Question

我创建了一个使用PHP和MySQL的表单。这里有人可以输入文本和上传图像。我的图像上传部分正在工作。图像存储在数据库中但文本没有存储在数据库中

我的代码如下：

的index.php

  <form action="upload.php" method="post" enctype="multipart/form-data">
      <input type="text" name="username">
       <input type="text" name="report">
        Select image to upload:
        <input type="file" name="image"/>
        <input type="submit" name="submit" value="UPLOAD"/>
    </form>

upload.php的

<?php
if(isset($_POST["submit"])){
    $check = getimagesize($_FILES["image"]["tmp_name"]);

    if($check !== false){
        $image = $_FILES['image']['tmp_name'];
        $imgContent = addslashes(file_get_contents($image));

        /*
         * Insert image data into database
         */

        //DB details
        $dbHost = 'localhost';
        $dbUsername = 'root';
        $dbPassword = 'root';
        $dbName = 'testreport';

        //Create connection and select DB
        $db = new mysqli($dbHost, $dbUsername, $dbPassword, $dbName);

        // Check connection
        if($db->connect_error){
            die("Connection failed: " . $db->connect_error);
        }

       $username = mysqli_real_escape_string($link, $_REQUEST['username']);
        $report = mysqli_real_escape_string($link, $_REQUEST['report']);
        $dataTime = date("Y-m-d H:i:s");

        //Insert image content into database
        $insert = $db->query("INSERT into images (username,report,image, created) VALUES ('$username','$report','$imgContent', '$dataTime')");
        if($insert){
            echo "File uploaded successfully.";
        }else{
            echo "File upload failed, please try again.";
        } 
    }else{
        echo "Please select an image file to upload.";
    }
}
?>

Answer 1

我认为它应该是$ _POST ['']，因为表单方法是'post' 所以它将是：

   $username = mysqli_real_escape_string($_POST['username']);
   $report = mysqli_real_escape_string($_POST['report']);

Answer 2

你必须为mysqli_real_escape_string（）

定义这样的链接

$link = mysqli_connect("localhost","root","root","testreport");

Answer 3

尝试以下代码

$username = mysqli_real_escape_string($db, $_POST['username']);
$report = mysqli_real_escape_string($db, $_POST['report']);
$dataTime = date("Y-m-d H:i:s");

当您使用mysqli_real_escape_string时，您需要在第一个参数中传递mysqli对象。 mysqli_real_escape_string

PHP表单图像上传正在工作，但文本输入不存储在数据库中

3 个答案: