i want to display image from database and below is my code where empid is the value of textbox, it's showing the image name stored in database but i am not able to echo this picture. anyone can help in this regard.
<?php include('connect.php'); $result = $db->prepare("SELECT image FROM info WHERE empid= $empid"); $result->bindParam('$empid', $empid); $result->execute(); for($i=0; $rows = $result->fetch(); $i++){ echo $rows['image']; } ?> Change this line.
$result = $db->prepare("SELECT image FROM info WHERE empid= '". $empid ."'"); 
<img src="your_image_path/ ".$row['image']." ">its only show you the filename that you stored in your db . if you want to show image then first add folder location with image name and pass into the <img src="excatImageLocation">
Example -
$yourExactPath = "YourImageUploadFolderLocation".$rows['image']; // img/yourFilename.jpg; echo "<img src='$yourExactPath' /> 
You either concatenate the id onto the text string containing the query or use a parameter place holder and then bind a value to it. Not both, as you were doing.
The most secure way is to use parameters.
<?php include('connect.php'); $result = $db->prepare("SELECT image FROM info WHERE empid= :empid"); $result->bindParam(':empid', $empid, , PDO::PARAM_INT); $result->execute(); while ($row = $result->fetch(PDO::FETCH_ASSOC)){ echo $row['image']; } ?> 