Question

我正在尝试创建一个将用户数据插入数据库表并上传文件的表单，但是我不知道如何将文件添加到数据库中。表单数据输入没有问题，只是文件没有问题。任何帮助，将不胜感激。

PHP表单：

<form method="post" action="index.php" id="frm_add" enctype="multipart/form-data">
      <input type="hidden" value="add" name="action" id="action">
              <div class="form-group">
                <label for="medID" class="control-label">ID</label>
                <input type="hidden" class="form-control" id="medID" name="medID"/> 
              </div>
      <div class="form-group">
                <label for="emp_id" class="control-label">Employee No:</label>
                <input type="text" class="form-control" id="emp_id" name="emp_id"/>
              </div>
              <div class="form-group">
                <label for="name" class="control-label">Full Name:</label>
                <input type="text" class="form-control" id="name" name="name"/>
              </div>
      <div class="form-group">
                <label for="title" class="control-label">Job Title:</label>
                <input type="text" class="form-control" id="title" name="title"/>
              </div>
      <div class="form-group">
        <label for="documents" class="control-label">Supporting Documents:</label>
        <input type="file" name="documents" class="form-control">
              </div>
        <div class="modal-footer">
            <button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
            <button type="button" id="btn_add" class="btn btn-primary">Save</button>
        </div>          
</form>

insert.php

    <?php
    //include connection file 
    include_once("connection2.php");

    $db = new dbObj();
    $connString =  $db->getConnstring();

    $params = $_REQUEST;

    $action = isset($params['action']) != '' ? $params['action'] : '';
    $empCls = new Student($connString);

    switch($action) {
     case 'add':
        $empCls->insertStudent($params);
     break;
     case 'edit':
        $empCls->updateStudent($params);
     break;
     case 'delete':
        $empCls->deleteStudent($params);
     break;
     default:
     $empCls->getStudents($params);
     return;
    }

    class Student {
    protected $conn;
    protected $data = array();
    function __construct($connString) {
        $this->conn = $connString;
    }

    public function getStudents($params) {

        $this->data = $this->getRecords($params);

        echo json_encode($this->data);
    }
    function insertStudent($params) {
        $data = array();
        $sql = "INSERT INTO `employee` (emp_id, name, title, documents) VALUES('" . $params["emp_id"] . "', '" . $params["name"] . "','" . $params["title"] . "', '" . $params["documents"] . "');  "; 
echo $result = mysqli_query($this->conn, $sql) or die("error to insert student data");
}

我做错了什么，我需要添加什么文档？任何帮助将不胜感激。

谢谢

Answer 1

将文件保存在数据库内部毫无意义。仅获取文件名，

$info = pathinfo($_FILES['userFile']['name']);
$ext = $info['extension']; // get the extension of the file
$newname = "newname.".$ext; 

$target = 'images/'.$newname;
move_uploaded_file( $_FILES['userFile']['tmp_name'], $target);

上面的代码来自这里：How to upload & Save Files with Desired name

因此，您需要做的是将名称保存在数据库中，而不是仅使用带有表示“名称”的文件的路径访问该名称。

PHP插入表单数据并将文件上传到MySQL表

1 个答案: