fileUpload');
for (var i = 0; i < imgData.files.length; i++) {
data.append('my_file[]', imgData.files[i], imgData.files[i].name);
}
//now call ajax
$.ajax({
url: "upload.php",
type: "POST",
data: data,
enctype: 'multipart/form-data',
processData: false, // tell jQuery not to process the data
contentType: false // tell jQuery not to set contentType
}).done(function( data ) {
console.log("PHP Output:");
console.log( data );
alert("upload success!")
});
您的文件将被上传