How to save image source variable using php

Question

Is it has method to save <img src="XXXXX"> image path to database using PHP. Why src path doesn't pick in PHP . My PHP code is

<?php
include('conn.php');
if(!empty($_POST["submit1"]) || isset($_POST["submit1"])){
 if (isset($_POST['img1'])) {
        $filename1= $_POST['img1'];
    }
  if (isset($_POST['img2'])) {
        $filename2= $_POST['img2'];
    }
}
?>

And my html code is

    <html>
    <head>
    <title>Test 
    </title>
</head>
    <body>
     <form  action="phpscript/test.php" method="post" enctype="multipart/form-data">
            <div>Images </div> 
              <input type="file" id="file" name="file">
              <div class="row align-text-center">                                     
                    <div class="col-sm2">
                        
 <input type="image"  id="img1" name="img1"  src="img/1.jpg" width="150" height="150"  style="border: "/> 
    
                    </div>
     <div class="col-sm2">
    <input type="image" src="img/2.jpg" width="150" height="150" id="img2" name="img2" style="border: " /> 
                                             
                    </div>
                  <div class="col-sm2">
    <button class="btn " type="submit" id="submit1" name="submit1">Submit</button>  
    </form>
    </body>
    
    </html>

Actually I need to pick image path in php code.But in php code $filename1 and $filename2 values are NULL.Is it another method to get image path in php code?

Did you already use var_dump($_FILES); to see, what variables it has? — klediooo
– klediooo, Commented Nov 23, 2020 at 11:42
Why should any $_FILES['img1'] exist? Your form field with that name is not a file upload field, it is just a type="image" field. Those do not send any file names. — C3roe
– C3roe, Commented Nov 23, 2020 at 11:43
I think you've misunderstood what the element type="image"does. The src-attribute is for how the element should look, it won't be submitted with the form — M. Eriksson
– M. Eriksson, Commented Nov 23, 2020 at 11:43
$_POST['img1'] also wouldn't be the file, take a look at php.net/manual/en/features.file-upload.post-method.php — user3783243
– user3783243, Commented Nov 23, 2020 at 11:52
I change code $_FILES to $_POST.But it also doesn't work.I think<input type="image" src="XXX"> doesn't work as input on POST method. — Navindu Hasaral
– Navindu Hasaral, Commented Nov 23, 2020 at 12:31

Rokan Chowdhury Onick · Accepted Answer · 2020-11-23 13:18:31Z

1

You have a misconception about <input type="image"> this. This just defines a image as a submit button. Check out this as a reference https://www.w3schools.com/tags/att_input_type_image.asp Now to save image path in a PHP variable, you have to select that image as a file.

<input type="image"  id="img1" name="img1"  src="img/1.jpg" width="150" height="150"  style="border:"/>
<input type="image" src="img/2.jpg" width="150" height="150" id="img2" name="img2" style="border:" />

This just makes an output of images which are img/1.jpg and img/2.jpg. To get images path in a PHP variable you have to change these line like bellow:

<input type="file" name="img1">
<input type="file" name="img2">

Then you can select an image file. In PHP part change your code with this code:

<?php
    include('conn.php');
    if(!empty($_POST["submit1"]) || isset($_POST["submit1"])){
     if (isset($_FILES["img1"])) {
            $filename1= $_FILES["img1"]["name"];
        }
      if (isset($_FILES['img2'])) {
            $filename2= $_FILES["img2"]["name"];
        }
    }
?>

To know details about PHP file/image uploading check out this https://www.php.net/manual/en/features.file-upload.post-method.php

answered Nov 23, 2020 at 13:18

Rokan Chowdhury Onick

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2 Comments

Navindu Hasaral Over a year ago

can we use hidden input to use for this?

Rokan Chowdhury Onick Over a year ago

Your question is not clear. What do you want to do actually? If you just want to get the image path, you don't need any hidden input tag.

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