PHP MySQL 设置selectpicker选中多个

Question

我需要有关 Bootstrap 选择器集成的 php 脚本的帮助

我有用于显示 selectpicker 多个值的 php 脚本

<select id="skill" name="skill" class="selectpicker" multiple data-size="5" data-selected-text-format="count">
<?php
$skills = mysql_query("SELECT skill_id, skill_name FROM skill",$conn);
while($r_skills = mysql_fetch_assoc($skills)){
echo '<option value="'.$r_skills['skill_id'].'">'.$r_skills['skill_name'].'</option>
';
}
?>  
</select>

在我的另一张 table 上，我有用户和技能ID，技能ID内容:1,2,3

我已经将它分解为每个值，如下所示:

$q = ("SELECT nik, skill_id FROM dosen WHERE nik = '".$nik."'",$conn);
while($row_dosen = mysql_fetch_assoc($q)){

$pieces = explode(",", $row_dosen['skill_id']);

for($i = 0; $i < count($pieces) ; $i++){

$pp  = mysql_query("SELECT skill_id, skill_name FROM skill WHERE skill_id = '".$pieces[$i]."'",$conn);

while($p = mysql_fetch_assoc($pp)){
    $var_each = $p['skill_id']; 

}
}
}

问题是:我如何为我的多个选择器选项设置默认选定值，以使它们被选为表上的用户数据？

谢谢。我们将不胜感激您的帮助:)

Answer 1

怎么样:

<select id="skill" name="skill" class="selectpicker" multiple data-size="5" data-selected-text-format="count">
<?php
    $skills = mysql_query("SELECT skill_id, skill_name FROM skill",$conn);
    while($r_skills = mysql_fetch_assoc($skills)){
      echo '<option value="'.$r_skills['skill_id'].'"';
      if ($r_skills['skill_id'] == '1337 codingz')
          echo ' selected';
      echo '>'.$r_skills['skill_name'].'</option>';
}
?>  
</select>