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echo mysql_query($s);无任何值输出解决方法

WBOY
WBOYOriginal
2016-06-13 10:17:243197browse

echo mysql_query($s);无任何值输出
  $query="select * from user";
  $results=mysql_query($query);
  while(@$r=mysql_fetch_array($results)){ 
  echo htmtocode($r['userid']);
  echo " "?>
  delete
 

    }
  ?>
    if(@$_GET['action']=='delete'){
  $s="select from user where userid=$_GET[userid]";
  echo mysql_query($s);
  die();
  if(mysql_query($s)){
 
  exit(')');
  }
  }
  ?>

------解决方案--------------------

PHP code
<?php $query="select * from user";  $results=mysql_query($query);  if( ! $results)      die("SQL: {$query} error: " . mysql_error() );  while( $r=mysql_fetch_array($results) )  {        echo htmtocode($r['userid']);      echo " ";  ?>      <a href="?action=delete&userid=<php%20echo%20%24r%5B'userid'%5D;%20?>">delete</a>      <br>  <?php }  ?>  <?php if($_GET['action']=='delete'){     if( ! isset($_GET['userid']) )         exit('userid 为空,放弃操作');     $userid = trim( $_GET['userid'] );     // 我不明白删除操作用select做什么?     $s="select * from `user` where userid = '{$userid}'";     // echo 只对字符串和数字有效,调试的话用var_dump()     var_dump( $result = mysql_query($s) );      if( $result ){          exit('<script language="javascript">                alert("删除成功!");                self.location = "admin.php";                ');     }else         echo "mysql error:". mysql_error();  }?><br><font color="#e78608">------解决方案--------------------</font><br>if(@$_GET['action']=='delete'){<br>$s="select from user where userid=$_GET[userid]";<br>echo mysql_query($s);<br>die();<br>你写的这三句<br>改为这样<br>if(@$_GET['action']=='delete'){//确认这个条件满足,进入了循环<br>$s="select from user where userid=$_GET[userid]";//<br>echo $s."<br>";//输出语句内容,也查一下运行到此步没有,可以单独到数据库里查一下看有值没<br>echo mysql_query($s);//query返回的只是成功不成功并不是查得的内容<br>die();//语句到此句就中止了<br>要注意你的userid是不是int型,若是text或是char啥的两边需加引号''<br>到此你定会知道为啥没有输出内容了<br><font color="#e78608">------解决方案--------------------</font><br>echo 只能输出字符串和数字,<br>mysql_query();返回值是资源类型,当然输不出来<br>用var_dump就可以输出了 ,要看结果要使用函数处理返回的资源<br><font color="#e78608">------解决方案--------------------</font><br>返回false表示你的SQL语句有问题,没成功执行<br>给后面加上一句 echo mysql_error();<div class="clear">
                 
              
              
        
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