Question

我需要通过bootstrap selectpicker集成帮助我的php脚本

我有用于show selectpicker多值的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>

在我的另一张桌子上我有user和skill_id，对于skill_id内容：1,2,3

我把它分成了这样的每个值：

$q = ("SELECT nik, skill_id FROM dosen WHERE nik = '".$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：如何为我的多个selectpicker选项设置默认选择值，使它们在桌面上被选为我的用户数据？

谢谢。帮助将不胜感激：）

Answer 1

如下：

<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>

PHP MySQL设置选择selectpicker多个

1 个答案: