Question

代码非常自我解释。我正在尝试使用sql来检索随机记录及其值。但是，当我上传这个时，我从源代码中得到的只是

<a href="/.php">
            <div class="image">
              <img src="https://s3.amazonaws.com/images/" 
                   alt="" 
              />
            </div>
          </a>

这是我的代码：

<?php
// Get the number of rows in the table
$count = mysql_fetch_assoc(mysql_query('SELECT COUNT(thumbnailID) FROM images'));
// Use those to generate a random number
$count = floatval($count);
$rand = rand(1,$count);
// Select the two columns we need, and use limit to set the boundaries
$query = 'SELECT link, pic, alt FROM images LIMIT '.$rand.',1';
// Run the query
if(($result = mysql_query($query)) !== FALSE) {
    // Dump the result into two variables
    list($link, $pic, $alt) = mysql_fetch_assoc($result);
    // Echo out the result
          echo '
          <a href="/' . $link . '.php">
            <div class="image">
              <img src="https://s3.amazonaws.com/images/' . $pic . '" 
                   alt="' . $alt . '" 
              />
            </div>
          </a>';
}
?>

谢谢！

Answer 1

SELECT link, pic, alt FROM images ORDER BY RAND() LIMIT 1;

简化代码：

<?php

$query = 'SELECT link, pic, alt FROM images ORDER BY RAND() LIMIT 1;';
// Run the query
if(($result = mysql_query($query)) !== FALSE) {
 list($link, $pic, $alt) = mysql_fetch_assoc($result);
          echo '
          <a href="/' . $link . '.php">
            <div class="image">
              <img src="https://s3.amazonaws.com/images/' . $pic . '" 
                   alt="' . $alt . '" 
              />
            </div>
          </a>';
}
?>

Answer 2

首先考虑

$count = 10; // for example: number of rows found

$count = floatval($count);
$rand = rand(1,$count);

$rand的数字介于1到10之间 ...您需要 0..9 才能在MySQL限制中使用。因此，如果 $ rand点击10 ，您将无法获得任何结果，因为您的上一行是..LIMIT 9,1，而不是LIMIT 10,1。

试试这个

$sql = 'SELECT link, pic, alt FROM images ORDER BY RAND() LIMIT 1;';
$res = mysql_query($sql);
if ($row = mysql_fetch_array($res)) {
  echo '<a href="/' . $row['link'] . '.php">
          <div class="image">
            <img src="https://s3.amazonaws.com/images/' . $row['pic'] . '" alt="' . $row['alt'] . '" />
          </div>
        </a>';
  }

我的代码没有输出我的随机mysql查询。救命？

2 个答案: