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我的守则预示着我随机的神秘怀疑。 帮助?
原标题:My code isn t outputting my random mysql query. Help?
  • 时间:2011-04-02 17:27:39
  •  标签:
  • php
  • mysql
  • sql

该法典完全是自我解释的。 我试图利用 s取随机记录及其价值。 然而,当我上载时,我从来文方的法典来看,所有我都是这样。

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

Here s my code:

<?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> ;
}
?>

感谢!

最佳回答

第一次审议

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

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

$rand will be number between 1 and 10... You need 0..9 for using in MySQL limit. So if $rand hits 10, you ll got nothing because your last row is ..LIMIT 9,1, not LIMIT 10,1.

引言

$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> ;
  }
问题回答
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> ;
}
?>




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