PHP move_uploaded_file uploads only one file - javascript

I'm testing uploads on 3 files on my desktop test.xlsx, test - Copy.xlsx, and test - Copy (2).xlsx. My PHP script is breaking at the line where my move_uploaded_files statement is. I get the error move_uploaded_file(uploads/test - Copy.xlsx): failed to open stream: No such file or directory. BUT test - Copy (2).xlsx is uploaded when I check my uploads folder. My PHP is as follows:
foreach($_FILES['file']['name'] as $key=>$value){
$fileName = $_FILES['file']['name'][$key];
$fileTmpLoc = $_FILES['file']['tmp_name'][$key];
$fileErrorMsg = $_FILES['file']['error'][$key];
$type = $_FILES['file']['type'][$key];
if(){
//validation
} else{ //if validated
if(move_uploaded_file($fileTmpLoc, 'uploads/'.$fileName)){
//do more stuff
} else{
echo "Upload failed.";
}
}
}
The //do more stuff actually runs once for test - Copy (2).xlsx (the only file uploaded successfully). My JS script is below but I doubt that's the part that's breaking since when I print_r($_FILES) in PHP, all files appear in the output.
var upload = function(event){
var file = document.getElementById('file').files;
var data = new FormData();
for (i=0; i<file.length; i++){
data.append('file[]', file[i]);
}
var request = new XMLHttpRequest();
request.addEventListener('load', completeHandler, false);
request.addEventListener('error', errorHandler, false);
request.addEventListener('abort', abortHandler, false);
request.open('POST', 'php/excel_upload_read.php');
request.send(data);
};
What am I doing wrong?

Looks like you are using the multi-dimensional array in the wrong order. Try this:
foreach($_FILES['file'] as $key=>$value){
$fileName = $_FILES['file'][$key]['name'];
$fileTmpLoc = $_FILES['file'][$key]['tmp_name'];
$fileErrorMsg = $_FILES['file'][$key]['error'];
$type = $_FILES['file'][$key]['type'];
...
}

The problem is the way $_FILES global is structured, it does not let you iterate easily.
You can use this function it s meant to reorganize the data structure
function reArrayFiles(&$file_post) {
$file_ary = array();
$file_count = count($file_post['name']);
$file_keys = array_keys($file_post);
for ($i=0; $i<$file_count; $i++) {
foreach ($file_keys as $key) {
$file_ary[$i][$key] = $file_post[$key][$i];
}
}
return $file_ary;
}
Then you can use foreach easily:
if(isset($_FILES['file']['tmp_name'], $_POST)) {
$files = reArrayFiles($_FILES['file']);
foreach ($files as $file) {
$temp = $file['tmp_name'];
$path = 'uploads/'.$file['name'];
if(move_uploaded_file($temp, $path)){
echo 'Upload succesful';
}else{
echo 'failed to upload';
}
}
}else{
echo 'file not uploaded';
}
This function reArrayFiles was posted by someone in the PHP documentation:
http://php.net/manual/en/features.file-upload.multiple.php

Related

How do I get a value that is returned by a php script in Ajax?

