Passing input values using post

Question

I am trying to pass the value of <input id="ids" type="hidden" name="ids" value="<?php echo $key['product_ID'];?>"> to the exchange.php page. Lets say i have apple | 1 , orange | 2, pineapple | 3. However, whenever i try to submit the values to another page, I am getting the 3 as the echo value of $product. When i try to remove the type="hidden", I get the correct value but when I try to submit, it turns out to be a different value.

<form action="exchange.php" method="post">
    <div class="row">
        <?php $query="SELECT * FROM Product" ; $data=$ MySQLi_CON->query($query); 
        foreach ($data as $key ) { ?>
        <strong>Name: </strong>
        <?php echo $productname=$ key[ 'product_Name'];?>
        <input id="ids" type="hidden" name="ids" value="<?php echo $key['product_ID'];?>">
        <strong>Status: </strong>
        <strong>Action: </strong>
        <input type="submit" value="Exchange" name="exchange_submit" class="btn btn-info btn-xs">
        <input id="id" name="id" type="hidden" value="<?php echo $id; ?>">
        <?php } ?>
    </div>
</form>

exchange.php

<?php
$id = $_POST['id']; 
$product = $_POST['ids']; 
echo $id; 
echo $product; 
exit;
?>

id should be unique. Currently you are setting id="ids" inside foreach, please change it. — Touheed Khan
– Touheed Khan, Commented Apr 15, 2017 at 4:53
id and ids are two different thing. id is the current user ID, ids is the product ID — Synetrix
– Synetrix, Commented Apr 15, 2017 at 4:56
if you are selecting all the products then running foreach to iterate over them. That means you are going to multiple create input by the id="id" or id="ids" which is wrong. id attribute should be unique inside whole page. we cannot assign same id to any other html element. — Touheed Khan
– Touheed Khan, Commented Apr 15, 2017 at 5:02

Gopi Chand · Accepted Answer · 2017-04-15 04:32:36Z

1

 <?php echo $productname=$ key[ 'product_Name'];?>
//is it working not showing error

answered Apr 15, 2017 at 4:32

Gopi Chand

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4 Comments

Synetrix Over a year ago

yes, it is working. Not showing error. Everything is working except for the $product = $_POST['ids']; Doesn't seem to display the correct output

Gopi Chand Over a year ago

You want multiple ids or single product id??

Synetrix Over a year ago

single product id, apple | 1 , orange | 2, pineapple | 3

Synetrix Over a year ago

but whenever for example i click on apple. It gives me id of 3

Brijesh_yadav · Accepted Answer · 2017-04-15 04:59:04Z

0

Whenever you execute a foreach loop, it will store the last output of the Database table in the hidden input of yours, and that's why no matter any input you give, it will take 3 as the id of the product.Try using a tag also.

answered Apr 15, 2017 at 4:59

Brijesh_yadav

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1 Comment

Synetrix Over a year ago

Thanks for the suggestion, but the answer above solved my problem :)

Touheed Khan · Accepted Answer · 2017-04-15 04:59:07Z

0

You are using foreach for products so I consider you've more than one products.

but when you doing this you are overwriting the $_POST['ids'].

<input id="ids" type="hidden" name="ids" value="<?php echo $key['product_ID'];?>">

thats why its showing last product_ID present in table.

answered Apr 15, 2017 at 4:59

Touheed Khan

2,14918 silver badges27 bronze badges

2 Comments

Synetrix Over a year ago

Thanks for the suggestion, but the answer above solved my problem :)

Touheed Khan Over a year ago

@user3676202 no worries :)

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