javascript to send value to $_POST in PHP

Question

What I want to do is to use a confirm box where if the user clicks OK, to delete a row from my SQL database. I have a js function I use to send the value "delete" to a php function on the same file like this:

<script>
    function myFunction() {
        if (confirm("Are you sure you want to delete?") == true) {
            document.getElementById("delete").name = "delete";
        }else{
            return false;
        }
    }
</script>

The button where the user will click to delete an image is this:

<button onclick="myFunction()">Delete Image</button>

I send the value to the PHP function like this:

<input type="hidden" name="" value="delete" id="delete">

This is my PHP function:

if(isset($_POST['delete'])){
    $img_path=$_POST['ipath'];
    $imgid=$_POST['imgid'];
    $link = mysqli_connect($host, $username, $password, $db);
    $delete = "DELETE FROM images_info
                WHERE Image_Id = $imgid";
    $result3 = mysqli_query($link, $delete);
    echo "Image Deleted : $imgid";
    mysqli_close($link);
}

What am I doing wrong? I believe it's gotta do with the javascript

The html is inside <form action="newsearch.php" method="post">. However I opened and closed several <div>s inside this form. Will this make a difference? @DivyeshSavaliya — Yohan Blake
– Yohan Blake, Commented Mar 30, 2016 at 5:47
@Ohgodwhy, is there a simple way like this using ajax? I'm not familiar with ajax — Yohan Blake
– Yohan Blake, Commented Mar 30, 2016 at 5:48

Abhishek Sharma · Accepted Answer · 2019-09-07 16:43:36Z

6

in your html, input name attribute is empty,

<input type="hidden" name="" value="delete" id="delete">

it should be like

<input type="hidden" name="delete" value="delete" id="delete">

$_POST['delete'] will not be set until the name attribute is provided with the value 'delete'

edited Sep 7, 2019 at 16:43

answered Mar 30, 2016 at 5:55

Abhishek Sharma

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5 Comments

Yohan Blake Over a year ago

the whole point of adding the javascript was to use a confirm box to ask the user to confirm if he/she wants to delete. If I wanted to delete straightaway by clicking a button, your method works

Abhishek Sharma Over a year ago

and in your php code u can use if(isset($_POST['delete'] && $_POST['delete'] == 'delete' )) , if you want to be sure that the data is not modified on client side before submitting the form

Yohan Blake Over a year ago

I think the problem here is with the javascript that send the value to the php. name="delete" is not what I'm looking for

Evelyn Kokemoor Over a year ago

You can use JavaScript to control the flow of data, but ultimately you do have to send your data from the client to the server. For _POST this means using the elements in your <form>. You can use JavaScript all you like to control the form input, but make your the final data gets to the form elements' name and value fields.

Abhishek Sharma Over a year ago

can u be more specific what is happening actually? i have tried to run this in fiddle the issue i m facing is, if i click yes, it sets the name = 'delete' and you can't undo that if u again click cancel in confirm dialog. so if once clicked yes the name will be set to delete and the php code will delete the image for sure jsfiddle.net/mommso8c

Sona · Accepted Answer · 2016-03-30 06:24:05Z

2

Use it like this. here is PhpFiddle

PHP CODE

  <?php
     if(isset($_POST['delete']))
    {
        // PERFORM YOUR DELETE QUERY HERE
        print_r($_POST);
     }
   ?>

JAVA SCRIPT

      <script>
      function myFunction() {
      if (confirm("Are you sure you want to delete?") == true) {
         document.getElementById("delete").name = "delete";

       }else{
       return false;
       }
     }
   </script>

HTML

     <form method="post" action="">
         <button onclick="myFunction()">Delete Image</button>
         <input type="hidden" name="" value="delete" id="delete">
     </form>

edited Mar 30, 2016 at 6:24

answered Mar 30, 2016 at 5:49

Sona

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6 Comments

Divyesh Savaliya Over a year ago

how form will submit?

Sona Over a year ago

By clicking on Submit Button. which i have added below delete button , see html code.

Divyesh Savaliya Over a year ago

so user have to click two button? first, delete image then submit? if user directly click on submit how he get alert ?

Sona Over a year ago

you need to add another javascript on form submit but better to use ajax instead if you want this type of validation.

Yohan Blake Over a year ago

@Sona, can you give me an example how to use ajax for this? I'm not familiar with ajax

|

Evelyn Kokemoor · Accepted Answer · 2016-03-30 06:07:53Z

1

The variable $_POST is an associative array obtained from a form's <input> tags. The keys into $_POST are defined by the name attributes of an input tag, and the corresponding values in $_POST are defined by the value attribute in the same tag. Data in $_POST does not come from JavaScript unless you use an AJAX library like jQuery. If you want to control this data with pure JavaScript, you must directly set the attributes of your input tags.

You can keep the same JavaScript code you're using, just set the input tag's value attribute instead of its name, which should always be delete. Then in your PHP code, you can test the value of $_POST['delete'] to find out if the user really wanted to delete.

answered Mar 30, 2016 at 6:07

Evelyn Kokemoor

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Comments

zfj3ub94rf576hc4eegm · Accepted Answer · 2016-03-30 06:27:11Z

1

Doesn't seem like anyone wants to answer your question. If you'd like to do this with pure js, here ya go:

var url = 'YOUR URL POST DESTINATION HERE';
function myFunction(elem){
    var id = elem.target.id;
    var xmlHttp = new XMLHttpRequest();
    var params = 'delete=true&imgid=' + id;

    xmlHttp.open('POST', url, true);
    xmlHttp.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
    xmlHttp.send(params);

}

You will need to set imgid in your js programmatically. Execute myFunction() from your page like

<button onclick="myFunction(this)">Delete Image</button>

As a side note, it's a REALLY bad idea to let users post to a form and modify your DB. ANYONE can make a post request and delete your DB. But I decided to give you the benefit of the doubt and answer your question.

answered Mar 30, 2016 at 6:27

zfj3ub94rf576hc4eegm

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2 Comments

Yohan Blake Over a year ago

Thanks for the answer, and if it's not good to use POST to modify db data, what's the best way? Sorry I'm new

zfj3ub94rf576hc4eegm Over a year ago

You need to be very careful when building SQL queries with user input, even if it has been sanitized. Your PHP is vulnerable to SQL injection attacks. If I sent 1 OR 1=1, it would delete your entire images_info table. I recommend this tutorial on using PHP Data Objects instead.

Yohan Blake · Accepted Answer · 2016-03-31 00:07:41Z

1

Sorry for all this mess guys, the actual problem was that I had another form inside the form. I didn't know that forms cannot contain forms inside them. Removed the other form, and it works

answered Mar 31, 2016 at 0:07

Yohan Blake

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Comments

Mr Lister · Accepted Answer · 2016-04-04 14:37:05Z

1

Do not use onclick="myFunction()" in the <input> tag. Instead, ensure:

<form method ="post" onsubmit="return confirm("Are you sure you want to delete?")">
    <button name = "delete">Delete Image</button>
</form>

  <?php
 if(isset($_POST['delete']))
{
     $img_path=$_POST['ipath'];
     $imgid=$_POST['imgid'];
     $link = mysqli_connect($host, $username, $password, $db);
     $delete = "DELETE FROM images_info WHERE Image_Id = $imgid";
     $result3 = mysqli_query($link, $delete);
     echo "Image Deleted : $imgid";
     mysqli_close($link);
 }
   ?>

edited Apr 4, 2016 at 14:37

Mr Lister

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answered Mar 30, 2016 at 6:07

Manav

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1 Comment

rashedcs Over a year ago

Good job @Manav

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