3 replies to this topic

[Void]

NeedHelp, I've done this before. The table needs to be a blob.
This is the code to insert into DB

add.php (Where the user inserts data to be submitted and browses for pictures)

<form enctype="multipart/form-data" action="insert.php" method="POST" name=changer>

<input type="hidden" name="MAX_FILE_SIZE" value="102400">

<input type="file" name="photo" accept="image/jpeg">

<input type="Submit" value="Submit">

insert.php file

<?php

// Connect to the DB somwhere up here


// Read our photo

$sPic="";

if (!empty($_FILES['photo']['name'])) {

	$sPic = addslashes(FormatImage($_FILES['photo']['name'], $_FILES['photo']['tmp_name']));

}


$query = "INSERT INTO name (tb_name) VALUES ('$spic')"; 

$results = mysql_query($query) or die ("Error " . mysql_error());

mysql_close();


// This function formats the image and can resize if desired

function FormatImage($sPhotoFileName, $sTempFileName) {

	//$sPhotoFileName = $_FILES['photo']['name']; // get client side file name

	

	if ($sPhotoFileName) // file uploaded

	{	$aFileNameParts = explode(".", $sPhotoFileName);

		$sFileExtension = end($aFileNameParts); // part behind last dot

		if ($sFileExtension != "jpg"

			&& $sFileExtension != "JPEG"

			&& $sFileExtension != "JPG")

		{	die ("Choose a JPG for the photo");

		}

		$nPhotoSize = $_FILES['photo']['size']; // size of uploaded file

		if ($nPhotoSize == 0)

		{	die ("Sorry. The upload of $sPhotoFileName has failed.

				Search a photo smaller than 100K, using the button.");

		}

		if ($nPhotoSize > 102400)

		{	die ("Sorry.

			The file $sPhotoFileName is larger than 100K.

			Advice: reduce the photo using a drawing tool.");

		}

	}

	

	// read photo

	//$sTempFileName = $_FILES['photo']['tmp_name']; // temporary file at server side

	$oTempFile = fopen($sTempFileName, "r");

	$sBinaryPhoto = fread($oTempFile, fileSize($sTempFileName));

	

	// Try to read image

	$nOldErrorReporting = error_reporting(E_ALL & ~(E_WARNING)); // ingore warnings

	$oSourceImage = imagecreatefromstring($sBinaryPhoto); // try to create image

	//error_reporting($nOldErrorReporting);

	

	if (!$oSourceImage) // error, image is not a valid jpg

	{ die ("Sorry.

			It was not possible to read photo $sPhotoFileName.

			Choose another photo in JPG format.");

	}

	

	$nWidth = imagesx($oSourceImage); // get original source image width

	$nHeight = imagesy($oSourceImage); // and height

	

	// create small thumbnail

	$nDestinationWidth = $nWidth;

	$nDestinationHeight = $nHeight;

	//$oDestinationImage = imagecreatetruecolor($nDestinationWidth, $nDestinationHeight);

	$oDestinationImage = imagecreate($nDestinationWidth, $nDestinationHeight);

	

	// below resizes the image

	imagecopyresized(

			$oDestinationImage, $oSourceImage,

			0, 0, 0, 0,

			$nDestinationWidth, $nDestinationHeight,

			$nWidth, $nHeight); // resize the image

	

	

		ob_start(); // Start capturing stdout.

		imageJPEG($oDestinationImage); // As though output to browser.

		$sBinaryThumbnail = ob_get_contents(); // the raw jpeg image data.

		ob_end_clean(); // Dump the stdout so it does not screw other output.

		

		// Return our image

		return $sBinaryThumbnail;

}


?>

[Void]

This is the code to view an image that has been stored in a database. You have to use a seperate php and call it like an image.

Here is the HTML where it calls the php image

<img src="viewpicture.php?id=<? print "$id"; ?>" />

^^ This can be stuck in any HTML page.

You can remove the ID from this if you only want to select the one image or whatever. Here is the viewpicture

viewpicture.php

<?php

// Connect to DB up here


// Select our picture and display it

$query = "SELECT * FROM db WHERE ID='$id'";


// Run our query and display the image

$results = mysql_query($query);

$row = mysql_fetch_array($results);

$sJpg = $row[##] // Replace ## with the row number


// Send header info

header("Content-type: image/jpeg"); // act as a jpg file to browser


// Show it

print $sJpg;


?>