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如何在 MySQL 中更新一张图片

phpmysqlimageimage-uploading
3
我有一个上传图片到文件夹并将名称插入mysql表格的php页面。现在我想创建一个页面来更新图片并删除目录文件夹中的旧图片或用新图片替换旧图片。
以下是代码，它无法更新图片或其他字段。
 <?php
// Start a session for error reporting
session_start();

// Call our connection file
require("includes/conn.php");



// Set some constants

// This variable is the path to the image folder where all the images are going to be stored
// Note that there is a trailing forward slash
$TARGET_PATH = "images/";

// Get our POSTed variables
$name = $_POST['name'];
$description  = $_POST['description '];
$price = $_POST['price'];
$image = $_FILES['image'];
$serial = $_POST['serial'];

// Sanitize our inputs
$name = mysql_real_escape_string($name);
$description = mysql_real_escape_string($description);
$price = mysql_real_escape_string($price);
$image['name'] = mysql_real_escape_string($image['name']);

// Build our target path full string.  This is where the file will be moved do
// i.e.  images/picture.jpg
$TARGET_PATH .= $image['name'];


// Here we check to see if a file with that name already exists
// You could get past filename problems by appending a timestamp to the filename and then continuing
if (file_exists($TARGET_PATH))
{
        $_SESSION['error'] = "A file with that name already exists";
        header("Location: updateproduct.php");
        exit;
}

// Lets attempt to move the file from its temporary directory to its new home
if (move_uploaded_file($image['tmp_name'], $TARGET_PATH))
{
        // NOTE: This is where a lot of people make mistakes.
        // We are *not* putting the image into the database; we are putting a reference to the file's location on the server
        $sql = "UPDATE products SET picture = '$image', description = '$description' ,price = '$price' ,name = '$name'  WHERE serial = '$serial'";


 $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error());
        header("Location: updateproduct.php");
        exit; 

}
else
{
        // A common cause of file moving failures is because of bad permissions on the directory attempting to be written to
        // Make sure you chmod the directory to be writeable
        $_SESSION['error'] = "Could not upload file.  Check read/write persmissions on the directory";
        header("Location: updateproduct.php");
        exit;
}
?>

这是一个表单。
   <?php require_once('Connections/shopping.php'); ?>
<?php
$colname_Recordset1 = "1";
if (isset($_POST['serial'])) {
  $colname_Recordset1 = (get_magic_quotes_gpc()) ? $_POST['serial'] : addslashes($_POST['serial']);
}
mysql_select_db($database_shopping, $shopping);
$query_Recordset1 = sprintf("SELECT * FROM products WHERE serial = %s", $colname_Recordset1);
$Recordset1 = mysql_query($query_Recordset1, $shopping) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1);
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<body>

<div align="center">
  <form method="post" name="form1" action="updateupload.php">
    <table align="center">
      <tr valign="baseline">
        <td nowrap align="right">Serial:</td>
        <td><?php echo $row_Recordset1['serial']; ?></td>
      </tr>
      <tr valign="baseline">
        <td nowrap align="right">Name:</td>
        <td><input type="text" name="name" value="<?php echo $row_Recordset1['name']; ?>" size="32"></td>
      </tr>
      <tr valign="baseline">
        <td nowrap align="right">Description:</td>
        <td><input type="text" name="description" value="<?php echo $row_Recordset1['description']; ?>" size="32"></td>
      </tr>
      <tr valign="baseline">
        <td nowrap align="right">Price:</td>
        <td><input type="text" name="price" value="<?php echo $row_Recordset1['price']; ?>" size="32"></td>
      </tr>
      <tr valign="baseline">
        <td nowrap align="right">Picture:</td>
        <td><input type="file" name="picture" value="<?php echo $row_Recordset1['picture']; ?>" size="32"></td>
      </tr>
      <tr valign="baseline">
        <td nowrap align="right">&nbsp;</td>
        <td><input name="submit" type="submit" value="Update record"></td>
      </tr>
    </table>
  </form>
</div>
</body>
</html>
<?php
mysql_free_result($Recordset1);
?>
请停止使用古老的mysql_*函数编写新代码。它们已不再维护，社区已经开始弃用过程。相反，您应该学习有关准备语句的知识，并使用PDO或MySQLi。 - Bono
我对php不太了解，我按照教程学会了插入部分，并尝试根据此来进行更新操作，但这并不是教程的一部分。@Bono - user1084949
2个回答
2
哦，我怎么会忘了呢？我会尝试一下并告诉你。 - user1084949
1
我添加了enctype="multipart/form-data"，但它没有起作用，请帮忙。 - user1084949
在这行代码"$image = $_FILES['image'];"中，如果字段名是'picture'，为什么你使用'image'？@user1084949 - jcho360
1

unlink 函数可能正是您想要的

if (file_exists($PATH_TO_IMAGE))
{
       unlink($PATH_TO_IMAGE);            
}
1
@user1084949，你可以从MySQL表中找到$PATH_TO_IMAGE，因为你已经将图片路径存储在表中。干杯 :) - Subail Sunny
@Subail Sunny，我将图像名称（product.gif）存储在表中而不是路径...图片本身保存在名为images的文件夹中。 - user1084949
1
@user1084949 如果所有图像都存储在同一个文件夹中，则 $PATH_TO_IMAGE = $folderpath.'/imagename'; 就可以完成工作。 - Subail Sunny
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