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Editorial Team
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Asked: 2026-06-19T00:14:46+00:00

I’m having difficulty populating a select box within a form to display existing forenames

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I’m having difficulty populating a select box within a form to display existing “forenames” of nurses from the “Nurses” table. Could anyone tell me what Im doing wrong? Thanks in advance!

Here is the form

 <form method="post" action="insert.php"> 
 <br>
 <tr><td align="left"><strong>Nurse Information</strong></td>
 </td>
 <tr> 
 <td>nurse_name</td>
       <td><select name="valuelist">
    <option value="valuelist" name="nurse_name" value='<?php echo $nurse_name; ?>'></option>

 </select></td>
 <tr>

The QUERY which should populate the nurse_forename: 

<html><head><title>Connect to Database</title></head><body>
<font size="4">Query gets Forename of nurse</font>
<br><br><font size="4">Choose a name</font><br><br> 

<form action="insert.php" method="post">
<select name="valuelist">;
<?php
$value=$_POST ["valuelist"];
$con = mysql_connect("localhost","root","") or die('Could not connect: ' . mysql_error());
 mysql_select_db("a&e", $con) or die('Could not select database.');

$fetch_nurse_name = mysql_query("SELECT DISTINCT $nurse_name FROM nurse");
$result = mysqli_query($con, $query) or die("Invalid query");

while($throw_nurse_name = mysqli_fetch_array($fetch_nurse_name)) {
echo '<option   value=\"'.$nurse_name['nurse_name'].'">'.$throw_nurse_name['nurse_name'].'</option>';
}
echo "</select>";

mysqli_close($con);
 ?>
<input type="submit" value="Submit">
</form></body></html>
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1 Answer

  1. Editorial Team

    Try this:

    <html><head><title>Connect to Database</title></head><body>
 <font size="4">Query gets Forename of nurse</font>
 <br><br><font size="4">Choose a name</font><br><br> 

 <form action="insert.php" method="post">
<select name="valuelist">;
<?php
$value=$_POST ["valuelist"];
$con = mysql_connect("localhost","root","") or die('Could not connect:'.mysql_error());
mysql_select_db("a&e", $con) or die('Could not select database.');

$fetch_nurse_name = mysql_query("SELECT DISTINCT Forename FROM nurse");


while($throw_nurse_name = mysql_fetch_array($fetch_nurse_name)) {
echo '<option   value=\"'.$throw_nurse_name[0].'">'.$throw_nurse_name[0].'</option>';
}
echo "</select>";


?>
<input type="submit" value="Submit">
</form></body></html>

    Dont use mysql and mysqli together….you should use mysqli or PDO, but not a mix of both 😉
    PS: Edited 😉

    Saludos.

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