Question

 $address_query = db_query('......');    
 $address1 ='';
 $address2 ='';
 $address3 ='';
 $address4 ='';
 $address5 ='';
 $state = '';
 $pincode ='';
 $addcategory ='';
 $residence ='';
 $adddaterpt ='';

 foreach($address_query as $adds) 
 {
$address1 .= $adds->addressline1;
$address2 .= $adds->addressline2;
$address3 .= $adds->addressline3;
$address4 .= $adds->addressline4;
$address5 .= $adds->addressline5;
$state .= $adds->state_name;
$pincode .= $adds->pincode;
$addcategory .= $adds->addcategory_name;
$residence .= $adds->residence_name;
$adddaterpt .= $adds->date_reported;
$addrptdate = substr($adddaterpt, 0, 2) . "-" . substr($adddaterpt, 2, 2) . "-" . substr($adddaterpt, 4, 4);


   $report.='<tr>
   <table id="address1">
             <tr><b>ADDRESS:</b>' . $address1 . $address2 . $address3 . $address4 . $address5 . $state . $pincode . '</tr>
             <tr><td><b>CATEGORY:</b>' . $addcategory . '</td> 
                 <td><b>RESIDENCE CODE:</b>' . $residence . '</td>
                 <td><b>DATE REPORTED:</b>' . $addrptdate . '</td>
             </tr>
         </table>
         </tr>';
   }
   $report.='</table>';

执行此代码时，我得到了foreach（）的最后一个值。其他值不会显示给我。

它在$ address_query中包含4行，但最后一行只显示。

请建议我，我的代码中有任何错误。

Answer 1

你好朋友实际上你不能在tr标签中使用表标签，并且你没有声明直接用于for-each语句的变量$ report。

试试此代码

$address_query = db_query('......');    
$adddaterpt ='';
$report = '<table>';
foreach($address_query as $adds) 
{
    $adddaterpt .= $adds->date_reported;
    $addrptdate = substr($adddaterpt, 0, 2) . "-" . substr($adddaterpt, 2, 2) . "-" . substr($adddaterpt, 4, 4);
    $report.='<tbody id="address1">
         <tr><td><b>ADDRESS:</b>' . $adds->addressline1 . $adds->addressline2 . $adds->addressline3 . $adds->addressline4 . $adds->addressline5 . $adds->state_name . $adds->pincode . '</td></tr>
         <tr><td><b>CATEGORY:</b>' . $adds->addcategory_name . '</td> 
             <td><b>RESIDENCE CODE:</b>' . $adds->residence_name . '</td>
             <td><b>DATE REPORTED:</b>' . $addrptdate . '</td>
         </tr>
     </tbody>';
}
$report.='</table>';

Answer 2

我并非100％确定我理解这个问题，但问题可能在于您的连接问题。

在当前代码中，每个循环继续添加到以下变量中。

    $address1
    $address2
    $address3
    $address4
    $address5
    $state
    $pincode
    $addcategory 
    $residence
    $adddaterpt

因此，在循环结束时，每个变量都堆叠有来自每个先前迭代的信息。

您需要为每个循环重置变量，或者如下例所示，完全删除连接。所以：

foreach($address_query as $adds) {
    $address1    = $adds->addressline1;
    $address2    = $adds->addressline2;
    $address3    = $adds->addressline3;
    $address4    = $adds->addressline4;
    $address5    = $adds->addressline5;
    $state       = $adds->state_name;
    $pincode     = $adds->pincode;
    $addcategory = $adds->addcategory_name;
    $residence   = $adds->residence_name;
    $adddaterpt  = $adds->date_reported;
    $addrptdate  = substr($adddaterpt, 0, 2) . "-" . substr($adddaterpt, 2, 2) . "-" . substr($adddaterpt, 4, 4);
}

希望这有帮助！如果没有，请更清楚你的问题。：）

Answer 3

you can do it dirctly as well why u r storing the data in variable just directly use it you are in foreach loop already...    
foreach($address_query as $adds) 
 {
   $report.='<tr>
   <table id="address1">
          <tr><b>ADDRESS:</b>' . $adds->addressline1. $adds->addressline2 .    $adds->addressline3 . $adds->addressline4 . $adds->addressline5 . $adds->state_name . $adds->pincode . '</tr>
          </table>
     </tr>';

}

在foreach最后一个值单独显示在输出中，其他值不显示在输出中？

3 个答案: