I have one form in HTML
I have a fetch
AJAX and I can send more than one form in a request to return just one alert message.
const has = document.querySelectorAll('.main_app_form_edit');
if (has) {
const newFormData = (new FormData());
newFormData.append('_token', document.querySelector('input[name="_token"]').value);
let i = 0;
document.querySelector('.test_submit')
.addEventListener('click', (e) => {
e.preventDefault();
const formCards = document.querySelectorAll('.main_app_form_edit');
const url = formCards[0].getAttribute('action');
const method = formCards[0].getAttribute('method');
for (element of formCards) {
const formConst = (new FormData(element));
// BEGIN FILE
const file = element[3].files[0];
const newObject = {
'tmp_name': file.mozFullPath,
'lastModifiedDate': file.lastModifiedDate,
'name': file.name,
'size': file.size,
'type': file.type
};
formConst.append('img', (new URLSearchParams(newObject)).toString());
formConst.delete('image');
// END FILE
newFormData.append(
'data_' + ++i,
JSON.stringify((new URLSearchParams(formConst)).toString())
);
}
getCards(method, url, 'multipart/form-data', newFormData);
});
}
In Laravel, when I try to get the file, I don't have a form to read the img
array as a file, because I don't have a fullPath
and when I tried to use monFullPath
, it did not work:
public static function createCardsPost(Request $request)
{
$forms = $request->except(['_token', 'img']);
foreach ($forms as $form) {
// PARSEANDO OS DADOS QUE DEVEM SER PARSEADOS
parse_str($form, $dataArray);
parse_str($dataArray['img'], $img);
$dataArray['img'] = (array)$img;
$requestArray = new Request($dataArray);
$dataArray = $requestArray->except(['"_token']);
$dataArray['user_id'] = 1;
$dataArray['folder_id'] = 1;
if ($requestArray->hasFile('img') && $requestArray->img->isValid()) {
$dataArray['img'] = $requestArray->file('img')->store('/cards');
}
var_dump($dataArray);
die();
\App\Card::create($dataArray);
}
header('Content-Type: application/json');
$callback['message'] = 'Pasta e cartões cadastrados com sucesso!';
return json_encode($callback);
}
How do I do that? How do I send a file array with a valid format to PHP?
I resolve that, solution:
Yesterday, I created a solution. I pass a unique identifier for the index file to your data set. Then I can find out which one is in which dataset.
ex : [[arrayDataSet_300], [FormDataFile_300000]]
I multiply the index for 1000 (100 * 1000) then I can send to Laravel more than one data set with a file with a valid format
// REGRA DE NEGÓCIO DOS CARDS
const createCards = async (folderId) => {
if (typeof folderId !== 'number') {
alert('Não foi possível cadastrar. Volte mais tarde!');
throw new Error({'folder_id': folderId});
}
const newFormData = new FormData();
const formCards = document.querySelectorAll('.main_app_form_edit');
const url = formCards[0].getAttribute('action');
const method = formCards[0].getAttribute('method');
let i = 1;
for (element of formCards) {
++i;
const formData = (new FormData(element));
formData.append('folder_id', folderId);
const file = element[3].files[0];
newFormData.append((i * 1000), file);
newFormData.append(i, (new URLSearchParams(formData)));
}
newFormData.append('_token', document.querySelector('input[name="_token"]').value)
getCards(method, url, 'multipart/form-data', newFormData);
}
// FIM DA REGRA DE NEGÓCIO DOS CARDS
php laravel side:
public static function createCardsPost(Request $request)
{
$forms = $request->except(['_token']);
for ($i = 2; $i <= (count($forms) !== 2 ? count($forms) - 2 : 2); $i++) {
// PARSEANDO OS DADOS QUE DEVEM SER PARSEADOS
parse_str($forms[$i], $dataArray);
$requestArray = new Request($dataArray);
$dataArray = $requestArray->except(['_token', 'image']);
/** -test--id- */
$dataArray['user_id'] = 1;
/** ---------- */
// ID DE RETORNO
$hrefId = $dataArray['folder_id'];
// SALVA O ARQUIVO SE HOUVER
$dataArray['img'] = Folder::saveFolder(((empty($forms[$i * 1000]) ? $forms[$i * 1000] : null)));
\App\Card::create($dataArray);
}
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