[RESOLVED] Passing this radio button info to new page.

MitchEvans

I have my users choosing from several radio buttons on a page(page1). When they click "submit" they should be given a table of results on another page(page2). Depending on which one they choose an X will appear in the correct box.
My problem is with the name of the radio buttons I think.

Code for page1 is:

<FORM method='POST' action='friday1.php'><table border=1><tr>
<td>Requirement Type</td><td>Currently In Place</td><td>Required-Essential</td><td>Required-Recommended</td></tr>";
echo"<tr><td><strong>1 - Operational</strong></td></tr>";
$productsql1 = "SELECT * FROM survey WHERE rtid=1";
$productResults1 = mysql_query($productsql1);
while ($productRow1 = mysql_fetch_array ($productResults1)){
$ProductName1 = $productRow1["requirement_txt"];

echo"
<tr>
<td>$ProductName1</td>
<td><INPUT type='radio' name='myradio[{$productRow1['id']}]' value='Currently In Place' CHECKED></td>
<td><INPUT type='radio' name='myradio[{$productRow1['id']}]' value='Required Essential'></td>
<td><INPUT type='radio' name='myradio[{$productRow1['id']}]' value='Required-Recommended'></td></tr>
</tr>";}

And for page2 (friday1.php):

if($POST['myradio'] == "Currently In Place") {
$s= "X";
}
elseif($POST['myradio'] == "Required Essential") {
$t = "X";
}
else {
$u= "X";
}
echo"
<table border='1'><tr>
<td>Requirement Type</td><td>Currently In Place</td><td>Required-Essential</td><td>Required-Recommended</td><td>i-Clean Solution</td></tr>";
echo"<tr><td><strong>1 - Operational</strong></td></tr>";
$productsql1 = "SELECT * FROM survey WHERE rtid=1";
$productResults1 = mysql_query($productsql1);
while ($productRow1 = mysql_fetch_array ($productResults1)){
$ProductName1 = $productRow1["requirement_txt"];

echo"
<tr>
<td>$ProductName1</td>
<td><center><strong>$s</strong></center></td>
<td><center><strong>$t</strong></center></td>
<td><center><strong>$u</strong></center></td>
<td>Module 1</td>
</tr>";}

jernhenrik

<INPUT type='radio' name='myradio[1]' value='Currently In Place' CHECKED>
<INPUT type='radio' name='myradio[1]' value='Required Essential'>
etc.

Will provide you with an array called myradio. In order to check what is in that array you would need to:

if ($_POST['myradio'][1] == 'something') {
	//do horrible stuff to poor defenseless variable.
}

And in order to make it easier on yourself, try and have a look at foreach()

e.g:

foreach ($_POST['myradio']  as $itemno => $radioselectionvalue) {
	if($radioselectionvalue == "Currently In Place") {
		$s = "X";
	}
	//etc.
}

hth

MitchEvans

I think I need to the $productRow1 as there are several sets of radio buttons that are being looped to give the correct output.
Code:

echo"<tr><td><strong>1 - Operational</strong></td></tr>";
$productsql1 = "SELECT * FROM survey WHERE rtid=1";
$productResults1 = mysql_query($productsql1);
while ($productRow1 = mysql_fetch_array ($productResults1)){
$ProductName1 = $productRow1["requirement_txt"];

MitchEvans

If I echo myradio on the next page I get a message saying "array"

MitchEvans

If I put this on the next page:

$productsql1 = "SELECT * FROM survey WHERE rtid=1";
$productResults1 = mysql_query($productsql1);
while ($productRow1 = mysql_fetch_array ($productResults1)){
$ProductName1 = $productRow1["requirement_txt"];

echo "<BR>You Chose " . $_POST['myradio'] . "\n";}

I get the output:

Array
Array
Array
Array
Array
Array

It's 6 times as there are 6 results to show.

shamly

use print_r($_POST['myradio']) and check the output