Convert image to blob php

Question

<?php
$file_name = $_FILES['files']['name'];
$tmp_name  = $_FILES['files']['tmp_name'];
$file_size = $_FILES['files']['size'];
$file_type = $_FILES['files']['type'];

// The codes written above work fine and have proper information.

$fp = fopen($tmp_name, 'r'); // This one crashes.
$file_content = fread($fp, $file_size) or die("Error: cannot read file");
$file_content = mysql_real_escape_string($file_content) or die("Error: cannot read file");
fclose($fp);

....

I'm a newbie to PHP stuff. I'm trying to store a jpg image as blob in a database but terribly struggling with it :( I tried many tutorials and read documents but still no luck. Any suggestions or tutorials that might help me out..?

Nội dung chính Show

Common Mime Types
Next Lesson PHP Tutorial

asked Oct 21, 2011 at 5:13

2

When opening binary files with fopen(), use the rb mode, ie

$fp = fopen($tmp_name, 'rb');

Alternatively, you may simply use file_get_contents(), eg

$file_content = file_get_contents($tmp_name);

To enable better error reporting, place this at the top of your script

ini_set('display_errors', 'On');
error_reporting(E_ALL);

answered Oct 21, 2011 at 5:43

PhilPhil

147k21 gold badges229 silver badges230 bronze badges

1

For the mimetype, see Mimey or another lib like it.
Or this shortcut might work for some file types like png and jpg: $mimetype = "image/".pathinfo($path,PATHINFO_EXTENSION);

$path = //path/to/file.jpg (or whatever)

$mimer = new \Mimey\MimeTypes();
$mimetype = $mimer->getMimeType(pathinfo($path,PATHINFO_EXTENSION));

$source = file_get_contents($path);
$base64 = base64_encode($source);
$blob = 'data:'.$mimetype.';base64,'.$base64;

$html = '<img src="'.$blob.'" alt="Description for differently abled humans." />';
echo $html;

Common Mime Types

Mdn has a list of common mime types.

For your convenience here are the (probably) most common image ones:

.gif: image/gif
.jpg or .jpeg: image/jpeg
.png: image/png

Table of contents

How to convert jpg image to proper blob data type using php
Php : Convert a blob into an image file
MySQL BLOB using PHP
How to convert the JPEG image to blob?
How to Display Blob Image in PHP from Database

How to convert jpg image to proper blob data type using php

<?php
$file_name = $_FILES['files']['name'];
$tmp_name  = $_FILES['files']['tmp_name'];
$file_size = $_FILES['files']['size'];
$file_type = $_FILES['files']['type'];

// The codes written above work fine and have proper information.

$fp = fopen($tmp_name, 'r'); // This one crashes.
$file_content = fread($fp, $file_size) or die("Error: cannot read file");
$file_content = mysql_real_escape_string($file_content) or die("Error: cannot read file");
fclose($fp);

....

$fp = fopen($tmp_name, 'rb');

$file_content = file_get_contents($tmp_name);

ini_set('display_errors', 'On');
error_reporting(E_ALL);

Php : Convert a blob into an image file

$image = imagecreatefromstring($blob); 

ob_start(); //You could also just output the $image via header() and bypass this buffer capture.
imagejpeg($image, null, 80);
$data = ob_get_contents();
ob_end_clean();
echo '<img src="data:image/jpg;base64,' .  base64_encode($data)  . '" />';

$image = new Imagick();
$image->readimageblob($blob);
echo '<img src="data:image/png;base64,' .  base64_encode($image->getimageblob())  . '" />';

$image = new Gmagick();
$image->readimageblob($blob);
echo '<img src="data:image/png;base64,' .  base64_encode($image->getimageblob())  . '" />';

$path = "/tmp/images";
$sql = "SELECT image_name, image_content FROM images";

$result = mysql_query($sql, $db_con);
if (!$result) {
   $message  = 'Invalid query: ' . mysql_error() . "\n";
   $message .= 'Whole query: ' . $sql;
   die($message);
}

while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
   $image = $row["image_contents"];
   $name = $row["image_name"];

   // option 1
   $file = fopen($path."/".$name,"w");
   echo "File name: ".$path."$name\n";
   fwrite($file, base64_decode($image));
   fclose($file);

   // option 2 (oneliner)
   // file_put_contents($path."/".$name, base64_decode($image));
}

<?php
if(isset($_REQUEST['id']))
{
   // get the file with the id from database
      include "dbconfig.php";
      $dbconn = mysql_connect($dbhost, $dbusr, $dbpass) or die("Error Occurred-".mysql_error());
      mysql_select_db($dbname, $dbconn) or die("Unable to select database");

      $id    = $_ REQUEST ['id'];
      $query = "SELECT `img_name`, `img_type`, `img_size`, `img_data`
                       FROM img_tbl WHERE id = ‘$id’";

      $result = mysql_query($query) or die(mysql_error());
      list($name, $type, $size, $content) = mysql_fetch_array($result);

      header("Content-length: $size");
      header("Content-type: $type");
      print $content;

      mysql_close($dbconn);
}
?>

MySQL BLOB using PHP

CREATE TABLE IF NOT EXISTS `output_images` (
  `imageId` tinyint(3) NOT NULL AUTO_INCREMENT,
  `imageType` varchar(25) NOT NULL DEFAULT '',
  `imageData` mediumblob NOT NULL,
  PRIMARY KEY (`imageId`)
)

<?php
if (count($_FILES) > 0) {
    if (is_uploaded_file($_FILES['userImage']['tmp_name'])) {
        require_once "db.php";
        $imgData = addslashes(file_get_contents($_FILES['userImage']['tmp_name']));
        $imageProperties = getimageSize($_FILES['userImage']['tmp_name']);
        
        $sql = "INSERT INTO output_images(imageType ,imageData)
	VALUES('{$imageProperties['mime']}', '{$imgData}')";
        $current_id = mysqli_query($conn, $sql) or die("<b>Error:</b> Problem on Image Insert<br/>" . mysqli_error($conn));
        if (isset($current_id)) {
            header("Location: listImages.php");
        }
    }
}
?>
<HTML>
<HEAD>
<TITLE>Upload Image to MySQL BLOB</TITLE>
<link href="imageStyles.css" rel="stylesheet" type="text/css" />
</HEAD>
<BODY>
    <form name="frmImage" enctype="multipart/form-data" action=""
        method="post" class="frmImageUpload">
        <label>Upload Image File:</label><br /> <input name="userImage"
            type="file" class="inputFile" /> <input type="submit"
            value="Submit" class="btnSubmit" />
    </form>
    </div>
</BODY>
</HTML>

<?php
    require_once "db.php";
    if(isset($_GET['image_id'])) {
        $sql = "SELECT imageType,imageData FROM output_images WHERE imageId=" . $_GET['image_id'];
		$result = mysqli_query($conn, $sql) or die("<b>Error:</b> Problem on Retrieving Image BLOB<br/>" . mysqli_error($conn));
		$row = mysqli_fetch_array($result);
		header("Content-type: " . $row["imageType"]);
        echo $row["imageData"];
	}
	mysqli_close($conn);
?>

<img src="imageView.php?image_id=<?php echo $row["imageId"]; ?>" />

<?php
    require_once "db.php";
    $sql = "SELECT imageId FROM output_images ORDER BY imageId DESC"; 
    $result = mysqli_query($conn, $sql);
?>
<HTML>
<HEAD>
<TITLE>List BLOB Images</TITLE>
<link href="imageStyles.css" rel="stylesheet" type="text/css" />
</HEAD>
<BODY>
<?php
	while($row = mysqli_fetch_array($result)) {
	?>
		<img src="imageView.php?image_id=<?php echo $row["imageId"]; ?>" /><br/>
	
<?php		
	}
    mysqli_close($conn);
?>
</BODY>
</HTML>

How to convert the JPEG image to blob?

FileStram imgStream = File.OpenRead("C:\\photos\me.jpg");
byte[] blob = new byte[imgStream.Length];
imgStream.Read( blob, 0, (int)imgStream.Length);

imgStream.Dispose( );

How to Display Blob Image in PHP from Database

<html>
<head>
<title>Display Blob Image in PHP</title>
</head>
<body>
    <form enctype="multipart/form-data" action="upload.php" method="post">
        <label>Upload the image file:</label><br /> 
        <input name="userImage" type="file" /> 
        <input type="submit" value="Upload" />
    </form>
    </div>
</body>
</html>

<?php
if(isset($_POST["submit"])){

    $b = getimagesize($_FILES["userImage"]["tmp_name"]);

    //Check if the user has selected an image
    if($b !== false){
        //Get the contents of the image
        $file = $_FILES['userImage']['tmp_name'];
        $image = addslashes(file_get_contents($file));
        
        $host     = 'localhost';
        $username = 'root';
        $password = ' ';
        $db     = 'test';
        
        //Create the connection and select the database
        $db = new mysqli($host, $username, $password, $db);
        
        // Check the connection
        if($db->connect_error){
            die("Connexion error: " . $db->connect_error);
        }
        
        //Insert the image into the database
        $query = $db->query("INSERT into gallery (image) VALUES ('$image')");
        if($query){
            echo "File uploaded successfully.";
        }else{
            echo "File upload failed.";
        } 
    }else{
        echo "Please select an image to upload.";
    }
}
?>

<?php
if(!empty($_GET['id'])){

    $host     = 'localhost';
    $username = 'root';
    $password = ' ';
    $db     = 'test';
    
    //Create the connection and select the database
    $db = new mysqli($host, $username, $password, $db);
    
    // Check the connection
    if($db->connect_error){
        die("Connexion error: " . $db->connect_error);
    }
    
    //Get the image from the database
    $res = $db->query("SELECT image FROM gallery WHERE id = {$_GET['id']}");
    
    if($res->num_rows > 0){
        $img = $res->fetch_assoc();
        
        //Render the image
        header("Content-type: image/jpg"); 
        echo $img['image']; 
    }else{
        echo 'Image not found...';
    }
}
?>

Convert BLOB image then save to file.

<img src="data:image/jpeg;base64,<?php echo base64_encode($content); ?>" width="100" />

	public function testImageConvert()
{
	$query = $this->db->query("SELECT Picture FROM `ItemPicture` LIMIT 1");

	if($query->num_rows() > 0)
	{
		$row = $query->row();

		file_put_contents('test.jpg',$row->Picture);
	}
	return false;
}