Possessing radio buttons ??

[NevadaJones]

I have removed extra code from the input form I posted here. There are three sets of radio buttons each set has the unique name as shown. Because the input buttons I posted are checked they have the indicated value. This information is submitted to the process form by the submit button.

<form id='Update' name='Update' action='process.php' method='POST'>

<input type='radio' name='id[9]' value'1'checked> 1

<input type='radio' name='id[10]' value'1'checked> 1

<input type='radio' name='id[11]' value'3'checked> 3

<input type='submit' name='admin' value='Up Date Status'>
</form>

NOW how do I get the process form to handle what is submitted? With the code I have I can print each radio button name. What do I have to do to display and INPUT into a database that the radio name 9 has a value of 1 and so forth. Thanks for any more code you can give.

foreach ($_POST['id'] as $id=>$value){
echo $_POST['id'] . "&nbsp; &nbsp;" . $id . "<p>";
}

 

[kaptlid]

it depends on what you need to do? do you need to edit a value in the database or insert a new row? If you need to insert a new row you would do something like:

foreach ($_POST['id'] as $id=>$value){
echo $_POST['id'] . "&nbsp; &nbsp;" . $id . "<p>";

//query should be inside foreach loop
mysql_query("insert into table (id9,id10,id11) values ($_POST[id9],$_POST[id10],$_POST[id11])");

}

I am guessing that this is what you need because you have multiple id[9] values, etc...

 

[NevadaJones]

Thank you kaptlid. I had a major problem in my radio button input lines by leaving out the = for values'1', etc.

Now the values are being properly passed and I can tell by the echo statement I included for testing. I need to improve my query statement because it is not UPDATEing the table completely.

$array = $_POST['id'];
foreach($array as $key => $value)
{
echo $key.' is: '.$value . '<br>';

	$SendQuery = "UPDATE $TableName 
		SET status = '$value'
		WHERE id = '$key'";
}

 

[kaptlid]

if you are inputting numbers into an INT field type make sure NOT to have the single quotes.

Why do you have a dollar sign before TableName?

mysql_query("update tablename set status = '$value' where id = $key");

 

[NevadaJones]

A value is assign to $TableName outside of the query. My script is working now. This link.

I had not included my mysqli_query() statement inside the brackets with the for loop.

Thanks for the replies.