未定義的索引爲mysql結果

Question

<?php 


$result = mysqli_query($conn,"SELECT * FROM news ORDER BY date DESC 
LIMIT 5")or die(mysql_error()); 



if (!$result) { 
    printf("Error: %s\n", mysqli_error($conn)); 
    exit(); 
} 
echo "<table align='left' CELLPADDING='50' >"; 

while($row = mysqli_fetch_array($result)) 
{ 

    if ($row['photoid'] == NULL){ 

$date_string = $row['date']; 

$date = strtotime($date_string); 
$date = date('m/d/y', $date); 

echo "<tr>"; 
echo "<td style='width: 750px'>" .$row['title'] . "</td>"; 
echo "</tr>"; 
echo "<tr>"; 
echo "<td style='width: 700px'>" . $row['news'] . "</td>"; 
echo "<td style='width: 100px'>" . $date . "</td>"; 
echo "</tr>"; 
    } 
    else 
    { 
     $result = mysqli_query($conn,"SELECT news.title,news.date,news.news,photo.photo FROM news, photo WHERE news.photoid = photo.photoid ORDER BY date DESC") or die(mysql_error()); 
     $row = mysqli_fetch_array($result); 
     $date_string = $row['date']; 

$date = strtotime($date_string); 
$date = date('m/d/y', $date); 

echo "<tr>"; 
echo "<td style='width: 750px'>" . $row['title'] . "</td>"; 
echo "</tr>"; 
echo "<tr>"; 
echo "<td style='width: 700px'>" . $row['news'] . "</td>"; 
echo "<td style='width: 100px'>" . $date . "</td>"; 
echo "</tr>"; 
echo "<td style='width: 800px'>" . '<img height="300" width="300" src="data:image/jpeg;base64,'.base64_encode($row["photo"]).'" >' . "</td>"; 
    } 
} 
echo "</table>"; 

mysqli_close($conn); 
?> 

因此，網站顯示新聞欄，代碼從數據庫中獲取新聞數據，如果有照片（如果新聞有照片去），然後它添加消息與桌上的照片。如果沒有照片將消息添加到沒有人的表格中。簡單。目前，用於測試我有兩個新聞文章都與照片，第一個完美的作品，則第二錯誤

Undefined index: photoid in 

對於線if ($row['photoid'] == NULL){。有一張照片。

Answer 1

在檢查值之前，您需要檢查是否設置了變量/鍵。試試這個：

 
if (isset($row['photoid']) && $row['photoid'] == NULL){