As the tittle says i'm coding a web app as part of a school project. My goal is for someone to upload a json file and save some data of it in a table on Mysql for further functionallity in the app.
My question is how exaclty can you pass a JSON file to PHP and then parse it from there as to store the wanted data to the DB. I tried sending it with the help of Jquery fileupload as the json files may be quite large and on the php side i used the function file_get_contents but i had no luck with it.
Here is my javascript code :
$(document).ready(function () {
$("#submitupload").click(function(){
var files = $("#files");
$("#uploadedfile").fileupload({
url: 'upload.php',
dataType: 'json',
autoUpload: false
}).on('fileuploadadd', function (e, data) {
var fileTypeAllowed = /.\.(json)$/i;
var fileName = data.originalFiles[0]['name'];
var fileSize = data.originalFiles[0]['size'];
console.log(data);
if (!fileTypeAllowed.test(fileName)){
$("#error").html('Only json files are allowed');
}else
data.submit();
}).on('fileuploaddone', function (e , data){
var msg = data.jqXHR.responseJSON.msg;
$("#error").html(msg);
}).on('fileuploadprogress', function (e,data){
var progress = parseInt(data.loaded / data.total * 100, 10 );
$("#progress").html("Completed: " + progress + "%");
})
})
})
And here is the PHP :
<?php
include_once ('connection.php');
if (isset($_FILES['uploadingfile'])){
$file = $_FILES['uploadingfile'];
$data = file_get_contents($file);
$array = json_decode($data, true );
foreach( $array as $row){
$sql = "INSERT INTO locations(timestamp) VALUES ('".$row["timestampMs"]."')";
mysqli_query($conn, $sql);
}
$msg = array("msg" => "times ok ");
exit(json_encode($msg));
}
Noticed the error in file_get_contents() that says that the $file variable is an array not a string so i tried to pass the $_FILES variable as an argument with no luck again.
Is this the correct way to way to do it and if yes what am i missing or should i use another approach?
Thanks for the long read and your time in advance ! Sorry if any of this sounds stupid but im new to PHP .
$_FILES['uploadingfile'] is an array with several pieces of information about the uploaded file. So you need to use:
$file = $_FILES['uploadingfile']['tmp_name'];
to get the filename where the data is stored.
See Handling File Uploads for full details.