I need to get an array $data from the backend to the frontend in order to organize it for viewing. I can only find examples using jQuery Ajax.
<!DOCTYPE html>
<html>
<style>
.table_class{
display:none;
}
</style>
<script>
async function uploadFile() {
let formData = new FormData();
formData.append("file", fileupload.files[0]);
console.log(formData);
await fetch('file_ingest.php', {
method: "POST",
body: formData,
});
var test = document.getElementById("test");
test.innerText = ('The file has been uploaded successfully.');
//var returned_vals = JSON.parse(formData.responseText);
//console.log(returned_vals);
}
</script>
<body>
<input id="fileupload" type="file" name="fileupload" />
<button id="upload-button" onclick="uploadFile()"> Upload </button>
<p id = "test"></p>
</body>
</html>
I need to get an array $data from the backend to the frontend in order to organize it for viewing. I can only find examples using jQuery Ajax.
<?php
set_time_limit(120);
/* Get the name of the uploaded file */
$filename = $_FILES['file']['name'];
/* Choose where to save the uploaded file */
$location = "uploads/".$filename;
if ( move_uploaded_file($_FILES['file']['tmp_name'], $location) ) {
echo 'Success';
} else {
echo 'Failure';
$CSVfp = fopen("uploads/${filename}", "r");
if($CSVfp !== FALSE) {
while(! feoF($CSVfp)) {
$data = fgetcsv ($CSVfp, 1000, ",");
print json_encode($data);
//echo $data;
}
}
fclose($CSVfp);
<?php
set_time_limit(120); // if you need this you may have other problems
/* Get the name of the uploaded file */
$filename = $_FILES['file']['name'];
/* Choose where to save the uploaded file */
$location = "uploads/".$filename;
$response = [];
/* Save the uploaded file to the local filesystem */
if ( move_uploaded_file($_FILES['file']['tmp_name'], $location) ) {
$response['status'] = 'Files Save Success';
// read the file here, if it failed the upload and move
// there is no point trying to read it
if (($fp = fopen($location, "r")) !== FALSE) {
while (($data = fgetcsv($fp, 1000, ",")) !== FALSE) {
$response['data'][] = $data
}
fclose($fp);
}
} else {
$response['status'] = 'File Save Failure';
}
// now you can return either an error or a success and data
echo json_encode($response);
Now you need to checnge the javascript to look for the status in the right place and unload the row data from the the right place

Stream php output to javascript callback

I am making a javascript XMLHttpRequest() request to execute some php in the background. Now I know the callback will only receive the first response from php. How can I get output to send back what php is echoing as it is happening to show it on the main page as php is running in the background?
javascript
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
var phpResponse = this.responseText
// i know this will only be the first thing echoed by php
}
};
xhttp.open("GET", "scan.php?scanRoot=yes", true);
xhttp.send();
php run in the back ground
function rootDirFileSort($scanDir,$getID3){
//Sort stray files in root directory. Put them in a dir named by artist
$printFix = str_repeat(" ", 8157);
$files = array_diff(scandir($scanDir), array('.', '..'));
//Get files in dir. Only get flac and mp3 files
$FilesArray = array_diff(scandir($scanDir), array('.', '..'));
$FilesArray = preg_grep('~\.(flac|mp3)$~', $FilesArray);
$files = array_values($FilesArray);
foreach ($files as $file){
$file_Basename = basename($file);
// Get the metadata from file. Returns as an array
$metaDataArray = $getID3->analyze($scanDir.$file);
//Check for ID3v1 metadata container
if (isset($metaDataArray['tags']['id3v1']['artist'][0])){
$artist = $metaDataArray['tags']['id3v1']['artist'][0];
$artist = str_replace(array('.',',','?','/','\\','$'), array('','','','','',''), ucwords($artist));
}else{
$artist = '';
}
//Check for vorbis comment metadata container
if(isset($metaDataArray['tags']['vorbiscomment']['artist'][0])){
$artist = $metaDataArray['tags']['vorbiscomment']['artist'][0]; echo $artist;
$artist = str_replace(array('.',',','?','/','\\','$'), array('','','','','',''), ucwords($artist));
}else{
$artist = '';
}
//If the artist meta tag is present move the file
if(!empty($artist)){
//Check if diretory exists before making one
if(!is_dir($scanDir.$artist)){
//If file DOES NOT exist
mkdir($scanDir.$artist, 0755, true);
rename($scanDir.$file, $scanDir.$artist.'\\'.$file);
echo $file.'<br>'.$artist.'<br>Moved To: '.$scanDir.$artist.'\\'.$file.'<br><br>'.$printFix;
}else{
//If file exists
rename($scanDir.$file, $scanDir.$artist.'\\'.$file);
echo $file.'<br>'.$artist.'<br>Moved To: '.$scanDir.$artist.'\\'.$file.'<br><br>'.$printFix;
}
}else{
//Put failures in an array to show which failed when done
$notProcessed[] = array('musicFile'=>$file_Basename);
}
}
//Show all the failures if they exist
if(count($notProcessed) >= 1){
echo '<strong>***** These items failed as they don\'t contain ID3v1, ID3v2 or Vorbis Comment meta container. Manually process directories. *****</strong><br><br>';
foreach($notProcessed as $value){
echo '<strong>File:</strong> '.$value['musicFile'].'<br><br>'.$printFix;
}
}
echo '<br><br>------------------------- Root File Scan Finished! -------------------------'.$printFix;
}

this code uploads csv file only but i want xls too?

<?php
include 'model.php';
$rs=new database();
if(isset($_POST["Import"])){
echo $filename=$_FILES["file"]["tmp_name"];
if($_FILES["file"]["size"] > 0)
{
$file = fopen($filename, "r");
while (($emapData = fgetcsv($file, 10000, ",")) !== FALSE)
{
$res=$rs->insert($emapData[0],$emapData[1],$emapData[2],$emapData[3],$emapData[4],$emapData[5]);
$result=mysql_fetch_array($res);
if(! $result )
{
echo "<script type=\"text/javascript\">
alert(\"Invalid File:Please Upload CSV File.\");
window.location = \"result.php?msg=valid\"
</script>";
}
}
fclose($file);
echo "<script type=\"text/javascript\">
alert(\"CSV File has been successfully Imported.\");
window.location = \"result.php?msg=valid\"
</script>";
mysql_close($conn);
}
}
?>
this code only uploads csv file but i want to upload xls too with this code. if possible i want to upload all format of excel . and the rest of code is working fine and also i dont want to change the method.
Download PHPExcel
https://github.com/PHPOffice/PHPExcel
and create this function
function getDataFromExcel($filename)
{
$excel = PHPExcel_IOFactory::load($filename);
$sheet = $excel->getSheet(0);
$highestRow = $sheet->getHighestRow();
$sheetData = $sheet->toArray(null, true, true, true);
return $sheetData;
}
It will return data in array
if you want to know the type of file use this method
function getFileType($key)
{
//Define type
$type = 'unknow';
if(isset($_FILES[$key])) {
$file = $_FILES[$key];
$fileType = $file['type'];
if (strrpos($fileType, 'csv')) {
$type = 'csv';
} else if (($fileType == 'application/vnd.openxmlformats-officedocument.spreadsheetml.sheet') || ($fileType == 'application/vnd.ms-excel')) {
$type = 'excel';
}
}
return $type;
}

AJAX won't execute all of the php code

I want a progressbar to be shown while a video-file is being uploaded. I'm using JS and AJAX to track progress of the file-upload php, and send the file. The php file contains a ffmpeg command that grabs some frames from the video. Essentially the php file creates a random folder, and puts both the video and the frames in it. when the php is called by if(isset($_FILES['file'])) it works just fine. But when i try to use AJAX it only uploads the video, but ignores the ffmpeg command and mysql_query.
Javascript:
function uploadFile(){
var file = _("file1").files[0];
var formdata = new FormData();
formdata.append("file1", file);
var ajax = new XMLHttpRequest();
ajax.upload.addEventListener("progress", progressHandler, false);
ajax.open("POST", "upload.php");
ajax.send(formdata);
Upload.php
$name = $_FILES['file1']['name'];
$type = explode(".",$name);
$type = end($type);
$tmp = $_FILES['file1']['tmp_name'];
$uploadOk = 1;
$exp = explode(".",$name);
$filename = rand().$exp[0];
$path = "videos/" . $filename . "/" . $name;
$ffmpeg = "/usr/local/bin/ffmpeg";
$size = "320x180";
if($type != 'mp4' && $type != 'MP4'){
$message = "Only mp4 format is supported!";
$uploadOk = 0;
}else{
mkdir("videos/".$filename);
// ffmpeg function
for($num = 1; $num <= 15; $num++){
$interval = $num * 3;
shell_exec("$ffmpeg -i $tmp -an -ss $interval -s $size /videos/$filename/thumb$num.png");}
move_uploaded_file($tmp, $path);
mysql_query("INSERT INTO table (name, url) VALUES('$name', '$path')");
Why dosen't AJAX execute the complete php?

