submit button to input value to database without form action PHP

Question

I would like click on submit and the value input in the field to be stored in database.

However, I do not want to use a form action. Is it possible to do it without creating form action with PHP?

<tr>
<form method="post">
<tr>
    <td>
        <label for="Item name"><b>Finish Product:</b></label>
    </td>
    <td>
        <input id="finish_product" type="text" maxlength="100" style="width:100px"name="finish_product" required>
    </td>
</tr>
<tr>
    <td>
        <input type="submit" value="Save" id="submit" />
    </td>
</tr>

<?php
if(isset($_POST['submit']))
{
var_dump($_POST); exit;

$SQL = "INSERT INTO bom (finish_product) VALUES ('$finish_product')";
$result = mysql_query($SQL);
}?>
</tr>

basically you doesn't want form action but you want pass data into post method ....? — Maksud Mansuri
– Maksud Mansuri, Commented Jan 13, 2016 at 12:12

Sanjay Kumar N S · Accepted Answer · 2016-01-13 12:16:31Z

Use Jquery ajax to do this. Try this: HTML

<script src="//code.jquery.com/jquery-1.11.2.min.js"></script>
<body>
<table>
<tr>
<td><label for="Item name"><b>Finish Product:</b></label></td>
<td><input id="finish_product" type="text" maxlength="100" style="width:100px" name="finish_product" required></td>
</tr>
</table>
<input type="button" value="Save" id="submit" />
    <script>
    $(document).ready(function(){
       $('#submit').click(function(){
          $.ajax({
                url: '1.php',
                data: {finish_product: $('#finish_product').val()},
                success: function (result) {
                    alert(result)
                }
            });
       });
    });
    </script>
</body>

PHP (1.php)

Note: Use mysqli_query since mysql_query is depricated in latest versions. And use bind param instead of directly appending values to query.

<?php
if (!empty($_GET)) {
    $SQL = "INSERT INTO bom (finish_product) VALUES ('".$_GET['finish_product:']."')";
    $result = mysql_query($SQL);
    echo 1;
} else {
    echo -1;
}
?>

Ben · Accepted Answer · 2016-01-13 12:14:46Z

0

This isn't a form, this is just a series of table elements with various inputs and a submit button. Clean up the code - there's no closing tr tag, and where is the submit button supposed to be?
Add a form element around it.
You need an method attribute to the form - in this case, "post". Without the action attribute, it will default to the same page.

<!-- add form tag -->
<form method="post">
    <tr>
        <td>
            <label for="Item name"><b>Finish Product:</b></label>
        </td>
        <td>
            <input id="finish_product" type="text" maxlength="100" style="width:100px"name="finish_product" required>
        </td>
    </tr>
    <!-- added new row for submit button - you might want to have the td element span two columns? -->
    <tr>
        <td>
            <input type="submit" value="Save" id="submit" />
        </td>
    </tr>
</form>
<-- end of form -->

<?php
if(isset($_POST['submit']))
{
    //see what is being POSTED - comment out this line when you're happy with the code!
    var_dump($_POST); exit;

    $SQL = "INSERT INTO bom (finish_product) VALUES ('$finish_product')";
    $result = mysql_query($SQL);
}

answered Jan 13, 2016 at 12:14

Ben

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1 Comment

Janice Over a year ago

Hi, I have tried but nothing is added to my database after I input Table at the field. Why is this happening?

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