Php 通过ajax上传文件

我正在尝试使用ajax上传文件

form enctype="myltipart/form-data" id="pastTest-form" method="POST" action="upload.php">
            <div class="feedback-form-inputs col-5">
                <div class="input-field">
                    <select id="type" required>
                        <option value="quiz">Quiz</option>
                        <option value="midterm">Midterm</option>
                        <option value="final">Final</option>
                    </select>
                </div>
                <div class="input-field">
                    <input type="text" id="professor"/>
                </div>
                <div class="input-field">
                    <input type="text" id="name"/>
                </div>
                <div class="input-field">
                    <input type="file" id="uploaded_file" name="file" accept="" required />
                </div>
            </div><!-- END feedback-form-inputs -->
            <div class="clear"></div>
                <input type="submit" value="submit" onclick="submit() />
                <div id="upload-status"> </div>
            </form>

因此，我将formData发送到php。但此时，我无法从表单数据中获取文件并将其上载到服务器。 这是我的php

// Ajax calls this code to add a past test
if (isset($_FILES['file']){
    $file = $_FILES['file'];
            $path = 'files/'.$type.'/'.$fileName;
            $moveResult = move_uploaded_file($file, $path);

        if ($moveResult != true) {
            echo "ERROR: File not uploaded. Try again.";
            //unlink($fileTmpLoc); // Remove the uploaded file from the PHP temp folder
            exit();
        }


    $path = 'files/'.$type.'/'.$fileName;
    $sql = "INSERT into past_papers VALUES ('$name', '$type', '$cid', '$professor','$path')";
    $query = mysqli_query($db_conx,$sql);
    if (mysqli_affected_rows($db_conx)>0){
        echo "success";
        exit();
        }   
    else {
        echo "failed sql";
        exit();
    }
}
?>

---

此外，我想抓取文件的输入，并一起处理它们。上传文件，并将输入插入数据库。

我能找到的最简单的一个。：）

jQuery代码

HTML代码

您应该使用搜索功能，大量的帖子都与此相关。例如，您的上传处理完全无效。您需要首先检查

$\u FILES['file']['error']

，以查看上载是否已意外执行并成功。你只是简单地假设没有什么会出错，而这恰恰是错误的做法。

// Ajax calls this code to add a past test
if (isset($_FILES['file']){
    $file = $_FILES['file'];
            $path = 'files/'.$type.'/'.$fileName;
            $moveResult = move_uploaded_file($file, $path);

        if ($moveResult != true) {
            echo "ERROR: File not uploaded. Try again.";
            //unlink($fileTmpLoc); // Remove the uploaded file from the PHP temp folder
            exit();
        }


    $path = 'files/'.$type.'/'.$fileName;
    $sql = "INSERT into past_papers VALUES ('$name', '$type', '$cid', '$professor','$path')";
    $query = mysqli_query($db_conx,$sql);
    if (mysqli_affected_rows($db_conx)>0){
        echo "success";
        exit();
        }   
    else {
        echo "failed sql";
        exit();
    }
}
?>

$("#form-id").on('submit',(function(e) {
    e.preventDefault();
    $.ajax({
        url: "file-to-call.php",
        type: "POST",
        data: new FormData(this),
        cache: false,
        processData: false,
        success: function(data) {
            //handle success
        }
   });
}));

<form name='form1' method='post' enctype='multipart/form-data' id='form-id'>
    <input type='submit' id='input' value='Upload' />
</form>