Hi,

I store images in DB properly... no issues using the
following code....

<?php

// store.php3 - by Florian Dittmer <[EMAIL PROTECTED]>
// Example php script to demonstrate the storing of
binary files into
// an sql database. More information can be found at
http://www.phpbuilder.com/
?>

<html>
<head><title>Store binary data into SQL
Database</title></head>
<body>

<?php
// code that will be executed if the form has been
submitted:

if ($submit) {

        // Include constants for database connectivity
        include("constants.php");

        //      The Database server connection function
        $connection     =
mysql_connect("$host","$db_user","$db_pass") or
die("Couldn't connect to the MySql Database Server");
        
        //      Database selection from the DB server
        $db     =       mysql_select_db("$db_name",$connection) or
die("Couldn't select the required Database.");

    $data = addslashes(fread(fopen($banner_image,
"r"), filesize($banner_image)));

    $result     =       MYSQL_QUERY("INSERT INTO banner_details
(text,banner_image,filename,filesize,filetype) ".
        "VALUES
('$banner_description','$data','$form_data_name','$form_data_size','$form_data_type')")
or die("Couldnt insert banner in the Db");

        echo "Banner uploaded";

        $id= mysql_insert_id();
    print "<p>This file has the following Database ID:
<b>$id</b>";

} else {

    // else show the form to submit new data:
?>

    <form method="post" action="<?php echo $PHP_SELF;
?>" enctype="multipart/form-data">
    Banner Text:<br>
    <input type="text" name="banner_description" 
size="40">
    <input type="hidden" name="MAX_FILE_SIZE"
value="1000000">
    <br>Image to upload/store in database:<br>
    <input type="file" name="banner_image"  size="40">
    <p><input type="submit" name="submit"
value="submit">
    </form>


<?php

        }

?>

</body>
</html>

------------

I use the foll file to reterieve it...

<?
/*
        Aim:            To fetch random image and its corresponding
text from the database.

        Working:        Generate a random number within the range of
1 and number of rows present in the DB and then fetch
the appropriate ROW ID.

        Includes:       constants.php for DB connectivity

*/
        // Include constants for database connectivity
        include("constants.php");

        //      The Database server connection function
        $connection     =
mysql_connect("$host","$db_user","$db_pass") or
die("Couldn't connect to the MySql Database Server");
        
        //      Database selection from the DB server
        $db     =       mysql_select_db("$db_name",$connection) or
die("Couldn't select the required Database.");

/*
Fetching the number of rows in the DB table
*/
        
        //      Reteriving the data from the banners table in the
DB
        $sql    =       "SELECT *"      .
                                "FROM   banner_details";

        //      Executing the Sql select statement
        $result =       mysql_query($sql, $connection) or
die("Couldn't Execute the query.");

        //      Getting the number of rows affected by the query
        $rows   =       mysql_num_rows($result);

        print("Rows in DB: $rows\n");

/*
Generating a Random number between 1 and the NUMBER OF
ROWS returned from the DB Table
*/

        //      Seeding the random number generator...
        //      this thing is usually used to SET an initial value
for the random number to be generated.
        mt_srand(doubleval(microtime()) *       100000000);

        //      The random number is generated...
        $random_number  =       mt_rand(1, $rows);

        print("The random number generated is
$random_number");
        echo "<BR><BR>";

/*
Fetching the actual data from the DB Table
*/
    $query = "select banner_image, text, filetype from
banner_details where id=$random_number";

    $result     =       mysql_query($query, $connection) or
die("Couldn't Execute the query.");

    $data = @MYSQL_RESULT($result,0,"banner_image");
    $text = @MYSQL_RESULT($result,0,"text");
        $type = @MYSQL_RESULT($result,0,"filetype");

        echo "The banner is ";

        Header("Content-type: $type");

        echo $data;


        echo "The assocaited text is : $text";


?>

--------------

And this for calling it...

<?php

echo "<img src=rotate.php>";

?>

I get an Img with a cross (broken)...

Whats the error?

Am i missing something?

Pls help

-Monil

=====
Best Regards,
Monil Chheda(INDIA)
http://domains.eliteral.com
===========================
===========================


                
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