php - 在选择元素中使用数组输出的mysql查询

Question

我正在尝试为表单中的几个 html 选项元素编写一个通用函数。每个元素的选项列表都会有所不同，所以我想要做的是根据在我的 mysql 表中引用它们的“id”显示选项，这将是一个数组。该表有 2 列，第一个“id”只是一个数字，第二个“prod_option”是我想要显示的值。请有人可以解释我的代码有什么问题吗？

这是我在 select 元素中调用函数的地方。我的连接没有问题。

//This should select 'prod_option' for id's 1,2 & 3 for example
<td><div class="formInput"><label>Cap or Vac:</label>
<select name="pack_req" id="pack_req[]" 
value="**<?php $newProd->packRequirements(1,2,3); ?>**" /></select></div></td>


//function in class
public function packRequirements($values){
$con = $this->connection;
$itms = array($values);

$sql = $con->query("SELECT `prod_option` 
FROM `packaging_requirements` 
WHERE `id`    IN('".implode("', '", $itms)."')"); 

while($r = $sql->fetch_array(MYSQLI_ASSOC)):
echo '<option value="'$itms['prod_option']'"</option>';  
endwhile;

}

Answer 1

您可能希望将您的功能放在选择标签之间，例如：

<select><?php $newProd->packRequirements(1,2,3); ?></select>

代替：

<select value="**<?php $newProd->packRequirements(1,2,3); ?>**"></select>

Answer 2

您的 HTML 似乎是错误的。选择是这样的：

<select name="myname">
    <option value="42">Everything</option>
</select>

但是相反，您的代码会生成

<select name="pack_req" id="pack_req[]" value="**<option value=".

Answer 3

you have to write like this:

<select name='selectName'><?php newProd->packRequirements(array(1,2,3)) ?></select>



public function packRequirements($itms = array()) {
   $con = $this->connection;

   $sql = $con->query("SELECT `prod_option` FROM `packaging_requirements` WHERE `id` IN('" . implode(",", $itms) . "')"); 

  while($r = $sql->fetch_array(MYSQLI_ASSOC)) {
     echo '<option value="' . $r['prod_option'] . '">' . $r['prod_option'] . '</option>';
  }
}

Answer 4

<td><div class="formInput"><label>Cap or Vac:</label>
<select name="pack_req" id="pack_req[]">
**<?php $newProd->packRequirements(1,2,3); ?>**</select></div></td>


//function in class
public function packRequirements($values){
$con = $this->connection;
$itms = array($values);

$sql = $con->query("SELECT `prod_option` 
FROM `packaging_requirements` 
WHERE `id`    IN('".implode("', '", $itms)."')"); 

while($r = $sql->fetch_array(MYSQLI_ASSOC)):
echo '<option value=".$itms['prod_option'].">'.$itms['prod_option'].'</option>';  
endwhile;

}

Answer 5

尝试这个 ：

public function packRequirements($values = array())
{
   $result ='';
   $string =  implode(',', $values);

   $sql = $con->query("SELECT `prod_option` 
                      FROM `packaging_requirements` 
                      WHERE `id`    IN('".$string."')");
   while($r = $sql->fetch_array(MYSQLI_ASSOC)){
        $result .= '<option value=".$r['prod_option'].">'.$r['prod_option'].'</option>';  
   }
 return $result;

}

<select name="pack_req" id="pack_req[]">
**<?php echo $newProd->packRequirements(arra(1,2,3)); ?>**
</select>