azra Posted April 19, 2015 Share Posted April 19, 2015 1. HTML FORM #for user to enter the data <html> <title>reg</title> <style type="text/css"> body { background-color: rgb(200,200,200); color: white; padding: 20px; font-family: Arial, Verdana, sans-serif;} h4 { background-color: DarkCyan; padding: inherit;} h3 { background-color: #ee3e80; padding: inherit;} p { background-color: white; color: rgb(100,100,90); padding: inherit;} </style> <form method="POST" action="login_back.php" enctype="multipart/form-data"></br>  <font color="DarkCyan"> Choose a user name:</font> <input type="text" name="username"> </br></br>  <font color="DarkCyan"> First name:</font> <input type="text" name="firstname"/> </br></br>  <font color="DarkCyan"> Last name:</font><input type="text" name="lastname"/> </br></br>  <font color="DarkCyan"> File: <input type="file" name="image"></font> </br></br> <input type="submit" value="Save and Proceed"> </form> </html> ---------- 2 STORING IN DATABASE #backend processing to store and retrieve data from db <?php error_reporting(0); #echo "<body style='background-color:rgb(200,200,200)'>"; session_start(); #if( isset($_POST['username']) && isset($_FILES['image']) ) #{ $_SESSION['username']=$_POST['username']; $_SESSION['firstname']=$_POST['firstname']; $lastname=$_POST['lastname']; $file=$_FILES['image']['tmp_name']; $image_size=getimagesize($_FILES['image']['tmp_name']); if(!isset($file)) echo"please select an image"; else { #$image=$_FILES['image']['tmp_image']; //grabing the file content $image_name=$_FILES['image']['name']; //grabing image name $image_size=getimagesize($_FILES['image']['tmp_name']); //getting image size } echo "</br>"; #connection to db mysql_connect("localhost","root","")or die(mysql_error()); mysql_select_db("wordgraphic")or die(mysql_error()); #checking the available username $query = mysql_query("SELECT * FROM userdata WHERE username = '" . $_SESSION['username'] . "'" ); $ans=mysql_num_rows($query); if ($ans > 0) { echo "Username already in use please try another."; } else if($image_size==FALSE) { echo"That's not an image."; } else { #Insert data into mysql #1.Inserting user name & image into db $sql="INSERT INTO userdata(username, firstname, lastname, image)VALUES('" . $_SESSION['username'] . "', '" . $_SESSION['firstname'] . "', '$lastname','$image')"; $result1=mysql_query($sql); if($result1) { echo "</br>"; echo "Registration successful"; echo "</br>"; //displaying image $lastid=mysql_insert_id();//get the id of the last record echo "uploaded image is :"; echo "<img src='get.php?id=".$lastid."'>"; > this command has some mistake }#if insertion into db successful else { echo "Problem in database operation"; } }# else block of unique username n img }#end of isset ?> ---------- 3. GET.PHP #additional file that retrieve image from database <?php #connection to db mysql_connect("localhost","root","")or die(mysql_error()); mysql_select_db("wordgraphic")or die(mysql_error()); if(isset($_REQUEST['id']) ) > this block of code is not running { $mid=(int)($_REQUEST['id']); $image=mysql_query("SELECT * FROM userdata WHERE id=$mid") or die("Invalid query: " . mysql_error()); $image=mysql_fetch_assoc($image); $image=$image['image']; header("Content-type: image/jpeg"); echo $image; } else echo"error"; ?> Quote Link to comment https://forums.phpfreaks.com/topic/295689-unable-to-display-image-from-database/ Share on other sites More sharing options...
IThinkMyBrainHurts Posted April 19, 2015 Share Posted April 19, 2015 1. Use code tags for code in the forum! Your code is hard to read without formatting! 2. Where you say "this command has some mistake", how do you know, what is is it like when in the html source page? 3. At the end, where you write echo "error"; There is a missing }, which will make the whole script fail! 4. Turn error reporting on http://php.net/manual/en/function.error-reporting.php Quote Link to comment https://forums.phpfreaks.com/topic/295689-unable-to-display-image-from-database/#findComment-1509378 Share on other sites More sharing options...
azra Posted April 19, 2015 Author Share Posted April 19, 2015 1. Use code tags for code in the forum! Your code is hard to read without formatting! 2. Where you say "this command has some mistake", how do you know, what is is it like when in the html source page? 3. At the end, where you write echo "error"; There is a missing }, which will make the whole script fail! 4. Turn error reporting on http://php.net/manual/en/function.error-reporting.php 1.m new on online discussing i will try n repost sorry fr d inconvenience caused. 2.i know the command has some mistake because user name, image first name, last name are getting stored in db seamlessly but when m trying to display image from database i am facing prob and in the third segment of code isset() segment is not running hence, data is not getting transfered between pages . 3. else has only one statement under it i.e. echo so skipping {} will do as with most other languages. 4. turned on error reporting. thank you Quote Link to comment https://forums.phpfreaks.com/topic/295689-unable-to-display-image-from-database/#findComment-1509379 Share on other sites More sharing options...
IThinkMyBrainHurts Posted April 19, 2015 Share Posted April 19, 2015 2. Seamlessly, do you mean successfully or concatenated together? Looking at that code they shouldn't get concatenated! ... Have you tried manually calling the view image page? That code looks right to me! 3. Oops ... When testing the view page manually output some other text to verify, not the image BTW you could you edit the initial post so tha it uses code blocks, you'll be surprised how much easier it is to spot issues when properly indented... Quote Link to comment https://forums.phpfreaks.com/topic/295689-unable-to-display-image-from-database/#findComment-1509391 Share on other sites More sharing options...
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