value taking from db (mysql)

20 New Member

i want to display the path of file in browser that i am uploaded to the server
for later downloading but when iam trying to do this using the mysql_fetch_array()
it not working

an error like

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\xampp\session1\upload.php on line 47

here the code i used

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 <html>

<head></head>

<body>

<form enctype="multipart/form-data" action="#" method="POST">

<input type="hidden" name="MAX_FILE_SIZE" value="100000" />

Choose a file to upload: <input name="uploadedfile" type="file" /><br />

<input type="submit" value="Upload File" />

</form>

</body>

</html>

<?php

//include("dbinfo.inc.php");

function db_connect()

{

$dbHost = "localhost";

$dbUser = "root";

$dbDatabase = "ajeesh";

$db = mysql_connect("$dbHost", "$dbUser", "") or die ("Error connecting to database.");

mysql_select_db("$dbDatabase", $db) or die ("Couldn't select the database.");

}
 
echo db_connect();

$target_path = "uploads/";
 
$name=$_FILES['uploadedfile']['name'];

$size=$_FILES['uploadedfile']['size'];

$type=$_FILES['uploadedfile']['type'];

$path="uploads/".$name;

echo "-------------".$name;

echo "<br />-----------".$tmpname;

$size=$size/1024;

echo "<br />-------------".$size."kb"."<p></p>";

echo "-------------------".$type;

$target_path = $target_path . basename( $_FILES['uploadedfile']['name']); 
 
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) 

{

    echo "The file ".( $_FILES['uploadedfile']['name']). " has been uploaded";

    $sql="INSERT INTO upload "."(name,size,path) "."VALUES "."('$name','$size','$tmpname')";

    $result=mysql_query($sql);

    //echo $result;

    //downloading-----------------------------------------------------------------------------------------------------------

    $sqlc="select path from upoad where name='arrow.png'";

    $result=mysql_query($sqlc);
 
     while ($line = mysql_fetch_array($result)) {
 
    foreach ($line as $col_value) {

        echo "$col_value";

    }

}

    echo "********************************************************";

    echo $path;

}

 else

{

    echo "There was an error uploading the file, please try again!";

}
 
mysql_close(db_conect());
 
?>

Oct 5 '09 #1

Subscribe Reply

1769

ajeeshc

New Member

how to take a value from the database and shown to the client

Oct 5 '09 #2

TheServant

1,168

Recognized Expert Top Contributor

@ajeeshc
First of all, please use [code] tags around your code so we can read it.

You will notice that you are looking in the table "upoad" rather than "upload" so I am guessing that is the major problem.

Consider using mysql syntax by puting the commands in capitals to improve readability like:

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SELECT path FROM upload WHERE name='arrow.png'

It may seem silly for such a simple statement, but it's a practice which makes your code a lot easier to read for you and anyone else!

How you can search for this in the future is when developing, you can include something like:

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 $result=mysql_query($sqlc) or die('There has been an error: '.mysql_error());
 

Which will tell you (for example): table "upoad" could not be found.

Let us know if you still have problems and I will have a more thorough look.

Oct 5 '09 #3

TheServant

1,168

Recognized Expert Top Contributor

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 $raw_result = mysql_query("SELECT value FROM table WHERE something='$something'");

$result = mysql_fetch_array($raw_result)

Have a look in Google

I am pretty sure you know this from your other post, so could you be more specific what you don't know?

Oct 5 '09 #4

ajeeshc

New Member

i tried
but error showing
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\xampp\session1\upload.php on line 47

Oct 5 '09 #5

ajeeshc

New Member

i am very new to php i try to display the file that uploaded to the db in this the while loop is not working i cant figure it out any one please help ........

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 <html>

<head></head>

<body>

<form enctype="multipart/form-data" action="#" method="POST">

<input type="hidden" name="MAX_FILE_SIZE" value="100000" />

Choose a file to upload: <input name="uploadedfile" type="file" /><br />

<input type="submit" value="Upload File" />

</form>

</body>

</html>

<?php

//include("dbinfo.inc.php");
 
$dbHost = "localhost";

$dbUser = "root";

$dbDatabase = "ajeesh";

$db = mysql_connect("$dbHost", "$dbUser", "") or die ("Error connecting to database.");

mysql_select_db("$dbDatabase", $db) or die ("Couldn't select the database.");
 
$target_path = "uploads/";
 
$name=$_FILES['uploadedfile']['name'];

$size=$_FILES['uploadedfile']['size'];

$type=$_FILES['uploadedfile']['type'];

$path="uploads/".$name;

echo "-------------".$name;

//echo "<br />-----------".$tmpname;

$size=$size/1024;

echo "<br />-------------".$size."kb"."<p></p>";

echo "-------------------".$type."<p></p>";

$target_path = $target_path . basename( $_FILES['uploadedfile']['name']); 
 
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) 

{
 
    echo "The file ".( $_FILES['uploadedfile']['name']). " has been uploaded"."<p></p>";

    $sql="INSERT INTO upload "."(name,size,path) "."VALUES "."('$name','$size','$tmpname')";

    $result=mysql_query($sql);

    //echo $result;

    //downloading-----------------------------------------------------------------------------------------------------------

    $sqlc="select path from upload where name='arrow.png'";

    $resultc=mysql_query($sqlc);

    //$line=mysql_fetch_array($resultc);
 
     while ( $line=mysql_fetch_array($resultc))

      {
 
          echo "wwwwwwwwwwwwwwwwwwwwwww";

        echo "Name :{$line['name']} <br>" ;

            }
 
    echo "-------------".$path;

}
 
 else

{

    echo "There was an error uploading the file, please try again!";

}
 
mysql_close($db);
 
?>

Oct 5 '09 #6

Markus

6,050

Recognized Expert Expert

@ajeeshc
Please see reply #3. You have an error in your MySQL query that is preventing mysql_query() from returning a valid resource to be used in mysql_fetch_array() (as the error quite clearly explains). So, if you apply the technique suggested by TheServant, that is, the section on using 'or die(mysql_erro());', you'll be informed of the error in your query.

Mark.

Oct 5 '09 #7

TheServant

1,168

Recognized Expert Top Contributor

I'm confused, were these topics merged? I am sure they were two different ones before??

Oct 5 '09 #8

Atli

5,058

Recognized Expert Expert

@TheServant
Yes, this topic was posted in three different threads, which I merged into this one thread.

Oct 6 '09 #9

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