Issue with updating image info on MYSQL database using PHP - php

Question

I'm having problem with my code. I can't update my image.
Here is my full code:
index.php
//this link will post my data to next page
update
updatepage.php
//this page will get all data that want to update...
<?php
include("config.php");
$name=$_GET['code'];
$sql1 = "select * from imagename where name='$name'";
$result = mysql_query($sql1) or die(mysql_error());
$row=mysql_fetch_array($result);?>
<form name="form1" method="post" action="processupdatepage.php" class="formposition" enctype="multipart/form-data" >
<table>
<tr><td>
<input type="hidden" name="id" value="<?php echo $row['id']; ?>" /></td></tr>
<tr><td width="98">Image</td><td width="288">
//This is where I have problem, I can't get my image value but other data value work very well.....
<input type="file" name="image" value="<?php echo $row['image'];?>" /></td></tr><br/>
<tr><td>nane</td><td><input type="text" name="name" value="<?php echo $row['name']; ?>"/></td></tr><br/>
<tr><td></td> <td colspan="2">
<input type="submit" name="submit" value="Save" /> </td></tr>
</table>
</form>
OK. Now this my process file:
processupdatepage.php
<?php
include("config.php");
$id=$_POST['id'];
$image=$_FILES['image']['name'];
$name=$_POST['name'];
$target = "images/";
$target = $target . basename( $_FILES['image']['name']);
$query = "UPDATE imagename SET name='$name', image='$image' WHERE id='$id'";
$bb = mysql_query($query) or die(mysql_error());
if($bb)
{
//Writes the photo to the server
if(move_uploaded_file($_FILES['image']['tmp_name'], $target))
{
$sql001="UPDATE imagename SET image='$image' WHERE image='$image'";
mysql_query($sql001);
} else { }
header("Location:index.php");
}
else
{
echo "Could not be updated";
}
?>
I can update BUT must select image. If I'm not selecting an image, my image field in database will be empty BUT other data updates are OK...

Answer 1

The value of an <input type="file"> cannot be programatically changed. Only by triggering its default behaviour that pops up the select file dialogue.
What I see from you example code is that you want to associate the value to the file form control element straight as it was retrieved from the database. So why saving the same back. Still, if you want to do this, you can use a hidden element with the value, that will be saved. But if you don't want the user to change it, just don't render it into the form and leave it out from your update query so it will remain unchanged.
Also, you could add some logic when you process the post and leave the field 'image' out of your update query if no file was selected.

Answer 2

I think the problem is here:
<input type="file" name="image" value="<?php echo $row['image'];?>" />
You can upload the image by clicking the browse button. so what do you trying to do like that?
If you want to show the image already uploaded you can use :
$img_path = 'get image path from database entry';
<img src= "<?php echo $img_path; ?>"/>

Answer 3

Here is full code to upload an Image, this is tested and works but you need to have a directory called 'upload' (this code does not create the folder).
It is advisable to only store image path on database.
<?php
if(isset($_POST['submit'])){
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/png")
|| ($_FILES["file"]["type"] == "image/pjpeg"))
&& ($_FILES["file"]["size"] < 200000))
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
}
else
{
if (file_exists("upload/" . $_FILES["file"]["name"])){
echo $_FILES["file"]["name"] . " already exists. ";
}
else{
move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $_FILES["file"]["name"]);
echo "Stored in: " . "upload/" . $_FILES["file"]["name"];
// HERE YOU CAN WRITE YOU INSERT INTO MYSQL CODE
//SOMETHING LIKE THIS:
$path = 'assets/article_images/'.$article_id.'/'.$_FILES["image_file"]["name"];
$statement1 = "INSERT INTO image_article (article_id, image_path) values ('$article_id', '$path') ";
$query_result = mysql_query($statement1) or die($statement1."mysql error<br/><br/>".mysql_error());
if($query_result==1){
echo 'Image uploaded';
}
}
}
}
else
{
echo "Invalid file";
}
}
?>
<html>
<body>
<form action="index.php" method="post"
enctype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="file" id="file"><br>
<input type="submit" name="submit" value="Submit">
</form>
</body>
</html>