Mysql num rows(): Invalid argument

Question

Yuki H. 0 Light Poster

19 Years Ago

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/user/public_html/im2db/image.php on line 14

This is the error I am getting

Here is the code:
Image.php
Am I supposed to add a exit(); after
$result = mysql_query ($sql, $conn);
line?

<?php
// database connection
$conn = mysql_connect("localhost", "user", "pass") OR DIE (mysql_error());
@mysql_select_db ("user_images", $conn) OR DIE (mysql_error());
$sql    = "SELECT * FROM images WHERE image_id=".$_GET["iid"]; 

$result = mysql_query ($sql, $conn);

if (mysql_num_rows ($result)>0) {
  $row = @mysql_fetch_array ($result) or die (mysql_error());
  $image_type = $row["image_type"];
  $image = $row["image"];
  Header ("Content-type: $image_type");
  print $image;
}
?>

Warning: fread(): supplied argument is not a valid stream resource in /home/user/public_html/im2db/index.php on line 18
This is the error I get when I submit a blank entry.

<?php
// index.php - by Hermawan Haryanto <hermawan@dmonster.com>
// Example PHP Script, demonstrating Storing Image in Database
// Detailed Information can be found at http://www.codewalkers.com

// database connection
$conn = mysql_connect("localhost", "user", "pass") OR DIE (mysql_error());
@mysql_select_db ("user_images", $conn) OR DIE (mysql_error());

// Do this process if user has browse the file and click the submit button
if ($_FILES) {
  $image_types = Array ("image/bmp",
                        "image/jpeg",
                        "image/pjpeg",
                        "image/gif",
                        "image/x-png");

  $userfile  = addslashes (fread (fopen ($_FILES["userfile"]["tmp_name"], "r"), filesize ($_FILES["userfile"]["tmp_name"])));
  $file_name = $_FILES["userfile"]["name"];
  $file_size = $_FILES["userfile"]["size"];
  $file_type = $_FILES["userfile"]["type"];

  if (in_array (strtolower ($file_type), $image_types)) {
    $sql = "INSERT INTO images (image_type, image, image_size, image_name, image_date) ";
    $sql.= "VALUES (";
    $sql.= "'{$file_type}', '{$userfile}', '{$file_size}', '{$file_name}', NOW())";
    @mysql_query ($sql, $conn);
    Header("Location:".$_SERVER["PHP_SELF"]);
    exit();
  }
}

// Do this process of user has click a file name to view or remove
if ($_GET) {
  $iid = $_GET["iid"];
  $act = $_GET["act"];
  switch ($act) {
    case rem:
      $sql = "DELETE FROM images WHERE image_id=$iid";
      @mysql_query ($sql, $conn);
      Header("Location:./index.php");
      exit();
      break;
    default:
      print "<img src=\"image.php?iid=$iid\">";
      break;
  }
}

?>
<html>
<head>
<title>Storing Images in DB</title>
</head>
<body>
<form method="post" enctype="multipart/form-data">
Select Image File: <input type="file" name="userfile"  size="40"><input type="submit" value="submit">
</form>
<?php
  $sql = "SELECT * FROM images ORDER BY image_date DESC";
  $result = mysql_query ($sql, $conn);
  if (mysql_num_rows($result)>0) {
    while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
      $i++;
      $str .= $i.". ";
      $str .= "<a href=\"index.php?iid=".$row["image_id"]."\">".$row["image_name"]."</a> ";
      $str .= "[".$row["image_date"]."] ";
      $str .= "[".$row["image_size"]."] ";
      $str .= "[<a href=\"index.php?act=rem&iid=".$row["image_id"]."\">Remove</a>]<br>";
    }
    print $str;
  }
?>
</body>
</html>

mysql php

3 Contributors
4 Replies
216 Views
1 Day Discussion Span
Latest Post 19 Years Ago Latest Post by val542

Troy 35 Posting Whiz

19 Years Ago

For your mysql error, test your SQL statement. Right after your line that builds your SQL statement, do this:

echo $sql;
exit();

Examine the output to see if something is wrong with the SQL statement. Try running it directly against your database using phpMyAdmin or the command line client.

For your fopen error, as you know, it happens when you submit a blank form. Yet, the code says "Oh, a form was submitted, let's open the file the user uploaded." But you didn't select a file--you submitted a blank form. SO...you need to add some code to test the validity of the form. Change your "if ($_FILES) {" line with this:

if ($_FILES['userfile']['error'] == 0) {

Reply to this topic

Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.

Yuki H. 0 Light Poster · Answer 1 · 2005-07-05T10:29:42+00:00

I do not get "Try running it directly against your database using phpMyAdmin or the command line client."
When I add
if ($_FILES == 0) {
I get:
Warning: fread(): supplied argument is not a valid stream resource in /home/user/public_html/im2db/index.php on line 19
as an error but now it is visible once I go on the index page.
Plus the image does not show up, only as a red x

<?php
// database connection
$conn = mysql_connect("localhost", "user", "pass") OR DIE (mysql_error());
@mysql_select_db ("user_images", $conn) OR DIE (mysql_error());

// Do this process if user has browse the file and click the submit button
if ($_FILES['userfile']['error'] == 0) { 

  $image_types = Array ("image/bmp",
                        "image/jpeg",
                        "image/pjpeg",
                        "image/gif",
                        "image/x-png");

  $userfile  = addslashes (fread (fopen ($_FILES["userfile"]["tmp_name"], "r"), 

filesize ($_FILES["userfile"]["tmp_name"])));
  $file_name = $_FILES["userfile"]["name"];
  $file_size = $_FILES["userfile"]["size"];
  $file_type = $_FILES["userfile"]["type"];

  if (in_array (strtolower ($file_type), $image_types)) {
    $sql = "INSERT INTO images (image_type, image, image_size, image_name, image_date) 

";
    $sql.= "VALUES (";
    $sql.= "'{$file_type}', '{$userfile}', '{$file_size}', '{$file_name}', NOW())";
    @mysql_query ($sql, $conn);
    Header("Location:".$_SERVER["PHP_SELF"]);
    exit();
  }
}

// Do this process of user has click a file name to view or remove
if ($_GET) {
  $iid = $_GET["iid"];
  $act = $_GET["act"];
  switch ($act) {
    case rem:
      $sql = "DELETE FROM images WHERE image_id=$iid";
      @mysql_query ($sql, $conn);
      Header("Location:./index.php");
      exit();
      break;
    default:
      print "<img src=\"image.php?iid=$iid\">";
      break;
  }
}

?>
<html>
<head>
<title>Storing Images in DB</title>
</head>
<body>
<form method="post" enctype="multipart/form-data">
Select Image File: <input type="file" name="userfile"  size="40"><input type="submit" 

value="submit">
</form>
<?php
  $sql = "SELECT * FROM images ORDER BY image_date DESC";
  $result = mysql_query ($sql, $conn);
  if (mysql_num_rows($result)>0) {
    while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
      $i++;
      $str .= $i.". ";
      $str .= "<a href=\"index.php?iid=".$row["image_id"]."\">".$row

["image_name"]."</a> ";
      $str .= "[".$row["image_date"]."] ";
      $str .= "[".$row["image_size"]."] ";
      $str .= "[<a href=\"index.php?act=rem&iid=".$row["image_id"]."\">Remove</a>]

<br>";
    }
    print $str;
  }
?>
</body>
</html>

with images.php as a result: SELECT * FROM images WHERE image_id=
when I go to the page.

<?php

// database connection
$conn = mysql_connect("localhost", "user", "pass") OR DIE (mysql_error());
@mysql_select_db ("user_images", $conn) OR DIE (mysql_error());
$sql    = "SELECT * FROM images WHERE image_id=".$_GET["iid"]; 

$result = mysql_query ($sql, $conn);
echo $sql;
exit(); 
if (mysql_num_rows ($result)>0)

 {
  $row = @mysql_fetch_array ($result) or die (mysql_error());
  $image_type = $row["image_type"];
  $image = $row["image"];
  Header ("Content-type: $image_type");
  print $image;

}
?>

Thank you a lot for helping me a lot though =O.

Troy 35 Posting Whiz · Answer 2 · 2005-07-05T18:18:54+00:00

Troubleshooting 101
When troubleshooting code problems, the secret is baby-steps. When you have an error, it is always a single statement (or line) that is producing the error. Create a new file, and put in only enough code to produce your problem, then work until you solve it--one line at a time. Baby-steps.

In keeping with the baby-steps philosophy, let's attack only one problem at a time. Let's look at your SQL problem. Create a new file with only this code:

$conn = mysql_connect("localhost", "user", "pass") OR DIE (mysql_error()); 
@mysql_select_db ("user_images", $conn) OR DIE (mysql_error()); 

$sql	= "SELECT * FROM images WHERE image_id=".$_GET["iid"]; 
echo $sql; 

if (!$result = mysql_query ($sql, $conn)) {
  echo mysql_error();
} else {
  echo "<br />query successful";
}

Now, call this script directly within your browser using a URL such as http://myserver.com/myscript.php?iid=1

The script should display in your browser the SQL statement that it runs against the database. Examine that statement--do you see anything wrong with it? If not, then copy the statement from your browser window and run it directly against your database. What I mean by that is if you are working with a database, you must have some tool you use to create your tables--right? For many people with MySql, that tool is phpMyAdmin. Do you have access to use phpMyAdmin? If not, whatever tool you use to manage you database, should have a way to paste in a query and execute it. This way you can test that the SQL statement you are generating in your PHP code is actually a valid statement that your database can execute. Maybe it is a valid statement, but it does not return any rows because no rows match the query.

Once you work through this part, you can add more steps---one at a time--baby steps--until you have the whole script working.

val542 0 Newbie Poster · Answer 3 · 2005-07-06T02:43:08+00:00

always make a test condition before using
mysql_num_rows()
because if the table is empty so the process is stopped

$array[]=@mysql_num_rows(elements functions);

if (count($array)==0) //ur table is empty

that's all!!