image name not showing in database using PHP

Question

Mysql Database attribute image name and type

`Image varchar256`

I m trying to upload image name in mysql but issue is image is move to folder as per requirement but name of image is not showing in database. kindly help...

<form method="post" enctype="multipart/form-data" >
    <div class="content table-responsive table-full-width">
        <div class="title" style="margin-left: 7%;"> Product ID</div>
        <input  class="form-control"  type="text"  name="P_id" placeholder="Product ID" readonly/>
        <div class="title" style="margin-left: 7%;" >Product Name</div>
        <input class="form-control"  type="text" name="P_Name" placeholder="Product Name" required/>
        <div class="title" style="margin-left: 7%;">Descriptiom</div>
        <textarea class="form-control"  type="text" name="P_Description" placeholder="Description"required></textarea>
        <div class="title" style="margin-left: 7%;">Category</div>

        <!-- Category Add in Drop Down-->
        <select name="C_id" class="form-control" required>
        <option>Select Category</option>
            <?php
            $get_category= "select * from category";
            $run_category= mysqli_query($con,$get_category);
            while($row_category=mysqli_fetch_array($run_category))
            {
                $C_id=$row_category['C_id'];
                $C_Name=$row_category['C_Name'];
                echo "<option value='$C_id'>$C_Name</option>";
            }
            ?>
        </select><!-- End Conde-->

        <div class="title" style="margin-left: 7%;">Price</div>
        <input class="form-control"  type="text" name="P_Price" placeholder="Enter Price"required/>
        <div class="title" style="margin-left: 7%;">Quantity</div>
        <input class="form-control"  type="text" name="P_Quantity" placeholder="Enter Quantity"required/>
        <div class="title" style="margin-left: 7%;">Image</div>
        <input class="form-control"  type="file" name="image"  placeholder="Upload Image"required accept=""/>
        <div class="navbar-form">
            <input type="submit" class="btn btn-primary " name="Insert" value="Insert"style="margin-left: 7%;"/>
            <label name="Label" ></label>
        </div>
        <div class="navbar-form">
            <a href="Categoryadd.php" class="btn btn-primary"style="margin-left: 7%;">+Add Category</a>
            <a href="Brands.php" class="btn btn-primary">+Add Brand</a>
        </div>
    </div>
</form>

The Example which move image to folder but image name not showing in database

<?php
if(isset($_POST['Insert'])){
    $P_id=$_POST['P_id'];
    $P_Name=$_POST['P_Name'];
    $P_Description= $_POST['P_Description'];
    $CId=$_POST['C_id'];
    $P_Price=$_POST['P_Price'];
    $P_Quantity=$_POST['P_Quantity'];

    //image names
    $PImage = $_FILES['image']['name'];

      //image temp names

   $tempname = $_FILES["image"]["tmp_name"];

//for image upload on folder
    move_uploaded_file($tempname,"Images/$PImage");

    if($PName='' OR $PDescription='' OR $C_id='' OR $Price='' OR $Quantity='' OR $PImage='' )
    {
        echo"<script>alert('please fill fields')</script>";
    }
    else {

        $insert_Product = "INSERT INTO `product`(`P_id`, `P_Name`, `P_Description`, `Price`, `Quantity`, `C_Id`, `Image`) VALUES ('','$P_Name','$P_Description','$P_Price','$P_Quantity','$CId','$PImage')";

        $run_Product = mysqli_query($con, $insert_Product);

        if($run_Product){
            $label= "Label";
            $label= "Enter Data successfully";
            echo $label;
        }
    }
}
?>

How can i resolve my issue ?

You have inconsistencies with variable names. You declare `$P_Name` but you call `$PName`. Same for many other variables. — cchoe1, Nov 24 '17 at 06:26
`if ($PName='' OR ...` - You realize that you're actually _setting_ all the variables in that `if`-statement? You need `==` or `===` to _compare_ the variables. — M. Eriksson, Nov 24 '17 at 06:26
You are wide open to [SQL Injections](http://php.net/manual/en/security.database.sql-injection.php) and should really use [Prepared Statements](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php) instead of concatenating your queries. Specially since you're not escaping the user inputs at all! — M. Eriksson, Nov 24 '17 at 06:28
`mysqli_query($con, $insert_Product)` - Where do you set `$con`? — M. Eriksson, Nov 24 '17 at 06:30
There are many issues and questions about this code. I'm quite sure that if you checked your servers error log, you would get a bunch of helpful error and warning messages. A good idea is also to turn `display_errors` on in your local PHP environment. Read more here: [How do I get PHP errors to display?](http://stackoverflow.com/questions/1053424/how-do-i-get-php-errors-to-display) — M. Eriksson, Nov 24 '17 at 06:33
my problem is solved thanx to all of you helping me.... thanx alot — Umair Jutt, Nov 24 '17 at 11:57

Raghav · Answer 1 · 2017-11-24T06:48:08.363

0

Try this Query..

       if(empty($P_Name) || empty($P_Description) || empty($P_Price) || empty($P_Quantity) || empty($PImage) || empty($CId)) 
                {
                    echo"<script>alert('please fill fields')</script>";
                }

                else {
                 $insert_Product = "INSERT INTO `product`(`P_id`, `P_Name`, `P_Description`, `Price`, `Quantity`, `C_Id`, `Image`) VALUES ('','$P_Name','$P_Description','$P_Price','$P_Quantity','$CId','$PImage')";
                 .
                 .
                 .
                 .

                 }

Hope this helps..

edited Nov 24 '17 at 06:48

answered Nov 24 '17 at 06:35

Raghav

388
1
12

That will most likely not solve all the OP's issues, since there are a bunch more in it. – M. Eriksson Nov 24 '17 at 06:45
Still wide open to SQL Injections. Never concatenate SQL queries. Always use prepared statements. – M. Eriksson Nov 24 '17 at 06:51

score -1 · Answer 2 · answered Nov 24 '17 at 06:39

This code is use to save image path into database.

In this first line get you image into file variable.
Second line use to get file extension.
Third line is used to change your image file name
Fourth line is use to move image into your given folder name

And the last line is getting image name with folder path

        $file      = $_FILES['image'];
        $extension = $file->getClientOriginalExtension();
        $imageName = 'anything'.time().'.'.$extension;
        $file->move(public_path('/foldername'), $imageName);
        $imageFile = '/foldername/'.$imageName;

Insert $imageFile into your database image column. It works for both.. image move to your folder and save image path into database.

You might want to mention that your code is actually using _Symfonys http-foundation_-library, which the OP never said is used. `$file->getClientOriginalExtension();` isn't a native PHP-method. Plus, this doesn't actually solve the OP's issues. — M. Eriksson, Nov 24 '17 at 06:47

image name not showing in database using PHP

Mysql Database attribute image name and type

2 Answers2