Question

我收到了insert.php个文件：

<?php

// Above is code for connection //
echo "succesful connection";

$image = addslashes(file_get_contents($_FILE['image']['tmp_name']));
$image_name = addslashes($_FILES['image']['name']);
$sql = "INSERT INTO `myTable` (`id`, `image`, `image_name`) VALUES ('1', '{$image}', '{$image_name}')";
if (!mysql_query($sql)) {
    echo "Something went wrong! :("; 
}

?>

我的form.php文件：

<form action="insert.php" method="POST" enctype="multipart/form-data">
    <label>File: </label><input type="file" name="image" />
    <input type="submit" />
</form>

当我想上传图片时出现错误：

“警告：file_get_contents（）[function.file-get-contents]：/ myDomain.com/insert.php中的文件名不能为空”

更新

我把它缩短为：

 <?php
 $content = base64_encode(file_get_contents($_FILES['image']['tmp_name']));
 $sql = "INSERT INTO `myTable` (`id`, `image`) VALUES ('1', '{$content}')";
 if (!mysql_query($sql)) {
 echo "Something went wrong! :("; 
 ?>

是添加到数据库中，但不是二进制形式。

Answer 1

始终添加检查以确保您确实获得了某些内容，

您可以使用is_uploaded_file试试这样：

<?php
  if(is_uploaded_file($_FILES['image']['tmp_name']){
     echo "succesful connection, we have an image!";
    //get the image name, no need for slashes
    $image_name  = basename( $_FILES['image']['name']);     
    //perform the the INSERT HERE
    //........
  }else{
      echo "Nothing was uploaded";
  }
?>

Answer 2

在大多数情况下，您不希望将文件存储在数据库中。你只想保存路径。

如果这是你想要的，这个脚本应该这样做。

// Get destination
$destination =  'path/to/folder' . $_FILES['image']['name'];

// Upload file
move_uploaded_file($_FILES['image']['tmp_name'], $destination);

// Get name
$image_name = addslashes($_FILES['image']['name']);

// Query
$sql = "INSERT INTO `myTable` (`id`, `image`, `image_name`) VALUES ('1', '{$destination}', '{$image_name}')";
if (!mysql_query($sql)) {
    echo "Something went wrong! :("; 
}

通过php将图像上传到mysql db

2 个答案: