I have a form with id: formC, on submit i call ajax:
var datiForm = new FormData();
var pos = [];
var i = 0;
posizioni.each(function () {
if($(this).find("input[type=checkbox]").is(":checked")){
pos[i] = $(this).find("input[type=checkbox]").data("id");
i++;
}
});
datiForm.append("nome",nome.val());
datiForm.append("cognome",cognome.val());
datiForm.append("email",email.val());
datiForm.append("telefono",telefono.val());
datiForm.append("dataNascita",dataNascita.val());
datiForm.append("titolo",titolo.val());
datiForm.append("ruolo",ruolo.find(":selected").data("ruolo"));
datiForm.append("sede",sede.find(":selected").data("sede"));
datiForm.append("posizione",pos);
datiForm.append("cvFile",$("#cvFile")[0].files[0]);
$.ajax({
type: "POST",
data: {datiForm: datiForm},
url: "saveCandidate.php",
processData: false,
contentType: false,
success: function (data) {
console.log(data);
},
error: function (data) {
var position = data;
}
});
I have a problem, on server $datiForm = $_POST["datiForm"];
is null why?
Moreover i have input file where i can select file pdf. I put it in FormData:
datiForm.append("cvFile",$("#cvFile")[0].files[0]);
Now on server i want to take file from $datiForm and save it into mysql as Blob is possible?
You specified the data field incorrectly, it should be just the form data object
data: datiForm,
also the way you add posizione
is not going to work, each entry in yrh array has to be added individually
posizioni.each(function () {
if($(this).find("input[type=checkbox]").is(":checked")){
datiForm.append("posizione["+i+"]", $(this).find("input[type=checkbox]").data("id"));
i++;
}
});
Now on server i want to take file from $datiForm and save it into mysql as Blob is possible?
Yes