Select option from dropdown menu with PHP and mySql [duplicate]

Question

I tried to create a simple select dropdown menu from MySQL database. However, it does not work on my code.

Here is my code:

            <?php
              mysql_select_db($database_conn, $conn);
              $query_RsCourse = "SELECT * FROM tbl_course ORDER BY courseid DESC";
              $RsCourse = mysql_query($query_RsCourse, $conn) or die(mysql_error());
              $totalRows_RsCourse = mysql_num_rows($RsCourse);
              $count=0;
              while ( $row = mysql_fetch_array($RsCourse, MYSQL_ASSOC)) {
              $courseid=$row["courseid"];
         $count++;  
              }
             ?>

            <tr>
            <td bgcolor="#CCCCCC">&nbsp;Course Name</td>
            <td bgcolor="#dfdfdf">&nbsp;<select name="courseid">
            <option value="" SELECTED>Selected Course</option>      
            <option value="<?php echo $courseid; ?>"><?php echo $row_RsCourse['$courseid']; ?></option>
            </select>
            </td>
            </tr>

Any advice will be appreciated!

You are overwriting $courseid in your while loop before building your select options, so you will only get the last value of $courseid — Sean
– Sean, Commented Nov 29, 2013 at 2:25

Joem · Accepted Answer · 2013-11-29 03:50:30Z

0

 <?php

 echo '<tr>
        <td bgcolor="#CCCCCC">&nbsp;Course Name</td>
        <td bgcolor="#dfdfdf">&nbsp;';

        mysql_select_db($database_conn, $conn);
        $query_RsCourse = "SELECT * FROM tbl_course ORDER BY courseid DESC";
        $RsCourse = mysql_query($query_RsCourse, $conn) or die(mysql_error());
        $totalRows_RsCourse = mysql_num_rows($RsCourse);
        if($totalRows_RsCourse)
        {
        echo '<select name="courseid"><option value="" SELECTED>Selected Course</option>';
            $count=0;

            while ($row = mysql_fetch_array($RsCourse, MYSQL_ASSOC))
            {
            $count++;
            echo '<option value="'.$row['courseid'].'">'.$row['courseid'].'</option>';
            }
            echo '</select>';
        }
        else
        {
        echo 'No courses to show yet.'; // no rows in tbl_course
        }
        echo '</td>
            </tr>';

?>

That was a mess, but hope you can go on from these new codes. Enjoy.

PS: this part >'.$row['courseid'].'</option> u can change to new one according to your table structure which one is not shown here.

<?php

 echo '<tr>
        <td bgcolor="#CCCCCC">&nbsp;Course Name</td>
        <td bgcolor="#dfdfdf">&nbsp;';

        mysql_select_db($database_conn, $conn);
        $query_RsCourse = "SELECT * FROM tbl_course ORDER BY courseid DESC";
        $RsCourse = mysql_query($query_RsCourse, $conn) or die(mysql_error());
        $totalRows_RsCourse = mysql_num_rows($RsCourse);


        echo '<select name="courseid"><option value="" SELECTED>Selected Course</option>';
            $count=0;

            while ($row = mysql_fetch_array($RsCourse, MYSQL_ASSOC))
            {
            $count++;
            echo '<option value="'.$row['courseid'].'">'.$row['courseid'].'</option>';
            }
            echo '</select></td>
            </tr>';

?>

edited Nov 29, 2013 at 3:50

answered Nov 29, 2013 at 2:43

Joem

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6 Comments

Mahou Over a year ago

Yep. I tried to echo "<option>$courseid</option>"; it is able to show up on screen. However, it is unable to come up on select drop down menu.

Joem Over a year ago

Yeah I expected that, that is why I gave you that code. Well, do you have ANY rows/data in your tbl_course????? Can you see any data via phpmyadmin in there???

Mahou Over a year ago

Is it able to do the code seperately with while command? I have form to post in the other php pager. Thanks.

Joem Over a year ago

This code is operational Mahou. If you see "No courses to show yet" that means: a) you dont have any data in "tbl_course", or b) it is mysql query error. But here is code bellow with clean "while".:P

Mahou Over a year ago

Thanks for your reply. I copied your code and pasta to create new one. It is working fine. However, I tried to work on seperate code from your code. It does not work. The table is including data. I am trying to fix up this issue now.

|

capcj · Accepted Answer · 2018-08-29 12:07:59Z

You can store everything in a buffer and print at once in the select below:

    <?php
      mysql_select_db($database_conn, $conn);
      $query_RsCourse = "SELECT * FROM tbl_course ORDER BY courseid DESC";
      $RsCourse = mysql_query($query_RsCourse, $conn) or die(mysql_error());
      $coursesHtml = "";
      while ( $row = mysql_fetch_array($RsCourse, MYSQL_ASSOC)) {
        $coursesHtml .= "<option value='{$row["courseid"]}'>{$row["coursename"]}</option>";
      }
     ?>

    <tr>
    <td bgcolor="#CCCCCC">Course Name</td>
    <td bgcolor="#dfdfdf">
    <select name="courseid">
    <option value="" SELECTED>Selected Course</option>      
    <?= $coursesHtml  ?>
    </select>
    </td>
    </tr>

PS: Avoid use   , style your html well with css using padding-left: 5px; or other features;

PS2: You should not show your page/form with tables structure, use divs with flexbox.

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