im trying to put a drop down list generated from mysql in html page , the thing is this code work fine in a php page but it doesnt in html page , I've put it in a php tags yet nothing was shown , I would appreciate any help here is my code
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Add Qustions</title>
</head>
<body>
<form action="saveQ.php" method="post" >
<br />
<?php $connectdb = mysql_connect('localhost','root','sara') or die ("Not Connect");
if (!$connectdb)
{
die('Could not connect :'. mysql_errno());
}
$selestdb = mysql_select_db('iexa', $connectdb) or die ("not selected database");
$qu = mysql_query("SELECT ID,Name FROM Course ORDER BY ID asc") or die ("mysql error");
echo "Choose the course that you want to create the test for : <br /> ";
echo " <select name='courseID' >Course</option>";
echo "<option value=0>Course </option>";
$curvalue=2;
while ($row = mysql_fetch_assoc($qu)){
echo '<option value="' .$row[ID].'">' .$row[ID].' ' .$row[Name].'</option>';
$curvalue = $curvalue+1;
}
echo "</select> "; ?>
<br />
/// some other unrelated code here
<input type="submit" value="Save Question" />
</form>
</body>
</html>
Maybe because you are doing it like this:
echo " <select name='courseID' >Course</option>";
Change it to:
echo "<select name='courseID'>";
I personally don't like to echo out html in PHP so i use this method, in any case your issue is html based and not php based...
<?php
$selectdb = mysql_select_db('iexa', $connectdb) or die ("not selected database");
$qu = mysql_query("SELECT ID,Name FROM Course ORDER BY ID asc") or die ("mysql error");
?>
<p>Choose the course that you want to create the test for:</p><br />
<select name="courseID">
<option value="0">0 Course</option>
<option value="1">1 Course</option>
<?php while( $row = mysql_fetch_assoc($qu) ): ?>
<option value="<?= $row['ID']; ?>"><?php echo $row['ID'].' '.$row['Name']; ?></option>
<?php endwhile; ?>
</select>
In short, you have an error here: echo " <select name='courseID' >Course</option>";.
It should be: echo " <select name='courseID' >Course</select>";
Also, you dont need to set $curvalue = $curvalue+1; You can simply do $curvalue++; IE:
<?php
while ($row = mysql_fetch_assoc($qu)){
<select name="courseID">
<option value="<?= $row['ID']; ?>"><?php echo $row['ID'].' '.$row['Name']; ?></option>
</select>
$curvalue++;
}
Also note that its considered best practice to not mix DOM with Server side code. IE you should put the while statement and select boxes in a seperate php view file that iterates through things, rather than using echo. This way there is less server processing and more client side processing.
I think this script is what you wanted to do:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Add Qustions</title>
</head>
<body>
<form action="saveQ.php" method="post">
<br />
<?php
$connectdb = mysql_connect('localhost','root','sara') or die ("Not Connect");
if (!$connectdb) die('Could not connect :'. mysql_errno());
echo " Choose the course that you want to create the test for: <br />
<select name=\"courseID\">
<option value=\"0\">Course </option>
";
$selestdb = mysql_select_db('iexa', $connectdb) or die ("not selected database");
$qu = mysql_query("SELECT ID,Name FROM Course ORDER BY ID asc") or die ("mysql error");
while ($row = mysql_fetch_assoc($qu))
echo " <option value=\"{$row["ID"]}\">{$row["ID"]} {$row["Name"]}</option>
";
echo " </select> ";
?>
<br />
<!-- some other unrelated code here -->
<input type="submit" value="Save Question" />
</form>
</body>
</html>
Your script's name should end with a PHP extension, like .php or .phtml.