这个是主页代码
<!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=gbk" />
<title>无标题文档</title>
</head>
<script language="javascript" type="text/javascript">
var xmlHttp;
function S_xmlhttprequest(){
if(window.ActiveXObject){
xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
}else if(window.XMLHttpRequest){
xmlHttp = new XMLHttpRequest();
}
}
function checktxtUserName()
{var txtUserName = document.form1.txtUserName.value;
var tip = document.getElementById("txtUserNameTip");
var imgwrong="<img src='img/wrong.gif' align='absmiddle' hspace='2'>"
if(txtUserName.length < 2 || txtUserName.length >30)
{
tip.innerHTML =" " + imgwrong + "用户名长度不能少于1个字或大于15个字";
//tip.style.color="#000000";
}
else
{
var f=document.form1.txtUserName.value;
S_xmlhttprequest();
xmlHttp.open("GET","for.php?id="+f/*escape(f)*/,true);
xmlHttp.onreadystatechange = checkpage;
xmlHttp.send(null);
}
}
function checkpage(){
if(xmlHttp.readyState == 4){
if(xmlHttp.status == 200){
var bychk = xmlHttp.responseText;
document.getElementById('txtUserNameTip').innerHTML = bychk;
}
}
}
</script>
<body>
<form name="form1" action="" method="post" enctype="text/plain">
用户名:
<input type="text" name="txtUserName" onblur="checktxtUserName()"/>
<div id="txtUserNameTip"></div>
</form>
</body>
</html>这是for.php代码
<?php
if($_GET[id]){
$conn=mysql_connect('localhost','root','');
mysql_select_db('user',$conn);
mysql_query("set names 'gbk'");
$sql="SELECT * from user where 'txtUserName'='$_GET[id]'";
$q=mysql_query($sql);
//$result = mysql_query("select count(*) from user");
//$row = mysql_num_rows($result);
if(is_array(mysql_fetch_row($q))){
echo "wrong";
}else
{
echo"ok";
}
}
?>为什么无论输入的是否和数据库里的重复都显示OK
而且如果把wrong,ok改成重名,可以注册,输出部分就显示乱码
<!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=gbk" />
<title>无标题文档</title>
</head>
<script language="javascript" type="text/javascript">
var xmlHttp;
function S_xmlhttprequest(){
if(window.ActiveXObject){
xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
}else if(window.XMLHttpRequest){
xmlHttp = new XMLHttpRequest();
}
}
function checktxtUserName()
{var txtUserName = document.form1.txtUserName.value;
var tip = document.getElementById("txtUserNameTip");
var imgwrong="<img src='img/wrong.gif' align='absmiddle' hspace='2'>"
if(txtUserName.length < 2 || txtUserName.length >30)
{
tip.innerHTML =" " + imgwrong + "用户名长度不能少于1个字或大于15个字";
//tip.style.color="#000000";
}
else
{
var f=document.form1.txtUserName.value;
S_xmlhttprequest();
xmlHttp.open("GET","for.php?id="+f/*escape(f)*/,true);
xmlHttp.onreadystatechange = checkpage;
xmlHttp.send(null);
}
}
function checkpage(){
if(xmlHttp.readyState == 4){
if(xmlHttp.status == 200){
var bychk = xmlHttp.responseText;
document.getElementById('txtUserNameTip').innerHTML = bychk;
}
}
}
</script>
<body>
<form name="form1" action="" method="post" enctype="text/plain">
用户名:
<input type="text" name="txtUserName" onblur="checktxtUserName()"/>
<div id="txtUserNameTip"></div>
</form>
</body>
</html>这是for.php代码
<?php
if($_GET[id]){
$conn=mysql_connect('localhost','root','');
mysql_select_db('user',$conn);
mysql_query("set names 'gbk'");
$sql="SELECT * from user where 'txtUserName'='$_GET[id]'";
$q=mysql_query($sql);
//$result = mysql_query("select count(*) from user");
//$row = mysql_num_rows($result);
if(is_array(mysql_fetch_row($q))){
echo "wrong";
}else
{
echo"ok";
}
}
?>为什么无论输入的是否和数据库里的重复都显示OK
而且如果把wrong,ok改成重名,可以注册,输出部分就显示乱码
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改成:
$sql="SELECT * from user where 'txtUserName'='".$_GET[id]".'";就可以了。
$sql="SELECT * from user where txtUserName='".$_GET[id]".'";调试操作数据库语句错误
最好的办法 是echo
$sql="SELECT * from user where 'txtUserName'='$_GET[id]'";//数据库的字段名不要+引号//去掉即可
$sql="SELECT * from user where txtUserName ='$_GET[id]'";