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);
?>

所以网站显示一个新闻栏，代码从数据库中获取新闻数据，如果有一个photoid（如果新闻有照片），那么它会将新闻添加到带有照片的表格中。如果没有照片，则在没有照片的情况下将新闻添加到桌面。简单。在测试的那一刻，我有两篇带有照片的新闻文章，第一篇完美地工作，然后是第二次错误

Undefined index: photoid in

对于行if ($row['photoid'] == NULL){。有一个photoid。

Answer 1

在检查值之前，您需要检查是否设置了变量/键。试试这个：

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

mysql结果的未定义索引

1 个答案: