php while loop displaying sql data incorrectly

Question

Ok, I have this while loop that basically grabs the data from an SQL DB and puts it into a dropdown menu, the problem is that a separate dropdown is created for each value. I need just one dropdown to display all the values.

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>

<body>

<?php

$conn = mysql_connect("localhost","root","")or die (mysql_error());

mysql_select_db("assignment_3", $conn);

$data = "select schoolName from schooltable";

$result = mysql_query($data, $conn) or die (mysql_error());

while ($row = mysql_fetch_assoc($result)) {
print "<select>";
print "<option value='' disabled='disabled' selected='selected'> Please Select your  Undergraduate School </option>";
print "<option value='1'>".$row['schoolName']."</option>";
print "</select>";
}

?>

</body>
</html>

John Conde · Accepted Answer · 2013-11-18 02:21:12Z

2

You need to move the print statements for the select element outside of the while loop:

print "<select>";
print "<option value='' disabled='disabled' selected='selected'> Please Select your  Undergraduate School </option>";
while ($row = mysql_fetch_assoc($result)) {
    print "<option>".$row['schoolName']."</option>";
}
print "</select>";

To save you another problem, you'll need to ditch the value attribute or else every option will submit the same value because they're all set to 1. Although a better option might be to print the id of the school instead:

$data = "select schoolId, schoolName from schooltable";

print '<option value="'.$row['schoolId'].'">'.$row['schoolName'].'</option>';

answered Nov 18, 2013 at 2:21

John Conde

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735Tesla · Accepted Answer · 2013-11-18 02:27:30Z

1

Each time you go around the loop again you have it create a new html element. It should be: Untitled Document

<body>

<?php

$conn = mysql_connect("localhost","root","")or die (mysql_error());

mysql_select_db("assignment_3", $conn);

$data = "select schoolName from schooltable";

$result = mysql_query($data, $conn) or die (mysql_error());
print "<select>";
while ($row = mysql_fetch_assoc($result)) {
print "<option value='' disabled='disabled' selected='selected'> Please Select your  Undergraduate School </option>";
print "<option value='1'>".$row['schoolName']."</option>";
}
print "</select>";
?>

</body>
</html>

answered Nov 18, 2013 at 2:27

735Tesla

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Franco Selem · Accepted Answer · 2013-11-18 02:59:16Z

0

This is how it should look, also, avoid using mysql_ prefixed functions. Either use mysqli_ of the OOP version of MySQLi.

$link = new mysqli( $host, $user, $password, $database ); // Connect to DB

$result = $link->query( "SELECT `schoolName` FROM `schoolTable`" );

echo "<select>";
echo "<option disabled="disabled" selected="selected">Please Select your undergraduate School</option>";

while ( $row = $result->fetch_assoc() ) {
    echo "<option>" . $row['schoolName'] . "</option>";
}

echo "</select>";

answered Nov 18, 2013 at 2:59

Franco Selem

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