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upload and post image with mysql


RCS

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Hi everyone, I'm trying to create a script where I can save form data e.g Make, Description, Price, Picture

after saving data to mysql I want to post it by id to a catalog page. I'm having an issue with getting image from database.

I can't seem to figure it out. If someone could help me it would be greatly appreciated.

Thank you very much in advance.

 

This is form

<html>

<body><form action="upload_file.php" method="POST"

enctype="multipart/form-data">

<p><label for="file">Filename:</label>

<input type="file" name="file" id="file" />

<input type="hidden" name="MAX_FILE_SIZE" value="100000000000000"></p>

<p><label>Make:</label>

<input type="text" name="make" id="make" /></p>

<p><label>Price:</label>

<input type="text" name="price" id="price" /></p>

<p><label>Category:</label>

<input type="text" name="category" id="category" /></p>

<p><label>Description:</label>

<textarea name="description" id="description" cols="45" rows="5"></textarea /></p>

 

 

<input type="submit" name="submit" value="Submit" />

</form></body>

</html>

 

This is upload page where everything gets saved to database

<?php

include("dbstuff.inc");

include("file_array.inc");

if(isset($_POST['upload']) && $_FILES['file']['size'] > 0)

{

$fp      = fopen($tmpName, 'r');

$content = fread($fp, filesize($tmpName));

$content = addslashes($content);

fclose($fp);

if(!get_magic_quotes_gpc())

{

    $fileName = addslashes($fileName);

}

}

$con = mysqli_connect($db_host, $db_user, $db_passwd, $db_name);

$query = "INSERT INTO upload(name, size, type, content, make, price, description)".

"VALUES ('$fileName', '$fileSize', '$fileType', '$content','$_POST[make]','$_POST[price]','$_POST[description]')";

$result=mysqli_query($con, $query) or die ("Could not complete query.");

mysql_close($con);

?>

<html>

<head><title>CMS</title></head>

<body><center>Information successfuly saved to database.</center></body>

</html>

 

This is page where data gets posted

<html>

<head>

<title>Download File From MySQL</title>

</head>

<body>

<?php

include("dbinfo.inc");

$con = mysqli_connect($db_host, $db_user, $db_passwd, $db_name);

$query = ("SELECT * FROM upload ORDER BY id");

$result = mysqli_query($con, $query) or die('Error, query failed');

echo "<table border='1'>

<tr>

<th>Make</th>

<th>Description</th>

<th>Price</th>

<th>Picture</th>

</tr>";

while($row = mysqli_fetch_array ($result))

  {

  echo "<tr>";

  echo "<td>" . $row['make'] . "</td>";

  echo "<td>" . $row['description'] . "</td>";

  echo "<td>" . $row['price'] . "</td>";

  echo "<td>" . $row['name'] . "</td>";

  echo "</tr>";

  }

echo "</table>";mysqli_close($con);

?>

<a href="download_file.php?id=<?php echo $id; ?>"><?php echo $name;?>[/url]

</body>

</html>

 

This is file_array

<?php

$fileName = $_FILES['file']['name'];

$tmpName  = $_FILES['file']['tmpName'];

$fileSize = $_FILES['file']['size'];

$fileType = $_FILES['file']['type'];

?>

 

Don't laugh to hard!!

 

Link to comment
https://forums.phpfreaks.com/topic/98842-upload-and-post-image-with-mysql/
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echo "<td><img src='http://www.refinedcomputersolutions.com/php_scripts/tmpName/'" . $row['name'] . "</td>"; and not echo "<td>" . $row['name'] . "</td>";

 

I tried that and just get a blank icon and yes image is in the directory path.

?????????????????????????????????????????????????? ??????????????????????

 

If someone could help me that would be great.

 

 

http://refinedcomputersolutions.com:8080/php_script/download_file.php

 

http://refinedcomputersolutions.com:8080/php_script/form.php

 

Just image info seems to be stored to database so I included a path to a directory where I saved the images.

http://www.refinedcomputersolutions.com/php_scripts/tmpName/

 

my database looks like this

 

CREATE TABLE upload (

id INT NOT NULL AUTO_INCREMENT,

name VARCHAR(30) NOT NULL,

type VARCHAR(30) NOT NULL,

size INT NOT NULL,

description VARCHAR(350) NOT NULL,

price VARCHAR(128) NOT NULL,

make VARCHAR(30) NOT NULL,

category VARCHAR(30) NOT NULL,

content MEDIUMBLOB NOT NULL,

PRIMARY KEY(id)

);

 

e.g

Make, "Opel"

Description, "4 Wheels, Very Big"

Price, 9995

Picture, 001.jpg the actual picture not the picture name.

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