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我在我的第二个脚本中遇到错误，说“错误！在尝试下载文件时不存在具有该 ID 的图像”。

我的第一个脚本

<?php
$username = $_POST["username"];
$password = $_POST["password"];
// Connect to the database
$dbLink = new mysqli('localhost', 'sqldata', 'sqldata', 'balhaf');
if(mysqli_connect_errno()) {
die("MySQL connection failed: ". mysqli_connect_error());
}
//for mysql injection (security reasons)

$username = mysqli_real_escape_string($dbLink, $username);

$password = mysqli_real_escape_string($dbLink, $password);

mysqli_select_db($dbLink,"balhaf2");


//checking if such data exist in our database and display result

$login = mysqli_query ($dbLink,"select * from users where USERNAME = '$username' and

PASSWORD = '$password'");

if(mysqli_num_rows($login) == 1) {

// Fetch the file information
$query = "select *  from users WHERE username = '".$dbLink->escape_string($username)."'";

$result = $dbLink->query($query);
$company = false;
//Now get the result information
$row = $result->fetch_object();  //will store the record in $row

//Access what you need
if($row) {
    $company = $row->company;  //variable name should match the field name in your database
    echo $company; //See if you get the value stored in the database
    }

    mysqli_select_db($dbLink,"balhaf");

   // Query for a list of all existing files
   $sql = "SELECT id, name, mime, size, created FROM $company";
   $result = $dbLink->query($sql);

   // Check if it was successfull
   if($result) {
   // Make sure there are some files in there
   if($result->num_rows == 0) {
    echo '<p>There are no files in the database</p>';
    }
    else {
    // Print the top of a table
    echo '<table border="1" align="center">
          <H2 align="center"> Report Table</H>

            <tr>
                <td><b>Name</b></td>
                <td><b>Mime</b></td>
                <td><b>Size (bytes)</b></td>
                <td><b>Created</b></td>
                <td><b>&nbsp;</b></td>
            </tr>';
       // Print each file
      while($row = $result->fetch_assoc()) {
        echo "
            <tr>
                <td>{$row['name']}</td>
                <td>{$row['mime']}</td>
                <td>{$row['size']}</td>
                <td>{$row['created']}</td>
                <td><a style='text-decoration:none;' href='get_file_work.php?id={$row['id']}&company=$company'>Download</a></td>
            </tr>";
    }

    // Close table
    echo '</table>';
}



// Free the result
$result->free();
}
else
{
echo 'Error! SQL query failed:';
echo "<pre>{$dbLink->error}</pre>";
}

// Close the mysql connection
$dbLink->close();
}

else {

echo "worng user"."</br>";
}
?>

我的第二个脚本 get_file_work.php 从第一个脚本中获取值，

<?php
error_reporting(E_ALL);
$company =$_GET['company']; // get value from the first script 
if(isset($_GET['id']))  // get value from first script
{
$id = intval($_GET['id']);
if($id <= 0)
{
die('The ID is invalid!');
}
else
{
$dbLink = new mysqli('localhost', 'sqldata', 'sqldata', 'balhaf');
if(mysqli_connect_errno())
{
die("MySQL connection failed: ". mysqli_connect_error());
}
$query = "SELECT mime, name, size, data FROM $company WHERE id = $id";
$result = $dbLink->query($query);
if($result)
{
if($result->num_rows == 1) {
$row = mysqli_fetch_assoc($result);
header("Content-Type: application/force-download");
header("Content-Disposition: attachment; filename=". $row['name']);
header("Content-Length: ".$row['size']);
header('Expires: 0');
header('Accept-Ranges: bytes');
header("Cache-control: private");
header('Pragma: private');
echo $row['data'];
}
if(!isset($row['data']))
{
echo "error no data";
}
else {
echo 'Error! No image exists with that ID.';
}
@mysqli_free_result($result);
}
else
{
echo "Error! Query failed: <pre>{$dbLink->error}</pre>";
}
@mysqli_close($dbLink);
}
}
else
{
echo 'Error! No ID was passed.';
}
?>
4

1 回答 1

0

您应该像这样修改代码：

if(!isset($row['data'])) { echo "error no data"; } 
else { echo 'Error! 
$row['data'];
于 2014-08-22T07:37:48.847 回答