Question

由于这个错误我感到很沮丧。当我在不使用任何框架的情况下编程php时，我的代码是可以的，但是当我在codeigniter中传输它时，它总是说严重性未定义的函数。请检查我发布的代码：

foreach($data['employee'] as $val)
    {
        if($val['Active']==0)
        {
            $status = @strtoupper('Resigned');
        }
        else
        {
            $status = @strtoupper('Active');
        }

        $str = @substr($val['WS_Code'],0,-2);
        if($str == "PR")
        {
            $line = @substr($val['WS_Code'],0,-1);
            $str2 = @substr($val['WS_Code'],-1);
            if($str2 == 'B')
            {
                $section = "BRIEF";
            }
            else
            {
                $section = "PANTY";
            }

        }
        elseif($str == "SP")
        {
            $line = @substr($val['WS_Code'],0,-1);
            $str2 = @substr($val['WS_Code'],-1);
            if($str2 == "B")
            {
                $section = "BRIEF";
            }
            else
            {
                $section = "PANTY";
            }
        }
        elseif($str == "BL")
        {
            $section = "BRIEF";
            $str2 = @substr($str,-3,2);
            $line = $str2;
        }
        elseif($str == "PL")
        {
            $section = "PANTY";
            $str2 = @substr($str,-3,2);
            $line = $str2;
        }

以下是代码的其余部分以及我如何使用它：

<tr>
   <td><label class='control-label'>Section:</label></td>
   <td>
      <select class='form-control' name='cmbsection' id='cmbsection'>
         <option value='".$section."'>".$section."</option>
         ";
         $section_code = array();
         $section_code[] = $section;
         foreach($data['sectionlist'] as $value)
         {
         if(!in_array($value['sectionName'],$section_code))
         {
         echo "<option value='".$value['sectionName']."'>".$value['sectionName']."</option>";
         }
         }
         echo "
      </select>
   </td>
   <td><label class='control-label'>Line:</label></td>
   <td>
      <select class='form-control' name='cmbline' id='cmbline'>
         ";
         $line_code = array();
         $line_code[] = $line;
         echo "
         <option value='$line'>$line</option>
         ";
         foreach($data['linelist'] as $value)
         {
         if(!in_array($value['lineName'],$line_code))
         {
         echo "<option value='".$value['lineName']."'>".$value['lineName']."</option>";
         }
         }
         echo "
      </select>
   </td>
</tr>

请注意，我正在回显控制器内的表格，错误始终指向$section代码中的$line和if(!in_array($value['lineName'],$line_code))。

Answer 1

我不会称这是一个很好的解决方案，但它至少会确保变量$line和$section存在。

您的问题是您在WS_Code获得了一个您没有处理的代码。如果您只看到已编码的代码，则只创建2个变量，您必须拥有一个您没有预期和编码的新代码。

正确的解决方案是找出您在此代码中未获得的WS_Code内容。

事实上，正确的解决方案是创建一个包含code和full_desc的表格，您可以查询这些表格，以便将这些代码扩展为您想要使用的完整描述，并确保当创建新代码时，表格会相应更新

$line = 'new unexpected code recieved';
$section = 'new unexpected section recieved';

foreach($data['employee'] as $val)
    {
        if($val['Active']==0) {
            $status = strtoupper('Resigned');
        } else {
            $status = strtoupper('Active');
        }

        $str = substr($val['WS_Code'],0,-2);
        if($str == "PR") {
            $line = substr($val['WS_Code'],0,-1);
            $str2 = substr($val['WS_Code'],-1);
            if($str2 == 'B') {
                $section = "BRIEF";
            } else {
                $section = "PANTY";
            }

        }
        elseif($str == "SP") {
            $line = substr($val['WS_Code'],0,-1);
            $str2 = substr($val['WS_Code'],-1);
            if($str2 == "B") {
                $section = "BRIEF";
            } else {
                $section = "PANTY";
            }
        }
        elseif($str == "BL") {
            $section = "BRIEF";
            $str2 = substr($str,-3,2);
            $line = $str2;
        }
        elseif($str == "PL") {
            $section = "PANTY";
            $str2 = substr($str,-3,2);
            $line = $str2;
        }

Answer 2

请使用以下代码：

if(isset($line)){
   echo "<option value='$line'>$line</option>";
}

foreach($data['linelist'] as $value){
    if(isset($value['lineName']) && isset($line_code)) {
        if (!in_array($value['lineName'], $line_code)) {
            echo "<option value='" . $value['lineName'] . "'>" . $value['lineName'] . "</option>";
       }
   }
}

Codeigniter严重性：注意 - ＆gt;未定义的变量

2 个答案: