来自mysql数据库的下拉菜单

Question

i need to populate a dropdown menu with date from database, this is what i've done so far

<div class="col-md-6">
    <?php
        $query_user_group = mysqli_prepare ($conn, "
SELECT group_name
     , group_id 
  FROM user_group_join 
  LEFT
  JOIN user_group 
    ON user_group_join . group_join_id = user_group . group_id 
 WHERE user_join_id = ?
");
       mysqli_stmt_bind_param($query_user_group, 'i', $client_id);
       mysqli_stmt_execute($query_user_group);
       mysqli_stmt_bind_result($query_user_group, $group_name, $group_id);
       mysqli_stmt_fetch ($query_user_group);
       mysqli_stmt_close($query_user_group);

    ?>

                        <div class="form-group">
                        <label class="control-label">Condominio in gestione*</label>
                        <select class="bs-select form-control" name="usergroup">
                        <option value="<?php echo $group_id;?> " selected="selected"><?php echo $group_name;?></option>

                        <?php 

                        $select_group_query= mysqli_prepare($conn, "SELECT group_id, group_name FROM user_group");  
                        mysqli_stmt_execute($select_group_query);
                        mysqli_stmt_bind_result($select_group_query, $idgruppo, $nomegruppo);


                            while(mysqli_stmt_fetch($select_group_query))

                                {     

                                    echo "<option value= '".$idgruppo."'>" . $nomegruppo . "</option>";

                                }


                        ?>  

                        </select>

                        <span class="help-block"> Assicurati di aver creato una scheda condominio! <br>Per inserire un nuovo condominio <a href="admin_create_new_group.php">Clicca Qui</a></span>



                        </div>
</div>

The problem is that the dropdown shows the selected value twice in the dropdown, any idea how i can sort it out? This is a screenshot of the error enter image description here Many thanks

Answer 1

You can use DISTINCT in your SELECT query, like this:

SELECT DISTINCT group_id, group_name FROM ...

Also, you don't need this line at all,

<option value="<?php echo $group_id;?> " selected="selected"><?php echo $group_name;?></option>

You can use the selected attribute of option tag to achieve the desired result.

So your dropdown list code should be like this:

<select class="bs-select form-control" name="usergroup">
<?php 
    $select_group_query= mysqli_prepare($conn, "SELECT DISTINCT group_id, group_name FROM user_group");  
    mysqli_stmt_execute($select_group_query);
    mysqli_stmt_bind_result($select_group_query, $idgruppo, $nomegruppo);
    while(mysqli_stmt_fetch($select_group_query)){ 
        $output = "<option value= '".$idgruppo."'";
        if($idgruppo == $group_id){
            $output .= " selected='selected'";
        }
        $output .= ">" . $nomegruppo . "</option>";
        echo $output;
    }
?>  
</select>