Question

function select_History_Date($link){
    $status = '';
    $sql = "SELECT Student_Id, Loaner_Laptop_Id, HDD_Id, "
        . "Phone_Id, date_returned " 
        . "FROM equipment_history WHERE Loaner_Laptop_Id "
        . "IS NOT NULL OR HDD_Id IS NOT NULL OR Phone_Id IS NOT NULL";
   $ret = mysqli_query($link, $sql);
   $row1 = mysqli_fetch_array($ret, MYSQLI_ASSOC);    
     if (!$ret) {
      die('Could not execute select statement:' . mysqli_errno($link));
     } else {
     while ($row1 = mysqli_fetch_array($ret, MYSQLI_ASSOC)) {
     $status .= '<tr>';
     if(is_Null($row1)){
        $status .= '<td><a href="ReturnLaptop.php?laptopId=' . 
                 $row1['Loaner_Laptops_Id'] . 'studentId='. 
                 $row1['Student_Id']. '">'. 'Return'. '</td>';
         } else{
        $status .= '<td>' . $row1['date_returned'] . '</td>';
        }            
        }
        $status .= '</tr>';
        return $status;
 }
}

我如何设想它应该是打印href链接，但它只是打印空白我错过了什么？或者这根本不可能做到。

Answer 1

您正在检查$row1是否为null，但它永远不会是null，因为您要在while中进行分配。如果您在while，那已经意味着它不是null。

我敢打赌你错过了if(is_null($row1))的索引，例如if(is_null($row1['value_to_check']))。

Answer 2

你来自db Loaner_Laptop_Id

$sql = "SELECT Student_Id, Loaner_Laptop_Id, HDD_Id, "
        . "Phone_Id, date_returned " 
        . "FROM equipment_history WHERE Loaner_Laptop_Id "
        . "IS NOT NULL OR HDD_Id IS NOT NULL OR Phone_Id IS NOT NULL";

使用$ row1 ['Loaner_Laptop s _ Id']检查密钥：）

没有显示我的参考链接

2 个答案: