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Uploading images to a database - help needed

  • gromituk2k
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Post July 3rd, 2009, 1:54 am

I am trying to upload images to a MySQL database, along with some other data.

The HTML page, containing the upload from looks like this:
Code: [ Select ]
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
 
<body>
<form method="post" action="insert.php" enctype="multipart/form-data">
    <p>Your name:<br>
    <input type="text" name="form_visitor"  size="40">
    <br>Loved ones name:<br>
    <input type="text" name="form_lovedone"  size="40">
    <input type="hidden" name="MAX_FILE_SIZE" value="1000000">
    <br>File to upload/store in database:<br>
    <input type="file" name="form_data"  size="40">
    <br>Comments:<br>
    <input type="text" name="form_comment" size="50">
    </p>
    <input type="submit" name="submit" value="submit">
</form>
</body>
</html>
 
  1. <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
  2. <html xmlns="http://www.w3.org/1999/xhtml">
  3. <head>
  4. <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
  5. <title>Untitled Document</title>
  6. </head>
  7.  
  8. <body>
  9. <form method="post" action="insert.php" enctype="multipart/form-data">
  10.     <p>Your name:<br>
  11.     <input type="text" name="form_visitor"  size="40">
  12.     <br>Loved ones name:<br>
  13.     <input type="text" name="form_lovedone"  size="40">
  14.     <input type="hidden" name="MAX_FILE_SIZE" value="1000000">
  15.     <br>File to upload/store in database:<br>
  16.     <input type="file" name="form_data"  size="40">
  17.     <br>Comments:<br>
  18.     <input type="text" name="form_comment" size="50">
  19.     </p>
  20.     <input type="submit" name="submit" value="submit">
  21. </form>
  22. </body>
  23. </html>
  24.  


And the PHP file for adding the data is like it:
Code: [ Select ]
<?php
$con = mysql_connect("localhost","user","pass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
 
mysql_select_db("database_name", $con);
 
if(!$data = addslashes(@fread(@fopen($dir.'/'.$file, "r"), @filesize($dir.'/'.$file)))){
   die('Cannot create $data');
}
 
@fclose($dir.'/'.$file);
 
$sql="INSERT INTO commentbook (visitor , lovedone, bin_data, filename, filesize, filetype, comment, id)
VALUES
('$_POST[form_visitor]','$_POST[form_lovedone]','$_POST[data]','$_POST[form_data_name]','$_POST[form_data_size]','$_POST[form_data_type]','$_POST[form_comment]','')";
 
if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";
 
mysql_close($con)
?>
  1. <?php
  2. $con = mysql_connect("localhost","user","pass");
  3. if (!$con)
  4.   {
  5.   die('Could not connect: ' . mysql_error());
  6.   }
  7.  
  8. mysql_select_db("database_name", $con);
  9.  
  10. if(!$data = addslashes(@fread(@fopen($dir.'/'.$file, "r"), @filesize($dir.'/'.$file)))){
  11.    die('Cannot create $data');
  12. }
  13.  
  14. @fclose($dir.'/'.$file);
  15.  
  16. $sql="INSERT INTO commentbook (visitor , lovedone, bin_data, filename, filesize, filetype, comment, id)
  17. VALUES
  18. ('$_POST[form_visitor]','$_POST[form_lovedone]','$_POST[data]','$_POST[form_data_name]','$_POST[form_data_size]','$_POST[form_data_type]','$_POST[form_comment]','')";
  19.  
  20. if (!mysql_query($sql,$con))
  21.   {
  22.   die('Error: ' . mysql_error());
  23.   }
  24. echo "1 record added";
  25.  
  26. mysql_close($con)
  27. ?>


I have had the form uploading all data except the image, but currently it is not uploading anything :(

Any ideas?
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Post July 3rd, 2009, 1:54 am

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Post July 3rd, 2009, 11:53 am

I wrote a tutorial on this

Check out http://www.rabiddog.co.za/tutorial.php? ... usingMySQL

Gimme a shout if you need any help
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  • gromituk2k
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Post July 6th, 2009, 1:19 am

I seem to be getting an unexpected " on the following line;

VALUES("" . addslashes($dataRead) "",

Any ideas?
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Post July 6th, 2009, 2:34 am

have you done any work with PHP?

Code: [ Select ]
 
$query = "INSERT INTO ImageTable(imageData,imageType)
          VALUES(\"" . addslashes($dataRead) . "\",
                 \"" . mysql_escape_string($fType) . "\");";
 
  1.  
  2. $query = "INSERT INTO ImageTable(imageData,imageType)
  3.           VALUES(\"" . addslashes($dataRead) . "\",
  4.                  \"" . mysql_escape_string($fType) . "\");";
  5.  


I had the slashes and dots in there, just seems that the manager I wrote initially took them out. So I have fixed the tutorial. Thanks for pointing it out
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  • gromituk2k
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Post July 6th, 2009, 3:07 am

Hmm still not working, not even getting any errors now, just not uploading to the database.

This is the PHP that I am using:

Code: [ Select ]
 
<?php
$server = "localhost";
$username = "username";
$password = "password";
 
$dbHandle = mysql_connect($server,$username,$password)
            or die("unable to conect to database server");
 
mysql_select_db("database_name")
                or die ("Couldn't use Database");
 
 
$upload= (empty($_FILES['image'])) ? 0 : $_FILES['image'];
$fileName = $upload['tmp_name'];
$fType = $upload['type'];
$size = $upload['size'];
$timer = $size * 1024;
set_time_limit($timer);
 
$open = fopen($fileName,"r");
$data_read = fread($open, $size);
fclose($open);
$query = "INSERT INTO ImageTable(imageData,imageType)
          VALUES(\"" . addslashes($dataRead) . "\",
                 \"" . mysql_escape_string($fType) . "\");";
?>
 
  1.  
  2. <?php
  3. $server = "localhost";
  4. $username = "username";
  5. $password = "password";
  6.  
  7. $dbHandle = mysql_connect($server,$username,$password)
  8.             or die("unable to conect to database server");
  9.  
  10. mysql_select_db("database_name")
  11.                 or die ("Couldn't use Database");
  12.  
  13.  
  14. $upload= (empty($_FILES['image'])) ? 0 : $_FILES['image'];
  15. $fileName = $upload['tmp_name'];
  16. $fType = $upload['type'];
  17. $size = $upload['size'];
  18. $timer = $size * 1024;
  19. set_time_limit($timer);
  20.  
  21. $open = fopen($fileName,"r");
  22. $data_read = fread($open, $size);
  23. fclose($open);
  24. $query = "INSERT INTO ImageTable(imageData,imageType)
  25.           VALUES(\"" . addslashes($dataRead) . "\",
  26.                  \"" . mysql_escape_string($fType) . "\");";
  27. ?>
  28.  
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Post July 6th, 2009, 3:38 am

Can you echo out some data so we can debug this thing? Prehaps echo things like file size etc.

Also please make sure the they $POST index keys match the names you have given them in the form. I can see you have retained $POST['image'] where it should be $POST['form_data'].

You are going to have to adjust the script you found in the tutorial page to fit your requirements. :)

If you read the entire tutorial you would notice that the table structure is

Code: [ Select ]
[ImageTable]
id (bigint, primary key)
imageData (blob)
imageType (varchar(50))
  1. [ImageTable]
  2. id (bigint, primary key)
  3. imageData (blob)
  4. imageType (varchar(50))


and the form structure is
Code: [ Select ]
< form method="post" action="uploadImage.php"
enctype="multipart/form-data">
< input type="file" name="image" />
< input type="submit" value=" Upload " />
< /form>
  1. < form method="post" action="uploadImage.php"
  2. enctype="multipart/form-data">
  3. < input type="file" name="image" />
  4. < input type="submit" value=" Upload " />
  5. < /form>
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  • gromituk2k
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Post July 6th, 2009, 4:45 am

Hmm, I tried echoing the data and I am getting results with that, something like this;
size - 4388
temp name - /tmp/phpLl1740
type - image/gif

I have also changed something in the php file, as I noticed a couple of problems, so now using:
Code: [ Select ]
<?php
$server = "localhost";
$username = "username";
$password = "pass";

$dbHandle = mysql_connect($server,$username,$password)
            or die("Unable to conect to database server");

mysql_select_db("datbase-name")
                or die ("Couldn't use Database");

$upload= (empty($_FILES['image'])) ? 0 : $_FILES['image'];
$fileName = $upload['tmp_name'];
$fType = $upload['type'];
$size = $upload['size'];
$timer = $size * 1024;
set_time_limit($timer);

$open = fopen($fileName,"r");
$data_read = fread($open, $size);
fclose($open);
$query = "INSERT INTO commentbook(imageData,imageType)
         VALUES(\"" . addslashes($data_read) . "\",
                \"" . mysql_escape_string($fType) . "\");";
        
echo "size - $size";
echo "</br>";
echo "temp name - $fileName";
echo "</br>";
echo "type - $fType";

?>
  1. <?php
  2. $server = "localhost";
  3. $username = "username";
  4. $password = "pass";
  6. $dbHandle = mysql_connect($server,$username,$password)
  7.             or die("Unable to conect to database server");
  9. mysql_select_db("datbase-name")
  10.                 or die ("Couldn't use Database");
  12. $upload= (empty($_FILES['image'])) ? 0 : $_FILES['image'];
  13. $fileName = $upload['tmp_name'];
  14. $fType = $upload['type'];
  15. $size = $upload['size'];
  16. $timer = $size * 1024;
  17. set_time_limit($timer);
  19. $open = fopen($fileName,"r");
  20. $data_read = fread($open, $size);
  21. fclose($open);
  22. $query = "INSERT INTO commentbook(imageData,imageType)
  23.          VALUES(\"" . addslashes($data_read) . "\",
  24.                 \"" . mysql_escape_string($fType) . "\");";
  25.         
  26. echo "size - $size";
  27. echo "</br>";
  28. echo "temp name - $fileName";
  29. echo "</br>";
  30. echo "type - $fType";
  32. ?>


The databse is set out the same as yours, apart from the table name, which I have changed. So there seems to be a problem with the inserting into the table???
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Post July 6th, 2009, 4:58 am

HAHAHHAHAHA I just noticed now you not executing the query! You have to execute the query against the database to put it into the db :) the $query variable is just a formatted sql statement.

if you check the tutorial page it links to a page that tells you how to execute queries against the mysql database
Code: [ Select ]
// execute your query against the database here.
// Connecting to a MySql Database
  1. // execute your query against the database here.
  2. // Connecting to a MySql Database


http://www.rabiddog.co.za/tutorial.php? ... qlDatabase
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Post July 6th, 2009, 5:39 am

Can you tell I am a little new to this :D

Thanks that seems to of worked fine.

I have now added a few other fields to the database and added these to the form. I have tried adding these to the INSERT query, the dtabase entries and the form entries, like this:

Code: [ Select ]
$query = "INSERT INTO commentbook(imageData,imageType,visitor,lovedone,comment)
         VALUES(\"" . addslashes($data_read) . "\",
                \"" . mysql_escape_string($fType) . "\",
                visitor,
                lovedone,
                comment);";
  1. $query = "INSERT INTO commentbook(imageData,imageType,visitor,lovedone,comment)
  2.          VALUES(\"" . addslashes($data_read) . "\",
  3.                 \"" . mysql_escape_string($fType) . "\",
  4.                 visitor,
  5.                 lovedone,
  6.                 comment);";


But no luck, the images are still getting added, but the other information is not.
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Post July 6th, 2009, 6:00 am

Code: [ Select ]
$query = "INSERT INTO commentbook(imageData,imageType,visitor,lovedone,comment)
         VALUES(\"" . addslashes($data_read) . "\",
                \"" . mysql_escape_string($fType) . "\",
                visitor,
                lovedone,
                comment);";
  1. $query = "INSERT INTO commentbook(imageData,imageType,visitor,lovedone,comment)
  2.          VALUES(\"" . addslashes($data_read) . "\",
  3.                 \"" . mysql_escape_string($fType) . "\",
  4.                 visitor,
  5.                 lovedone,
  6.                 comment);";


where are you assigning values to the visitor, lovedone and comment you want to insert? Prehaps I can help by explaining the structure of the query

Code: [ Select ]
$query = "INSERT INTO commentbook(imageData,imageType,visitor,lovedone,comment)
         VALUES(\"" . addslashes($data_read) . "\",
                \"" . mysql_escape_string($fType) . "\",
                visitor,
                lovedone,
                comment);";
//commentbook(imageData,imageType,visitor,lovedone,comment)
//commentbook is the name of the table
//imageData, imageType, visitor, lovedone, comment are the names of
//the columns that data is going to be inserted into

//Anything between the VALUES brackets needs to be assigned a value
//so $data_read contains the byte array, $fType contains the file type value
// visitor should be declared something to the effect of $visitor etc
  1. $query = "INSERT INTO commentbook(imageData,imageType,visitor,lovedone,comment)
  2.          VALUES(\"" . addslashes($data_read) . "\",
  3.                 \"" . mysql_escape_string($fType) . "\",
  4.                 visitor,
  5.                 lovedone,
  6.                 comment);";
  7. //commentbook(imageData,imageType,visitor,lovedone,comment)
  8. //commentbook is the name of the table
  9. //imageData, imageType, visitor, lovedone, comment are the names of
  10. //the columns that data is going to be inserted into
  12. //Anything between the VALUES brackets needs to be assigned a value
  13. //so $data_read contains the byte array, $fType contains the file type value
  14. // visitor should be declared something to the effect of $visitor etc


I can only help you if you post the form you are using so I can see the name values of your input

Just in closing on this post your query might be better off like this
Code: [ Select ]
$query = "INSERT INTO commentbook(imageData,imageType,visitor,lovedone,comment)
         VALUES(\"" . addslashes($data_read) . "\",
                \"" . mysql_escape_string($fType) . "\",
                \"" . mysql_escape_string($visitor) . "\",
                \"" . mysql_escape_string($lovedone) . "\",
                \"" . mysql_escape_string($comment) . "\");";
  1. $query = "INSERT INTO commentbook(imageData,imageType,visitor,lovedone,comment)
  2.          VALUES(\"" . addslashes($data_read) . "\",
  3.                 \"" . mysql_escape_string($fType) . "\",
  4.                 \"" . mysql_escape_string($visitor) . "\",
  5.                 \"" . mysql_escape_string($lovedone) . "\",
  6.                 \"" . mysql_escape_string($comment) . "\");";


I could go into detail about creating a formatted string template then returning a formatted string as opposed to dynamically concating the entire string but I figured I would keep it simple or now :) Try the above and see what happenes
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Post July 6th, 2009, 6:05 am

gromituk2k wrote:
I am trying to upload images to a MySQL database, along with some other data.

The HTML page, containing the upload from looks like this:
Code: [ Select ]
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
 
<body>
<form method="post" action="insert.php" enctype="multipart/form-data">
    <p>Your name:<br>
    <input type="text" name="form_visitor"  size="40">
    <br>Loved ones name:<br>
    <input type="text" name="form_lovedone"  size="40">
    <input type="hidden" name="MAX_FILE_SIZE" value="1000000">
    <br>File to upload/store in database:<br>
    <input type="file" name="form_data"  size="40">
    <br>Comments:<br>
    <input type="text" name="form_comment" size="50">
    </p>
    <input type="submit" name="submit" value="submit">
</form>
</body>
</html>
 
  1. <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
  2. <html xmlns="http://www.w3.org/1999/xhtml">
  3. <head>
  4. <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
  5. <title>Untitled Document</title>
  6. </head>
  7.  
  8. <body>
  9. <form method="post" action="insert.php" enctype="multipart/form-data">
  10.     <p>Your name:<br>
  11.     <input type="text" name="form_visitor"  size="40">
  12.     <br>Loved ones name:<br>
  13.     <input type="text" name="form_lovedone"  size="40">
  14.     <input type="hidden" name="MAX_FILE_SIZE" value="1000000">
  15.     <br>File to upload/store in database:<br>
  16.     <input type="file" name="form_data"  size="40">
  17.     <br>Comments:<br>
  18.     <input type="text" name="form_comment" size="50">
  19.     </p>
  20.     <input type="submit" name="submit" value="submit">
  21. </form>
  22. </body>
  23. </html>
  24.  


And the PHP file for adding the data is like it:
Code: [ Select ]
<?php
$con = mysql_connect("localhost","user","pass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
 
mysql_select_db("database_name", $con);
 
if(!$data = addslashes(@fread(@fopen($dir.'/'.$file, "r"), @filesize($dir.'/'.$file)))){
   die('Cannot create $data');
}
 
@fclose($dir.'/'.$file);
 
$sql="INSERT INTO commentbook (visitor , lovedone, bin_data, filename, filesize, filetype, comment, id)
VALUES
('$_POST[form_visitor]','$_POST[form_lovedone]','$_POST[data]','$_POST[form_data_name]','$_POST[form_data_size]','$_POST[form_data_type]','$_POST[form_comment]','')";
 
if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";
 
mysql_close($con)
?>
  1. <?php
  2. $con = mysql_connect("localhost","user","pass");
  3. if (!$con)
  4.   {
  5.   die('Could not connect: ' . mysql_error());
  6.   }
  7.  
  8. mysql_select_db("database_name", $con);
  9.  
  10. if(!$data = addslashes(@fread(@fopen($dir.'/'.$file, "r"), @filesize($dir.'/'.$file)))){
  11.    die('Cannot create $data');
  12. }
  13.  
  14. @fclose($dir.'/'.$file);
  15.  
  16. $sql="INSERT INTO commentbook (visitor , lovedone, bin_data, filename, filesize, filetype, comment, id)
  17. VALUES
  18. ('$_POST[form_visitor]','$_POST[form_lovedone]','$_POST[data]','$_POST[form_data_name]','$_POST[form_data_size]','$_POST[form_data_type]','$_POST[form_comment]','')";
  19.  
  20. if (!mysql_query($sql,$con))
  21.   {
  22.   die('Error: ' . mysql_error());
  23.   }
  24. echo "1 record added";
  25.  
  26. mysql_close($con)
  27. ?>


I have had the form uploading all data except the image, but currently it is not uploading anything :(

Any ideas?


There is a simple solution to this. I am not sure if I am hearing right or you are doing a totally different thing.

Can you tell me, what exactly do you want to do with the image?
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Post July 6th, 2009, 6:10 am

He is trying to upload an image to the server and write it to the database along with a few other variables. Writing the data to the database is the key here :)
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Post July 6th, 2009, 6:15 am

Okay, there is this 'lazy' work here but I do not think you and him would like it. :)

I am going to just stop here.

/*Is he going to just store the images in the database, Rabid Dog?

:)
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Post July 6th, 2009, 6:19 am

Seem to be getting the strangest of problems now. The extra fields that I have added, onlt the first character is getting added to the database.

The PHP I am using is:

Code: [ Select ]
 
<?php
 
$server = "localhost";
$username = "user";
$password = "pass";
 
$dbHandle = mysql_connect($server,$username,$password)
            or die("Unable to conect to database server");
 
mysql_select_db("databse-name")
                or die ("Couldn't use Database");
 
$upload= (empty($_FILES['image'])) ? 0 : $_FILES['image'];
$fileName = $upload['tmp_name'];
$fType = $upload['type'];
$size = $upload['size'];
$timer = $size * 1024;
set_time_limit($timer);
$visitor = $visitor['visitor'];
$lovedone = $lovedone['lovedone'];
$comment = $comment['comment'];
 
$open = fopen($fileName,"r");
$data_read = fread($open, $size);
fclose($open);
$query = "INSERT INTO commentbook(id,imageData,imageType,visitor,lovedone,comment)
          VALUES(\"" . addslashes($data_read) . "\",
                 \"" . mysql_escape_string($fType) . "\",
                 \"" . mysql_escape_string($fType) . "\",
                 \"" . mysql_escape_string($visitor) . "\",
                 \"" . mysql_escape_string($lovedone) . "\",
                 \"" . mysql_escape_string($comment) . "\");";
 
mysql_query($query, $dbHandle);
if(mysql_affected_rows($dbHandle) > 0){
  echo "Successfully inserted data </br>";
}else{
  echo mysql_error();
}
 
echo "size - $size";
echo "</br>";
echo "temp name - $fileName";
echo "</br>";
echo "type - $fType";
 
?>
  1.  
  2. <?php
  3.  
  4. $server = "localhost";
  5. $username = "user";
  6. $password = "pass";
  7.  
  8. $dbHandle = mysql_connect($server,$username,$password)
  9.             or die("Unable to conect to database server");
  10.  
  11. mysql_select_db("databse-name")
  12.                 or die ("Couldn't use Database");
  13.  
  14. $upload= (empty($_FILES['image'])) ? 0 : $_FILES['image'];
  15. $fileName = $upload['tmp_name'];
  16. $fType = $upload['type'];
  17. $size = $upload['size'];
  18. $timer = $size * 1024;
  19. set_time_limit($timer);
  20. $visitor = $visitor['visitor'];
  21. $lovedone = $lovedone['lovedone'];
  22. $comment = $comment['comment'];
  23.  
  24. $open = fopen($fileName,"r");
  25. $data_read = fread($open, $size);
  26. fclose($open);
  27. $query = "INSERT INTO commentbook(id,imageData,imageType,visitor,lovedone,comment)
  28.           VALUES(\"" . addslashes($data_read) . "\",
  29.                  \"" . mysql_escape_string($fType) . "\",
  30.                  \"" . mysql_escape_string($fType) . "\",
  31.                  \"" . mysql_escape_string($visitor) . "\",
  32.                  \"" . mysql_escape_string($lovedone) . "\",
  33.                  \"" . mysql_escape_string($comment) . "\");";
  34.  
  35. mysql_query($query, $dbHandle);
  36. if(mysql_affected_rows($dbHandle) > 0){
  37.   echo "Successfully inserted data </br>";
  38. }else{
  39.   echo mysql_error();
  40. }
  41.  
  42. echo "size - $size";
  43. echo "</br>";
  44. echo "temp name - $fileName";
  45. echo "</br>";
  46. echo "type - $fType";
  47.  
  48. ?>


And the HTML form is:

Code: [ Select ]
 
<form method="post" action="insert.php"
 enctype="multipart/form-data">
  <p>
  <input type="file" name="image" />
  </p>
  <p>
          Name:
      <input type="text" name="visitor" />
  </p>
  <p>
    Loved one:
    <input type="text" name="lovedone"/>
  </p>
  <p>
      Comment:
      <textarea name="comment" cols="45" rows="5"></textarea>
  </p>
  <p>
    <input type="submit" value=" Upload " />
  </p>
</form>
 
  1.  
  2. <form method="post" action="insert.php"
  3.  enctype="multipart/form-data">
  4.   <p>
  5.   <input type="file" name="image" />
  6.   </p>
  7.   <p>
  8.           Name:
  9.       <input type="text" name="visitor" />
  10.   </p>
  11.   <p>
  12.     Loved one:
  13.     <input type="text" name="lovedone"/>
  14.   </p>
  15.   <p>
  16.       Comment:
  17.       <textarea name="comment" cols="45" rows="5"></textarea>
  18.   </p>
  19.   <p>
  20.     <input type="submit" value=" Upload " />
  21.   </p>
  22. </form>
  23.  


The extra entries in the table are two varchar(50) and one longtext.
  • Rabid Dog
  • Web Master
  • Web Master
  • User avatar
  • Joined: May 21, 2004
  • Posts: 3229
  • Loc: South Africa
  • Status: Offline

Post July 6th, 2009, 6:23 am

George L. wrote:
Okay, there is this 'lazy' work here but I do not think you and him would like it. :)

I am going to just stop here.

/*Is he going to just store the images in the database, Rabid Dog?

:)


Yes he is :)
Watch me grow
  • Anonymous
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  • Joined: 25 Feb 2008
  • Posts: ?
  • Loc: Ozzuland
  • Status: Online

Post July 6th, 2009, 6:23 am

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