Image uploaded in database but not display on page in php?

Question

I have code for image upload and view in php and MySQL. After click on "Submit" button in "imageUpload.php" page image is stored in database. but not displaying in "listImages.php" page. I don't know what's the problem. I see "image not displaying when uploading in php" but its seems different solution for me. here is my code please have a look where i am wrong.

imageUpload.php :

<?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`)
   ) */    

if(count($_FILES) > 0) {
if(is_uploaded_file($_FILES['userImage']['tmp_name'])) {
mysqli_connect("localhost", "root", "");
mysqli_select_db ("test");
$imgData =addslashes(file_get_contents($_FILES['userImage']['tmp_name']));
$imageProperties = getimageSize($_FILES['userImage']['tmp_name']);

$sql = "INSERT INTO output_images(imageType ,imageData)
VALUES('{$imageProperties['mime']}', '{$imgData}')";
$current_id = mysqli_query($sql) or die("<b>Error:</b> Problem on Image Insert<br/>" . mysqli_error());
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>

listImages.php :

<?php
$conn = mysqli_connect("localhost", "root", "");
mysqli_select_db("test");
$sql = "SELECT imageId FROM output_images ORDER BY imageId DESC"; 
$result = mysqli_query($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>

imageView.php :

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

What happens if you visit imageView.php?image_id=1 in your browser? Do you see the image? (With a valid image_id) — Rolf Pedro Ernst
– Rolf Pedro Ernst, Commented Sep 22, 2016 at 9:16
No in listimages.php their is broken icons as seen in above image. when i see code view of listimages.php then in <img> their is "<img src="imageView.php?imageId=8" /><br/>" that means imageView.php getting id properly. — John120
– John120, Commented Sep 22, 2016 at 9:20
@Rolf Pedro Ernst I see "localhost/mysql_blob_using_php/mysql_blob/…" in browser but page is blank their is no image. — John120
– John120, Commented Sep 22, 2016 at 9:30
@John120 right click the broken image and inspect.. see whether the image echo correctly or not — FullStack
– FullStack, Commented Sep 22, 2016 at 9:46

FullStack · Accepted Answer · 2016-09-22 10:05:36Z

1

this would help you

<a href="imageView.php?image_id=<?php echo $row["imageId"]; ?>">
  <img src="<?php echo $row['imagedata']; ?>" alt="my picture" height="128" width="128" />
</a>

answered Sep 22, 2016 at 10:05

FullStack

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Comments

Senthil · Accepted Answer · 2016-09-22 10:15:01Z

1

it should be

    $conn=mysqli_connect("ur_servername_ex_localhost","ur_username","ur_password","ur_db");
    mysqli_query($conn, $sql);

answered Sep 22, 2016 at 10:15

Senthil

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