Home >Backend Development >PHP Tutorial >如何实现“当名字在数据库中不存在时，转到错误页面。”？
如何实现“当名字在数据库中不存在时，转到错误页面。”？
- WBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOYWBOriginal
- 2016-06-23 14:04:241274browse
各位朋友：这段代码是“查询密码第一步：输入会员名”，想实现的效果：当名字在数据库中不存在时，转到错误页面。
我这个代码老是出现：Warning: Cannot modify header information - headers already sent by (output started at F:\PHPSites\member\getpwd1-1.php:23) in F:\PHPSites\member\getpwd1-1.php on line 36
应该怎么改呢？非常感谢了！
<?php$totalRows_rsMember = 0;?><?php require_once('Connections/member.php'); ?><?php$colname_rsMember = "-1";if (isset($_POST['txtusername'])) { $colname_rsMember = (get_magic_quotes_gpc()) ? $_POST['txtusername'] : addslashes($_POST['txtusername']);}mysql_select_db($database_member, $member);$query_rsMember = sprintf("SELECT username FROM members WHERE username = '%s'", $colname_rsMember);$rsMember = mysql_query($query_rsMember, $member) or die(mysql_error());$row_rsMember = mysql_fetch_assoc($rsMember);$totalRows_rsMember = mysql_num_rows($rsMember);?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head><meta http-equiv="Content-Type" content="text/html; charset=gb2312" /><title>无标题文档</title><link href="style/style.css" rel="stylesheet" type="text/css" /><script language="javascript">function checkFormData () { if ( form1.txtUsername.value == "" ) { alert ( "用户名不能为空，请重新输入！" ); form1.txtUsername.focus(); return false; }}</script></head><body><?php include('menu.php'); ?><?phpif ($totalRows_rsMember == 0) { header ( "Location: message-1.php?error=3" ); exit;}?><form action="getpwd2-1.php" method="post" name="form1" id="form1" onsubmit="return checkFormData()"> <table width="256" border="1" align="center" rules="none"> <tr> <th colspan="2" bgcolor="#0066CC" class="title">第一步：输入会员名</th> </tr> <tr> <td width="72" height="32" align="right">用户名：</td> <td width="168" height="32"><input name="txtUsername" type="text" id="txtUsername" /></td> </tr> <tr> <td height="32" colspan="2" align="center"><input name="btnNext" type="submit" id="btnNext" value="下一步" /></td> </tr> </table></form></body></html><?phpmysql_free_result($rsMember);?>
回复讨论(解决方案)
在 getpwd1-1.php 的 23 行处有输出
导致 header ( "Location: message-1.php?error=3" ); 出错
先检查那个输出是否确实必要？是否能调整到别处
实在不行的话，可在 require_once('Connections/member.php'); 前加上
ob_start();
来回避
在 getpwd1-1.php 的 23 行处有输出
导致 header ( "Location: message-1.php?error=3" ); 出错
先检查那个输出是否确实必要？是否能调整到别处
实在不行的话，可在 require_once('Connections/member.php'); 前加上
ob_start();
来回避
……
版主大人：我贴上来的就是getpwd1-1.php中的所有代码，具体要怎么调整？
如果在 require_once('Connections/member.php'); 前加上ob_start();的话，就直接转到错误页面了，而不是“输入不存在的用户名再转到错误页面”了。请指点，非常感谢！
粗心了没核实文件名
把 16 到 33 行（含33行）的内容移到 39 行的后面
另外 10 到 14 行的内容应当在 7 行的条件分支里
粗心了没核实文件名
把 16 到 33 行（含33行）的内容移到 39 行的后面
另外 10 到 14 行的内容应当在 7 行的条件分支里
版主大人：我照着您指点的改了，如下，但还是直接就转到错误页面啊！
<?php$totalRows_rsMember = 0;?><?php require_once('Connections/member.php'); ?><?php$colname_rsMember = "-1";if (isset($_POST['txtusername'])) { $colname_rsMember = (get_magic_quotes_gpc()) ? $_POST['txtusername'] : addslashes($_POST['txtusername']); mysql_select_db($database_member, $member);$query_rsMember = sprintf("SELECT username FROM members WHERE username = '%s'", $colname_rsMember);$rsMember = mysql_query($query_rsMember, $member) or die(mysql_error());$row_rsMember = mysql_fetch_assoc($rsMember);$totalRows_rsMember = mysql_num_rows($rsMember);}?><?phpif ($totalRows_rsMember == 0) { header ( "Location: message-1.php?error=3" ); exit;}?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head><meta http-equiv="Content-Type" content="text/html; charset=gb2312" /><title>无标题文档</title><link href="style/style.css" rel="stylesheet" type="text/css" /><script language="javascript">function checkFormData () { if ( form1.txtUsername.value == "" ) { alert ( "用户名不能为空，请重新输入！" ); form1.txtUsername.focus(); return false; }}</script></head><body><?php include('menu.php'); ?><form action="getpwd2-1.php" method="post" name="form1" id="form1" onsubmit="return checkFormData()"> <table width="256" border="1" align="center" rules="none"> <tr> <th colspan="2" bgcolor="#0066CC" class="title">第一步：输入会员名</th> </tr> <tr> <td width="72" height="32" align="right">用户名：</td> <td width="168" height="32"><input name="txtUsername" type="text" id="txtUsername" /></td> </tr> <tr> <td height="32" colspan="2" align="center"><input name="btnNext" type="submit" id="btnNext" value="下一步" /></td> </tr> </table></form></body></html><?phpmysql_free_result($rsMember);?> 5 到 21 行改成
<?php$colname_rsMember = "-1";if (isset($_POST['txtusername'])) { $colname_rsMember = (get_magic_quotes_gpc()) ? $_POST['txtusername'] : addslashes($_POST['txtusername']); mysql_select_db($database_member, $member); $query_rsMember = sprintf("SELECT username FROM members WHERE username = '%s'", $colname_rsMember); $rsMember = mysql_query($query_rsMember, $member) or die(mysql_error()); $row_rsMember = mysql_fetch_assoc($rsMember); $totalRows_rsMember = mysql_num_rows($rsMember); if ($totalRows_rsMember == 0) { header ( "Location: message-1.php?error=3" ); exit; }}?>自己想一下，问什么要这样写？ 5 到 21 行改成
PHP code?123456789101112131415
原来可以这样，终于把这个逻辑理清楚了，多谢版主大人！