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我無法讓我的代碼正常工作。當用戶鍵入用戶名時,它會檢查數據庫以查看它是否存在,以及是否將<P id="checkusername">
設置爲「用戶名已接收」。如果沒有將其設置爲「未採取」。我看不出爲什麼我的代碼無法正常工作。Ajax檢查用戶名是否存在
1 register.html
<html>
<body>
<CENTER>
<form name="register" action="register.php" method="post">
Username: <input type="text" name="username" onkeyup="checkUsername(this.value)" />
<P id="checkusername">checker</P>
<input type="submit" value="Login" />
</form>
</CENTER>
<script type="text/javascript">
function checkUsername(){
var xmlhttp;
var username=document.forms["register"]["username"].value;
if(username.length==0){
document.getElementById("checkusername").innerHTML="Empty";
return;
}
if(window.XMLHttpRequest){
xmlhttp=new XMLHttpRequest();
}else{
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
var url = "login.php";
var params = "header=checkusername&username="+username+"&password="";
xmlhttp.open("POST", url, true);
xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xmlhttp.setRequestHeader("Content-length", params.length);
xmlhttp.setRequestHeader("Connection", "close");
xmlhttp.onreadystatechange = function() {//Call a function when the state changes.
if(xmlhttp.readyState == 4 && xmlhttp.status == 200){
document.getElementById("checkusername").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.send(params);
}
</script>
</body>
</html>
2 register.php
<?php
$header = $_POST["header"];
if(header=="checkusername"){
checkusername($_POST["username"]);
}else{
echo "No match: " . $header;
}
function connection(){
$con = mysql_connect(URL, username, password);
if(!$con){
die('Could not connect: ' . mysql_error());
}
return $con;
}
function checkusername($username){
$con = connection();
mysql_select_db(database, $con);
$result = mysql_query("SELECT * FROM users WHERE username = \"" . $username . "\";");
while($row = mysql_fetch_array($result)){
if(($row['username'] == $username)){
echo "Username Taken.<br/>";
return;
}
}
echo "Not Taken.";
}
mysql_close();
?>
限制你的問題,以更小的代碼示例。只有相關的部分。 –
在你的while循環php文件,你總是輸出「沒有采取」,它應該在IF條件下..加..爲什麼你需要循環查看用戶名是否匹配..你的SQL已經做到了..只是看看如果計數是0它沒有采取。採取其 – Kamal
SQL注入發現 – pomeh