我想编写一个登录脚本,用户输入电子邮件ID和密码,如果电子邮件和密码都正确,则会转到另一个页面.如果电子邮件和密码是正确的; 值存储在DB中.
这是整个纠正的工作代码:
<?php
if($_POST['submit']){
$email = protect($_POST['email']);
$password = protect($_POST['password']);
$md5password=MD5($password);
if(!$email || !$password){
echo '<span style="color: red;" /><center>You need to fill in your <b>User Name</b> and <b>Password</b>!</center></span>';
}else{
$res = mysql_query("SELECT * FROM `employer` WHERE `email` = '".$email."'");
$num = mysql_num_rows($res);
if($num == 0){
echo '<span style="color: red;" /><center>The <b>E Mail ID</b> you supplied does not exist!</center></span>';
}else{
$res = mysql_query("SELECT * FROM `employer` WHERE `email` = '".$email."' AND `password` = '".$md5password."'");
$num = mysql_num_rows($res);
if($num == 0){ …我正在运行以下代码,但它给了我以下错误.我读了几篇关于SO的文章,仍然没有用.我得到的错误是:
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/xyz/public_html/13/beta/signup.php on line 49
Warning: Cannot modify header information - headers already sent by (output started at /home/xyz/public_html/13/beta/signup.php:49) in /home/xyz/public_html/13/beta/signup.php on line 69
我的PHP剧本:(我已标记线条49&69)
if(isset($_POST['submit'])&&$_POST['submit']=='Login')
{
$err = array();
if(!$_POST['username'] || !$_POST['pass'])
$err[] = 'All the fields must be filled in!';
if(!count($err))
{
$_POST['username'] = mysql_real_escape_string($_POST['username']);
$_POST['pass'] = mysql_real_escape_string($_POST['pass']);
$_POST['remember'] = (int)$_POST['remember'];
$row = mysql_fetch_assoc(mysql_query("SELECT id,user,email,clg FROM users WHERE user='{$_POST['username']}' AND …我的c++代码中的一行是:
cout<<(i%3==0 ? "Hello\n" : i) ;//where `i` is an integer.
但我得到这个错误:
operands to ?: have different types 'const char*' and 'int
如何修改代码(最少字符)？
我有一个表viewer有id,ip,date_last_viewed和blog_id作为列.我首先检查具有相同IP和blog_id的特定条目是否存在.如果是,则更新日期.否则,它会插入一个新条目.
我的代码如下:
$search_ip = mysql_query("SELECT ip FROM viewer WHERE ip = '".$_SERVER['REMOTE_ADDR']."' AND blog_id= '".$b_id."' ");
if ($search_ip == false){
$insert_ip = mysql_query("INSERT INTO viewer (ip, blog_id, date_last_viewed) VALUES ('".$_SERVER['REMOTE_ADDR']."', '".$b_id."', NOW())");
}
else {
$update_ip = mysql_query("UPDATE viewer SET date_last_viewed = NOW() WHERE ip = '".$_SERVER['REMOTE_ADDR']."' AND blog_id='".$b_id."' ");
}
该表没有插入任何内容.我在这做错了什么？另外,由于我是PHP编程的新手,有人可以告诉我如何将上述代码修改为PDO吗？
这是我的代码:
for($i=1;$i<=100;$i++){
if($i%15==0) print "Divisible by 15";
else if($i%5==0) print "Divisible by 5";
else print ($i%3==0)? "Divisible by 3":$i;
print "\n";
}
它是一个非常简单的代码.我使用Java工作,虽然它在Perl中出错.错误是:
syntax error at line 2, near ") print"
Execution aborted due to compilation errors.
我是Perl的新手.我怎样才能让它运转起来？