CODEWITHSUNDEEP
phphtml
dansasu11
dansasu11

Reputation: 913

php dropdown list from mysql

I just started learning php last week and I am trying to populate values of a drop down list in my html form from mysql database. the dropdown shows but it only shows "choose" as an option. what am I doing wrong, any help would be march appreciated. Below is the problem part of my code.

<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">

 <?php
  require_once('connectvars.php');
  $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
  $query = "SELECT event_id, event_name FROM events";

  $results= mysqli_query( $dbc, $query);

  $options="";

 while ($row=mysql_fetch_array($result)) {

 $id=$row["event_id"];
 $event=$row["event_name"];
 $options ="<OPTION VALUE=\"$id\">".$event;
 } 
?>


 <SELECT NAME=eventid>
<OPTION VALUE=0>Choose
  <?php echo $options ?>
 </SELECT> 
</form>

Upvotes: 0

Views: 4891

Answers (5)

user1048311
user1048311

Reputation:

This is the mysql-driver solution.

<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<?php
    require_once('connectvars.php');
    $dbc = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
    $db = mysql_select_db(DB_NAME);
    $results= mysql_query("SELECT event_id, event_name FROM events");
?>
<select name="eventid">
    <option value="0">Choose</OPTION>
    <?php 
        while($row = mysql_fetch_array($results)) {
            echo '<option value="'.$row['event_id'].'">'. $row['event_name'].'</option>';
        }
    ?>
</select> 
</form>

Upvotes: 1

check123
check123

Reputation: 2009

Include correction after Benedikt Olek's answer. 

<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">

<?php
 require_once('connectvars.php');
 $dbc = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
 $query = "SELECT event_id, event_name FROM events";
 $db = mysql_select_db(DB_NAME);
 $results= mysql_query( $dbc, $query);
?>


<SELECT NAME=eventid>
<OPTION VALUE=0>Choose</OPTION>
  <?php 
     while($row = mysql_fetch_array($results)
        echo "<option value=\"" . $row['event_id'] . "\">" . $row['event_name'] . "</option>";
  ?>
</SELECT> 
</form>

Upvotes: 1

user1048311
user1048311

Reputation:

Since "Choose" is a hardcoded option it is always displayed. You are not getting more results because the Variable $options is empty. You can verify that by using var_dump($options).

The reason for your empty variable is the while-iteration. Your function mysql_fetch_array() never returns true.

That is because you are mixing up drivers. You are connecting via mysqli, hence mysql_fetch_array is not working.

You have two options: change to use the mysql-driver for connection or recode the iteration to make use of the mysqli-api.

Upvotes: 0

Zaina
Zaina

Reputation: 1

try this one :

 <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
 <SELECT NAME=eventid>
 <?php
  require_once('connectvars.php');
  $dbc = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
  mysql_select_db(DB_NAME);
  $query = "SELECT event_id, event_name FROM events";

  $results= mysql_query( $dbc, $query);

  $options="";

 while ($row=mysql_fetch_array($result)) {

 $id=$row["event_id"];
 $event=$row["event_name"];
echo "<OPTION VALUE=".$id\.">".$event."</option>";
 } 
?>

 </SELECT> 
</form>

Upvotes: 0

mishu
mishu

Reputation: 5397

change 

$options ="<OPTION VALUE=\"$id\">".$event;

to

$options .="<OPTION VALUE=\"$id\">".$event;

to concatenate instead of changing the value always.. also consider closing the option tag..

Upvotes: 1

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