i need to save info from multi dynamic select using to database - php

Question

I am trying to make a rating system using multi-select dynamically generated from rating text on the database. The final result should look like this: Rating Service but now I need to save selected option after save is clicked.
in the database there is two table the first one have 2 column the first is the id and the second is the text of the rate
the second table is where i wont to save the selection that the user chose using php
page code
<table class="table table-striped">
<thead>
<tr>
<th>#</th>
<th>About</th>
<th>Provider</th>
<th>Rating</th>
</tr>
</thead>
<?php
$query_rate_text = "SELECT * FROM rateing_text";
$selecting_rates = mysqli_query($con,$query_rate_text);
$i = 0;
while($row_rate = mysqli_fetch_assoc($selecting_rates)){
$rate_id= $row_rate['rate_id'];
$rate_text= $row_rate['rate_text'];
$i++
?>
<form name="rating_form">
<tbody>
<tr>
<td><?php echo $i ?></td>
<td> <?php echo $rate_text ?></td>
<td><span class="badge badge-danger">
<?php echo $provider_name ?></span></td>
<td>
<select name="p<?php echo $i ?>">
<option value="1">Very Bad</option>
<option value="2">Bad</option>
<option value="3">Good</option>
<option value="4">Very Good</option>
<option value="5">Excelent</option>
</select>
</td>
</tr>
</tbody>
<?php } ?>
<tfoot>
<tr>
<td colspan="3"><button type="submit" name="save_rate"> Save </button> </td>
</tr>
</tfoot>
`enter code here`
</form>
</table>
How can I get info form the select and save them to the database ?

Answer 1

Radio button is much better than <select>.
The rate id and selected value must be passed as a key=>value
You need to stop using Word Press syntax.
Learn to use Herdoc Syntax
Using SELECT * is a poor standard practice.
Fetch array and assign values with a list
This code is untested.
<?php
echo <<<EOT
<form name="rating_form" action="#">
<table class="table table-striped">
<tr><th>#</th><th>About</th><th>Provider</th><th>Rating</th></tr>
EOT;
$query_rate_text = "SELECT `rate_id`,`rate_text` FROM rateing_text WHERE 1 ORDER BY `rate_id`";
$selecting_rates = mysqli_query($con,$query_rate_text);
$i = 0;
while(list($rate_id, $rate_text) = mysqli_fetch_array($selecting_rates)){
$i++;
echo "<tr><td>$i</td><td>$rate_text</td><td>$provider_name</td><td>5<input type="radio" name=\"$rate_id\" value=\"5\" /> 4<input type="radio" name=\"$rate_id\" value=\"4\" /> 3<input type="radio" name=\"$rate_id\" value=\"3\" /> 2<input type="radio" name=\"$rate_id\" value=\"2\" /> 1<input type="radio" name=\"$rate_id\" value=\"1\" /></td></tr>\n";
}
echo '</table><button type="submit" name="save_rate"> Save </button></form>';
?>
UPDATE
thank you very much that's helped to understand new way's,one more
thing how can i get the data out from this radio 5 buttons to insert
them to the database to the table named rating which have let's say 3
column id and provider name and the rate it self , again thank you for
helping
The radio input type is a more user friendly way then the cumbersome select.
The way I setup the radio buttons with the rate_id as the "name" so the selected "value" is passed as a key=>value pair to the form's action script.
Your select name should also contain the rate_id
The problem your code has is, the action script will receive will receive keys of sequential numbers that have no meaning and difficult to know how many were posted by the form.
To pass the provide name add a hidden input type:
<input type="hidden" name="provider" value="$provider_name" />
And I would remove the provider from the table td.
Not knowing your rate_id convention I could not improve the radio button naming convention.
The way I would do it is when passing key=>value pairs I would begin the key "name" with a unique character where no other "name" in the form would start with that character.
For example if the rate_id is a numeric value I may prepend a 'k' to the numeric key value.
So instead of
name=\"$rate_id\"
I would use
name=\"k$rate_id\"
The in the receiving action script I would get the key=>values like this
$provider_name = $_GET['provider']
foreach($_GET as $key => $value)){
if(substr($key,0,1) == 'k'){
$rate_id = intval(substr($key,1));
$rating = intval($value);
$sql = "INSERT INTO `table` (`rate_id`, `provider_name`, `rating`) VALUES ($rate_id, '$provider_name', $rating)";
mysqli_query($link,$sql);
}
}