I need to handle inputs from form post and I have no idea, how to do it in php, because when I write for example $_POST["header"], it var_dumps null.
I am creating formData object and inserting all inputs from form. Then posting it with ajax.
Can you please help me? I need to handle "header", "content", "password" and files.
<form method="post" enctype="multipart/form-data" id="uploadFiles">
<label for="newsHeader" id="headerLabel">Nadpis</label>
<input type="text" name="newsHeader" id="newsHeader">
<label for="content" id="contentLabel">Text novinky</label>
<textarea name="content" id="content"></textarea>
<label for="files" id="filesLabel">Fotky</label>
<input type="file" name="files" id="files" accept="image/jpeg" multiple>
<label for="password" id="passwordLabel">Heslo pro upload</label>
<input type="text" name="password" id="password">
<button type='submit' id='uploadFilesSubmit'>NAHR??T</button>
</form>
$("#uploadFiles").submit(function(event){
event.preventDefault();
var formDataObj = new FormData(),
header = $("#newsHeader").val(),
content = $("#content").val(),
password = $("#password").val();
formDataObj.append("header", header);
formDataObj.append("content", content);
formDataObj.append("password", password);
$.each($("#files")[0].files, function(i, file) {
formDataObj.append('file', file);
});
console.log(Array.from(formDataObj));
$("#uploadFilesSubmit").html("<div class='buttonSubmitIcon'><i class='fas fa-sync'></i></div>");
$.ajax({
method: "POST",
url: "uploadNews.php",
data: {
formDataObj: formDataObj
},
dataType: 'json',
contentType: false,
processData: false,
success: function(results){
}, error: function(){
}
});
});
In uploadNews.php I have this:
exit(json_encode(var_dump($_POST["header"])));
It always returns "Undefined index: header", same as content or count($_FILES["file"]["name"])
All I want is to get somehow to posted values.. Thank you very much
You just to pass the actualformDataObj
variable via your$.ajax
. This is not the correctsyntax
to passFormData
via ajax =>formDataObj: formDataObj
A FormData itself is anobject
which stores your data so what you are doing is making anotherobject
on top of it when you pass it viadata
You can nowvar_dump(header)
orvar_dump($_FILES["file"]["name"])
to see everything coming to yourPHP
file.
Live Demo: (Change youjQuery
code to this below and it will just work fine)
$("#uploadFiles").submit(function(event) {
event.preventDefault();
var formDataObj = new FormData(),
header = $("#newsHeader").val(),
content = $("#content").val(),
password = $("#password").val();
formDataObj.append("header", header);
formDataObj.append("content", content);
formDataObj.append("password", password);
$.each($("#files")[0].files, function(i, file) {
formDataObj.append('file', file);
});
$("#uploadFilesSubmit").html("<div class='buttonSubmitIcon'><i class='fas fa-sync'></i></div>");
$.ajax({
method: "POST",
url: "uploadNews.php",
data: formDataObj, //just pass the form Data object.
dataType: 'json',
contentType: false,
processData: false,
success: function(results) {
},
error: function() {
}
});
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<form method="post" enctype="multipart/form-data" id="uploadFiles">
<label for="newsHeader" id="headerLabel">Nadpis</label>
<input type="text" name="newsHeader" id="newsHeader">
<label for="content" id="contentLabel">Text novinky</label>
<textarea name="content" id="content"></textarea>
<label for="files" id="filesLabel">Fotky</label>
<input type="file" name="files" id="files" accept="image/jpeg" multiple>
<label for="password" id="passwordLabel">Heslo pro upload</label>
<input type="text" name="password" id="password">
<button type='submit' id='uploadFilesSubmit'>NAHR??T</button>
</form>
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