Question

我一直在搜索SO的解决方案并尝试过，但我发现的答案中没有一个解决了我的问题。所以就是这样。

我正在开发一个简单的Web应用程序，其中包含输入详细信息和文件上载（.docx文件）。我使用FormData来处理表单数据，但服务器端操作仅适用于表单输入，如text, password, date等。

这是我的JS代码;评论的行是我试过但没有用的行。

var myForm = $("#add-thesis-form");
    var formData = new FormData(myForm);

    //var formData = myForm.serialize();

    $.ajax({
        type: "POST" ,
        url: "thesis-scripts/add_thesis.php",
        data: formData
        // processData: false,
        // contentType: false
    }).done(function(data){
        $('#addThesisModal').modal('hide');
        alert(data);
        //alert("Successfully saved a record to the database. ");
        showThesis();
    });

我的PHP代码（使用Meekro DB）

$title = $_POST["title"];
$description = $_POST["description"];
$categoryId = $_POST["category_id"];
$collegeId = $_POST["college_id"];
$departmentId = $_POST["department_id"];
$uploadedBy = $_SESSION["user"];
$uploadedAt = DB::queryFirstField("SELECT NOW();");
$year = $_POST["thesis_year"];
$keywords = $_POST["keywords"];
$authors = $_POST["authors"];

//FILE UPLOAD

$targ_dir = "../uploads/";
$targ_file = $targ_dir . basename($_FILES["thesis"]["name"]);
$flagOk = 1;
$tempFolder = $_FILES["thesis"]["tmp_name"];
move_uploaded_file($tempFolder, $targ_file);

//!FILE UPLOAD

$result = DB::insert('theses', array(
    "title" => $title,
    "description" => $description,
    "categoryId" => $categoryId,
    "collegeId" => $collegeId,
    "departmentId" => $departmentId,
    "uploadedBy" => $uploadedBy,
    "uploadedAt" => $uploadedAt,
    "pubyear" => $year,
    "filepath" => $targ_file,
    "views" => 0,
    "rating" => 0,
    "keywords" => $keywords,
    "authors" => $authors
));

我已经尝试了print_r($_FILES)并且那个有效，但是这个没有。请帮我。谢谢。设置contentType:false会导致服务器在所有Undefined index变量上抛出错误$_POST。

Answer 1

也许您的表单应包含enctype="multipart/form-data"

也许这会奏效：

 $.ajax({
        type: "POST" ,
        url: "thesis-scripts/add_thesis.php",
        data: "multipart/form-data"
     ...

使用表单输入AJAX无效的文件上传

1 个答案: