Need Help To Solve Images Disppeared After Again Form Submiitted? - php

Question

i have multiple image upload options in the form and they are working fine and update file name in the mysql database columns?
when i uploads image1 then image move on to the server and also it is showing next to html table columns now problem is that when i upload image with file option2 after form submit then image1 which is on the next to the image1 file option will be disappeared and in the mysql database image1 columns will be blank?
Multiple Images Uploading Functions
$id=$_REQUEST['id'];
if(isset($_POST['submit']))
{
if (!empty($_FILES['image']['name']))
{
$rnd_1 = rand(11111,99999);
$file_name= $rnd_1.'_'.$_FILES['image']["name"];
$file_path = "uploads/";
$image = new imgMark();
$image->font_path = "arial.ttf";
$image->font_size = 25;
$image->water_mark_text = "© www.edge.pk";
$image->color = 'CC003E';
$image->opacity = 50;
$image->rotation = 0;
if($image->convertImage('image', $file_name, $file_path))
$demo_image = $image->img_path;
}
if (!empty($_FILES['image1']['name']))
{
$rnd_1 = rand(11111,99999);
$file_name= $rnd_1.'_'.$_FILES['image1']["name"];
$file_path = "uploads/";
$image = new imgMark();
$image->font_path = "arial.ttf";
$image->font_size = 35;
$image->water_mark_text = "© www.edge.pk";
$image->color = 'CC003E';
$image->opacity = 50;
$image->rotation = 0;
if($image->convertImage('image1', $file_name, $file_path))
$demo_image2 = $image->img_path;
}
if (!empty($_FILES['image2']['name']))
{
$rnd_1 = rand(11111,99999);
$file_name= $rnd_1.'_'.$_FILES['image2']["name"];
$file_path = "uploads/";
$image = new imgMark();
$image->font_path = "arial.ttf";
$image->font_size = 35;
$image->water_mark_text = "© www.edge.pk";
$image->color = 'CC003E';
$image->opacity = 50;
$image->rotation = 0;
if($image->convertImage('image2', $file_name, $file_path))
$demo_image3 = $image->img_path;
}
Update Query
UPDATE products SET
image='$demo_image',addimage1='$demo_image2',addimage2='$demo_image3'
WHERE id='$id'
}
Select Images Query
$query1=mysql_query("select images,addimages1,addimages2 from products
where id='$id' ")or die("query");
$row2=mysql_fetch_array($query1);
<form method="post" enctype="multipart/form-data">
Image Upload Option1
<input type="file" name="image" id="image" />
<img src="<?php echo $image['image'] ?>" width="150" height="150" />
Image Upload Option2
<input type="file" name="image1" id="image1"/>
<img src="<?php echo $image['addimage1'] ?>" width="150" height="150" />
Image Upload Option3
<input type="file" name="image2" id="image2"/>
<img src="<?php echo $image['addimage2'] ?>" width="150" height="150" />
<input type="submit" class="bg" name="submit" />
</form>

Answer 1

The problem lays here:
UPDATE products SET
image='$demo_image',addimage1='$demo_image2',addimage2='$demo_image3'
WHERE id='$id'
}
From what I understood you first submit the form where you upload the first image and then you submit the form again where you upload the second image. The reason of this is that during your second upload the variable $demo_image will be blank because you are not sending any value of the input "image" during the resubmit. See here:
$demo_image = $image->img_path;
$image->img_path is blank, therefore $demo_image will be blank (or NULL? - not sure) as well.
There are many solutions of this problem. You can retrieve data from you db to the form and then resubmit, or test if your variables are blank (or NULL?) and then having different UPDATE commands for different variations, or retrieving data from MySQL just before you run UPDATE and many more.
I'd pick retrieving "old" data from MySQL and then just test if their NULL. If they are not NULL in MySQL, you will just resubmit the same data to your db. Like this:
if ($demo_image_from_db!=NULL) $demo_image = $demo_image_from_db;
It is possible that there is also a MySQL native function for updating only when NULL - I've heard of COALESCE but I don't exactly know how to use it - that would be definitely the simplest way to do it.