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php-如何在Mysql中更新图像

2019-12-01 08:15:21 作者：互联网

我有一个php页面,可将图像上传到一个文件夹,并将名称插入mysql表中.现在,我想创建一个页面,该页面将更新图片并删除目录文件夹中的旧图片,或者将旧图片替换为新图片.

这是不更新图像或其他字段的代码.

 <?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;
}
?>

这是表格

   <?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);
?>

解决方法:

我想你想念

enctype="multipart/form-data"

处理文件.

http://www.w3schools.com/php/php_file_upload.asp

标签：image,mysql,php,image-uploading
来源： https://codeday.me/bug/20191201/2079501.html