Radio buttons not showing values from database

**hsriat** · Feb 26 '08, 07:06 AM

What is IsSponsored?

**lozwaldo** · Feb 26 '08, 07:22 AM

Originally posted by hsriat

What is IsSponsored?

IsSponsored is a field in table

Has value of "1" or "0"

**hsriat** · Feb 26 '08, 07:52 AM

[php]$sponsored = array("";"check ed");
$not_sponsored = array("checked" ;"");
while($show = mysql_fetch_arr ay( $result )) {
// Print out the contents of each row into a table
echo "<tr> <td>";
echo $show['ChildID'];
echo "</td> <td >";
echo $show['givenName'];
echo "</td> <td>";
echo $show['gender'];
echo "</td> <td>";
echo $show['countryName'];
echo "</td> <td>";
echo $show['favouritePlay'];
echo "</td> <td>";

/*this box checked if IsSponsored = 1*/
echo "<input type=\"radio\" name=\"".$show['ChildID']." \" value=\"1\" ".$sponsore d[$show['IsSponsored ']]."></input>Is sponsored";

echo "</td> <td>";

/*this box checked if IsSponsored = 0*/
echo "<input type=\"radio\" name=\"".$show['ChildID']." ''\" value=\"0\" ".$not_sponsore d[$show['IsSponsored ']]."></input>Is not sponsored";

echo "</td> <td>";
echo $show['dateOfBirth'];
echo "</td> </tr>";
}[/php]

Check this... is this what you wanted or not?

**lozwaldo** · Feb 26 '08, 08:30 AM

Originally posted by hsriat

[php]$sponsored = array("";"check ed");
$not_sponsored = array("checked" ;"");
while($show = mysql_fetch_arr ay( $result )) {
// Print out the contents of each row into a table
echo "<tr> <td>";
echo $show['ChildID'];
echo "</td> <td >";
echo $show['givenName'];
echo "</td> <td>";
echo $show['gender'];
echo "</td> <td>";
echo $show['countryName'];
echo "</td> <td>";
echo $show['favouritePlay'];
echo "</td> <td>";

/*this box checked if IsSponsored = 1*/
echo "<input type=\"radio\" name=\"".$show['ChildID']." \" value=\"1\" ".$sponsore d[$show['IsSponsored ']]."></input>Is sponsored";

echo "</td> <td>";

/*this box checked if IsSponsored = 0*/
echo "<input type=\"radio\" name=\"".$show['ChildID']." ''\" value=\"0\" ".$not_sponsore d[$show['IsSponsored ']]."></input>Is not sponsored";

echo "</td> <td>";
echo $show['dateOfBirth'];
echo "</td> </tr>";
}[/php]

Check this... is this what you wanted or not?

Sorry but tried this and page wouldn't run.

**hsriat** · Feb 26 '08, 08:34 AM

Originally posted by lozwaldo

Sorry but tried this and page wouldn't run.

isSponsored is a field in database, right?

And is there any error?

**lozwaldo** · Feb 26 '08, 08:47 AM

Originally posted by hsriat

isSponsored is a field in database, right?

And is there any error?

"IsSponsore d" is a field in the database

This page is returning a query from another page. When this page tries to display the results the browser comes back with the generic "The page cannot be displayed" message. Here is all the code from the page.

many thanks

/////////////////////////////////////////////////////////////////////////////////////////////////////

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"

"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dt d">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Admin</title>
</head>

<body>

<?php
$link ="<br><br><a href='searchpag e.php'>Click here to go back to the search page.</a>";

$countryName = $_POST['countryName'];
$gender = $_POST['gender'];
$ChildID = $_POST['ChildID'];

//Just checking that values are being submitted properly
echo $countryName;
echo"<br>";
echo $gender;
echo"<br>";
echo $ChildID;
echo"<br>";

$db = mysql_connect(' localhost', 'xxxx', 'xxxxx');
mysql_select_db ("child",$db );

if ($ChildID != "")
{$query ="SELECT * FROM child.child
where (countryName = '$countryName') and (gender ='$gender') and

(ChildID ='$ChildID')";
}else{
$query ="SELECT * FROM child.child
where (countryName = '$countryName') and (gender ='$gender')";
}

$result = mysql_query($qu ery);
if (mysql_num_rows ($result) == 0) {
echo "sorry, no matching results.";
echo $link;
exit;
}

echo "<table border='1' >";
echo "<tr><th>ChildI D</th>
<th>givenName </th>
<th>gender</th>
<th>countryName </th>
<th>favouritePl ay</th>
<th>Sponsored </th>
<th>Not Sponsored</th>
<th>dateOfBirth </th>
</tr>";
// keeps getting the next row until there are no more to get

$sponsored = array("";"check ed");

$not_sponsored = array("checked" ;"");

while($show = mysql_fetch_arr ay( $result )) {

/*Print out the contents of each row into a table*/

echo "<tr> <td>";
echo $show['ChildID'];
echo "</td> <td >";
echo $show['givenName'];
echo "</td> <td>";
echo $show['gender'];
echo "</td> <td>";
echo $show['countryName'];
echo "</td> <td>";
echo $show['favouritePlay'];
echo "</td> <td>";

/*this box checked if IsSponsored = 1*/
echo "<input type=\"radio\" name=\"".$show['ChildID']." \" value=\"1\" ".$sponsore d[$show['IsSponsored ']]."></input>Is sponsored";
echo "</td> <td>";

/*this box checked if IsSponsored = 0*/
echo "<input type=\"radio\" name=\"".$show['ChildID']." ''\" value=\"0\" ".$not_sponsore d[$show['IsSponsored ']]."></input>Is not sponsored";
echo "</td> <td>";
echo $show['dateOfBirth'];
echo "</td> </tr>";
}

echo "</table>";

?>

</body>
</html>

**hsriat** · Feb 26 '08, 09:07 AM

The code seems to be fine, the problem may be with the file name or directory.

**lozwaldo** · Feb 26 '08, 09:14 AM

Originally posted by hsriat

The code seems to be fine, the problem may be with the file name or directory.

stupid IE
Firefox returned this error
"PHP Parse error: syntax error, unexpected ';', expecting ')' in C:\Inetpub\wwwr oot\practice\ch ildc.php on line 63"

am looking into this now

**hsriat** · Feb 26 '08, 09:18 AM

Originally posted by lozwaldo

stupid IE
Firefox returned this error
"PHP Parse error: syntax error, unexpected ';', expecting ')' in C:\Inetpub\wwwr oot\practice\ch ildc.php on line 63"

am looking into this now

oops...
My fault...
Array fields are separated with comma, not semicolon!
Like this...
$sponsored = array("","check ed");
$not_sponsored = array("checked" ,"");

**lozwaldo** · Feb 26 '08, 09:25 AM

Originally posted by hsriat

oops...
My fault...
Array fields are saperated by comma, not semicolon!
Like this...
$sponsored = array("","check ed");
$not_sponsored = array("checked" ,"");

sorry.......
another error

Undefined index: IsSponsored in C:\Inetpub\wwwr oot\practice\ch ildc.php on line 83

Undefined index: IsSponsored in C:\Inetpub\wwwr oot\practice\ch ildc.php on line 87

**hsriat** · Feb 26 '08, 09:29 AM

Originally posted by lozwaldo

sorry.......
another error

Undefined index: IsSponsored in C:\Inetpub\wwwr oot\practice\ch ildc.php on line 83

Undefined index: IsSponsored in C:\Inetpub\wwwr oot\practice\ch ildc.php on line 87

remove the space after IsSponsored (in line 83 and 87) and check its spelling against the one in the db.

**Markus** · Feb 26 '08, 09:30 AM

Use code tags
[php]
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"

"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dt d">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Admin</title>
</head>

<body>

<?php
$link ="<br><br><a href='searchpag e.php'>Click here to go back to the search page.</a>";

$countryName = $_POST['countryName'];
$gender = $_POST['gender'];
$ChildID = $_POST['ChildID'];

//Just checking that values are being submitted properly
echo $countryName;
echo"<br>";
echo $gender;
echo"<br>";
echo $ChildID;
echo"<br>";

$db = mysql_connect(' localhost', 'xxxx', 'xxxxx');
mysql_select_db ("child",$db );

if ($ChildID != "")
{$query ="SELECT * FROM child.child
where (countryName = '$countryName') and (gender ='$gender') and

(ChildID ='$ChildID')";
}else{
$query ="SELECT * FROM child.child
where (countryName = '$countryName') and (gender ='$gender')";
}

$result = mysql_query($qu ery);
if (mysql_num_rows ($result) == 0) {
echo "sorry, no matching results.";
echo $link;
exit;
}

echo "<table border='1' >";
echo "<tr><th>ChildI D</th>
<th>givenName </th>
<th>gender</th>
<th>countryName </th>
<th>favouritePl ay</th>
<th>Sponsored </th>
<th>Not Sponsored</th>
<th>dateOfBirth </th>
</tr>";
// keeps getting the next row until there are no more to get

$sponsored = array("";"check ed");

$not_sponsored = array("checked" ;"");

while($show = mysql_fetch_arr ay( $result )) {

/*Print out the contents of each row into a table*/

echo "<tr> <td>";
echo $show['ChildID'];
echo "</td> <td >";
echo $show['givenName'];
echo "</td> <td>";
echo $show['gender'];
echo "</td> <td>";
echo $show['countryName'];
echo "</td> <td>";
echo $show['favouritePlay'];
echo "</td> <td>";

/*this box checked if IsSponsored = 1*/
echo "<input type=\"radio\" name=\"".$show['ChildID']." \" value=\"1\" ".$sponsore d[$show['IsSponsored ']]."></input>Is sponsored";
echo "</td> <td>";

/*this box checked if IsSponsored = 0*/
echo "<input type=\"radio\" name=\"".$show['ChildID']." ''\" value=\"0\" ".$not_sponsore d[$show['IsSponsored ']]."></input>Is not sponsored";
echo "</td> <td>";
echo $show['dateOfBirth'];
echo "</td> </tr>";
}

echo "</table>";

?>

</body>
</html>
[/php]

**lozwaldo** · Feb 26 '08, 09:35 AM

Originally posted by hsriat

remove the space after IsSponsored (in line 83 and 87) and check its spelling against the one in the db.

removed space and works PERFECTLY in Firefox........ however get this error from IE6

Undefined index: countryName in C:\Inetpub\wwwr oot\practice\ch ildc.php on line 15 PHP
Undefined index: gender in C:\Inetpub\wwwr oot\practice\ch ildc.php on line 16 PHP
Undefined index: ChildID in C:\Inetpub\wwwr oot\practice\ch ildc.php on line 17

**Markus** · Feb 26 '08, 09:37 AM

Originally posted by lozwaldo

removed space and works PERFECTLY in Firefox........ however get this error from IE6

Undefined index: countryName in C:\Inetpub\wwwr oot\practice\ch ildc.php on line 15 PHP
Undefined index: gender in C:\Inetpub\wwwr oot\practice\ch ildc.php on line 16 PHP
Undefined index: ChildID in C:\Inetpub\wwwr oot\practice\ch ildc.php on line 17

It wont be the browser as the errors are given from the server.

Place an @ symbol before the post:
[php]
$_something = @$_POST['something'];
[/php]
So, if the post isnt there, it won't throw an error.

Radio buttons not showing values from database

Radio buttons not showing values from database

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