Need Help Handling a XML HTTP File Upload Request

So, I looked up a tutorial for uploading and sending files to a server with an XML HTTP Request. I followed the tutorial, however, I think I must be missing something. While the file appears to be uploaded and sent, nothing in the "handler" file is ever accessed. Is there a PHP function I need to write to process it? For context, here is what I wrote:
$(document).ready(function()
{
$('#upload-button').click(function(event)
{
$('#upload-button').removeClass("btn-danger");
});
$( "#report-form" ).submit(function( event )
{
var form = document.getElementById('report-form');
var fileSelect = document.getElementById('file-select');
var uploadButton = document.getElementById('upload-button');
event.preventDefault(); // Stop the event from sending the way it usually does.
uploadButton.value = 'Submitting...'; // Change text.
var files = fileSelect.files;
var maxfiles = <?php echo $config['Report_MaxFiles'] ?>;
var mfs = <?php echo $config['Report_MaxFileSize'] ?>;
if(files.length > maxfiles) // Make sure it's not uploading too many.
{
uploadButton.value = 'You uploaded too many files. The limit is ' + maxfiles + '.'; // Update button text.
$('#upload-button').addClass('btn-danger'); // Make the button red, if so.
return;
}
var formData = new FormData(); // Make a "form data" variable.
for (var i = 0; i < files.length; i++) {
var file = files[i];
// Add the file to the request.
if(file.size / 1000 > mfs)
{
uploadButton.value = 'One of the files is too big. The file size limit is ' + (mfs) + 'kb (' + (mfs / 1000) + 'mb).';
$('#upload-button').addClass('btn-danger');
return;
}
formData.append('files[]', file, file.name); // Not really sure what this does, to be honest,
// but I think it makes a file array.
}
var xhr = new XMLHttpRequest(); // Construct an XML HTTP Request
xhr.open('POST', 'assets/class/FileHandler.php', true); // Open a connection with my handler PHP file.
xhr.onload = function ()
{
if (xhr.status === 200)
{
uploadButton.value = 'Files Submitted!'; // NOTE: I do get this message.
}
else
{
uploadButton.value = 'An error occurred.';
$('#upload-button').addClass("btn-danger");
}
};
xhr.send(formData); // I think this is where it dies.
});
});
At the "send(formData)" line, I'm not actually sure if it's sending. Do I set up some sort of listener in FileHandler.php that is activated when the files are sent via XML HTTP request? Or more specifically, how to I save the uploaded files to the server using my FileHandler.php file?
EDIT: I haven't been able to come up with any other PHP code in the FileHandler.php file than this, which I thought might be called when the form is sent (but it isn't):
EDIT 2: Okay, now I have something, but it isn't working (didn't expect it to). I think I may be using the variables wrong:
<?php
$uploaddir = 'data/reports/uploads/' . $_POST['id'] . "/";
$uploadfile = $uploaddir . basename($_FILES['files']['name']);
echo "<script>console.log('RECEIVED');</script>";
echo '<pre>';
if (move_uploaded_file($_FILES['files']['tmp_name'], $uploadfile)) {
echo "File is valid, and was successfully uploaded.\n";
} else {
echo "Possible file upload attack!\n";
}
echo 'Here is some more debugging info:';
print_r($_FILES);
print "</pre>";
It's not saving the file to the directory, nor is it printing the script message. How do I get my report.php file to execute these things in FileHandler.php?
Thanks to the help and patience of #Florian Lefèvre, I got it fixed. :)
The problem was with the path. It wasn't locating the path to the folder data/uploads/ and wasn't making the directory. Here is what I did:
$uploaddir = '../../data/reports/uploads/' . $_POST['id'] . "/";
echo "NAME: " . $_FILES['files']['name'][0] . "\n";
foreach($_FILES['files']['name'] as $filenumber => $filename)
{
$uploadfile = $uploaddir . basename ($filename);
echo "UploadDir " . $uploaddir . "\n";
echo "UploadFile " . $uploadfile . "\n";
echo '<pre>';
echo "MKDir for UploadDir which is: ". $uploaddir . "\n";
mkdir ($uploaddir);
if (move_uploaded_file ($_FILES['files']['tmp_name'][$filenumber], $uploadfile))
{
echo "File is valid, and was successfully uploaded.\n";
}
else
{
echo "Possible file upload attack!\n";
}
echo 'Here is some more debugging info:';
print "</pre>";
}
var_dump ($_FILES);
I haven't gotten rid of some of the debug stuff yet, but that's the general solution.

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