Help with images

mjgdunne · April 30, 2008

Hi all, i have a form for uploading images into a mysql database, the form seems to work as the image does go to the database, my problem is how do i retrieve the image, when i query the database i only get a link, and my question is am i uploading the file correctly or do i need to change the code, any help or advice would be great thanks.

Here is my upload form:

<html>
<head>
<title>Car Rentals & Returns</title>
<meta http-equiv="Content-Type" content="text/html" />
<link href="style2.css" rel="stylesheet" type="text/css" />
<script type="text/javascript">
function validate_required(field,alerttxt)
{
with (field)
{
if (value==null||value=="")
  {alert(alerttxt);return false;}
else {return true}
}
}function validate_form(thisform)
{
with (thisform)
{
 if (validate_required(mymake,"Car Make must be selected!")==false)
  {mymake.focus();return false;}

 if (validate_required(mymodel,"Car Model must be filled out!")==false)
  {mymodel.focus();return false;}

 if (validate_required(myregistration,"Registration must be filled out!")==false)
  {myregistration.focus();return false;}

}
}
</script>





</head>
<body>
<form action="case.php"
onsubmit="return validate_form(this);"
method="post">

<div id="wrapper3">
<img src="images/cars.jpg" width="996" height="100"></a>

<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=100%>
<tr>
<td align="center">
<H1>Create new case</H1>
</td></tr>
<tr><td>
<h2>Please upload a new picture</h2>

<input type=hidden name=MAX_FILE_SIZE value=150000>

<input type=hidden name=completed value=1>

<br>Please choose an image to upload: <input type=file name=myimage><br>
<br>

Please enter the car make: <input name=mymake><br><br>

Please enter the car model: <input name=mymodel><br><br>


Please enter the car registration: <input name=myreg><br>
<br>




</TABLE>
<TABLE BGCOLOR="#F0F8FF" WIDTH=100%>
<TR><TD ALIGN="CENTER">
<input type="submit" value="Add Case"/><input type="reset" name="reset" value="Clear Form"/>
</form><form action="login_success.php" method="post">
<input type=submit value="Home"></form></TD></tr>
<TR><TD><BR></TD></TR>
</table>

</td></tr>
</body>
</html>

Here is my code that im using to retrieve the image:

<html>
<head>
<title>Car Rentals & Returns</title>
<meta http-equiv="Content-Type" content="text/html" />
<link href="style2.css" rel="stylesheet" type="text/css" />
</head>
<body>
<div id="wrapper3">
<img src="images/cars.jpg" width="996" height="100"></a>

<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=100%>
<tr>
<td align="center">
<H1>View Cases:</H1>
<form method="post" action="view_2.php">
</td>
</table>
<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=100%>
<tr>
<TD ALIGN=CENTER>



<?php
$host="localhost"; // Host name
$username="root"; // Mysql username
$password="root"; // Mysql password
$db_name="test"; // Database name
$tbl_name="reff"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "root", "root")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");


$result = mysql_query("SELECT * FROM reff ORDER BY make;");

$_SESSION['reff'] = array();
while ($row = mysql_fetch_assoc($result)) {
$_SESSION['reff'][] = $row;
}

foreach ($_SESSION['reff'] as $row) {



echo "<table border=\"1\" align=\"center\">";
echo "<tr><th>ID</th>";
	echo "<td>";
        echo $row['pid'];
echo "</td>";


echo "<tr><th>Make</th>";
	echo "<td>";
        echo $row['make'];
echo "</td>";


echo "<tr><th>Model</th>";
	echo "<td>";
        echo $row['model'];
echo "</td>";

echo "<tr><th>Registration</th>";
	echo "<td>";
    echo $row['registration'];
echo "</td>";

echo "<tr><th>Image</th>";
	echo "<td>";
    	echo $row['imgdata'];
echo "</td>";

echo "<tr><th></th>";
	echo "<td>";
    	echo "<button onclick=\"window.location.href='view_3.php?pid=".$row['pid']."'\">Details</button>";
echo "</td>";


echo "<tr>";
echo "<br>";
echo "<br>";
echo "</tr>";

}



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


if (mysql_num_rows($result) == 0) {
    echo "No records found";
    exit;
}
?>


<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=30%>
</form>
<tr><form method="post" action="login_success.php">
<td align="center"><input type="submit" value="Home"/></td>
</tr>
</table>
</table>
</td>
</tr>
</table>
</BODY>
</HTML>

The Little Guy · April 30, 2008

Are you saving the image in the database, or a link to the image?

mjgdunne · April 30, 2008

Hi i am saving the images to a longblob in the database.

The Little Guy · April 30, 2008

OK, you will need to make php page that creates an image...

So, you will need to pull the text from the database, then output the file:

image.php

<?php
$id = $_GET['id'];
$query = "SELECT * FROM databaseName WHERE id = '$id'";
$sql = mysql_query($query);
if(mysql_num_rows($sql) > 0){
     // Image was found
     header('Content-type: image/jpeg');
     $row = mysql_fetch_array($sql);
     echo $row['myLongBlob'];
}else{
     // There was no image
     header('Content-type: image/jpeg');
     readfile('/link/to/my/alterative.jpg');
}
?>

Next you will need to get the file using HTML:

...
<img src="image.php?id=1243" alt="My Image" />
...

it would all look something like the above (I didn't test the code, it may need some work).

mjgdunne · April 30, 2008

Hi i am getting the error:

Warning: Cannot modify header information - headers already sent by (output started at C:\xampp\htdocs\view_2.php:19) in C:\xampp\htdocs\view_2.php on line 43
C:\Documents and Settings\Michael Dunne\My Documents\My Pictures\06-02-2008\DSC00064.JPG

The Little Guy · April 30, 2008

want to post everything from line 19 and above (if the file is short, just post the whole thing).

mjgdunne · April 30, 2008

Hi this is from lines 19 - 46:

<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=100%>
<tr>
<TD ALIGN=CENTER>



<?php
$host="localhost"; // Host name
$username="root"; // Mysql username
$password="root"; // Mysql password
$db_name="test"; // Database name
$tbl_name="reff"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "root", "root")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

$pid = $_GET['pid'];
$query = "SELECT * FROM reff WHERE pid = '$pid'";

$sql = mysql_query($query);

if(mysql_num_rows($sql) > 0){
     // Image was found
     header('Content-type: image/jpeg');
     $row = mysql_fetch_array($sql);
     echo $row['imgdata'];
}else{

The Little Guy · April 30, 2008

the image.php file I posted above MUST but in its own file.

mjgdunne · April 30, 2008

Hi, i have heard that it is not a good idea to store images in a database as it can slow it down etc, so i was told to save the upload image to a location on the server. I have a script as below:

Here is the upload form:

<form enctype="multipart/form-data" action="upload.php" method="post" name="upload_file">
<input type="hidden" name="MAX_FILE_SIZE" value="5000000">
<input name="userfile" type="file">
<input type="submit" value="upload" name="file_uploaded">
</form>

And here is the upload.php:

<?php


if (is_uploaded_file($_FILES['userfile']['tmp_name']))
{
$file_realname = $_FILES['userfile']['name'];
copy($_FILES['userfile']['tmp_name'], realpath("upload").trim($file_realname));
}
else
{
echo "<b><font color=red>No file uploaded.</font></b><BR>No file available or file too big to upload.";
}
echo $file_realname;
?>

This code does work, but i need to be able to assign it an picture id and save the picture id in the database along with the other details so when you query for results the image and other details will be displayed.

Do you know how i would go about this?

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