我的管理面板和成员面板的登录在本地服务器上都能正常工作，但在Live服务器成员面板上工作不起作用。由于管理员和成员面板都使用相同的连接文件，因此它意味着连接文件正常工作。更多的是当我们填写错误的用户或密码时，它说
无效的用户或密码
但是当我们使用正确的用户或密码登录时，它会返回而不会显示错误。
我的登录文件上层php部分是：
<?php
include_once("../init.php");
$msg='';
?>
<?php
if(isset($_POST['click']))
{
$user = trim($_POST['user']);
$pass = trim($_POST['pass']);
if(($user =='' )|| ($pass=='')){
$msg ='Please enter username & password';
}else{
$npass = ($pass);
$qry = mysql_query("select * from user where user ='$user'");
if(mysql_num_rows($qry)==0) {
$msg ='Invalid UserName';
} else {
$res = mysql_fetch_array($qry);
if($res['pass']==$npass) {
$_SESSION['USE_USER'] = $res['user'];
$_SESSION['SID'] = $res['id'];
$_SESSION['USE_NAME'] = $res['fname'];
$_SESSION['USE_SPONSOR'] = $res['sponsor'];
$_SESSION['PACKAGE_AMT'] = $res['package_amt'];
$_SESSION['ADDRESS'] = $res['address'];
$_SESSION['PHONE'] = $res['phone'];
$_SESSION['JOIN_DATE'] = $res['join_date'];
header('location: main.php');
} else {
$msg ='Invalid Password';
}
}
}
}
?>
我的头文件main.php是
<?php
include_once("../init.php");
validation_check($_SESSION['SID'],MEM_HOME_ADMIN);
$msg='';
$dir ='../'.USER_PIC;
$sId = $_SESSION['SID'];
?>
会话从另一个名为function.php
的文件启动<?php
function logout($destinationPath)
{
if(count($_SESSION))
{
foreach($_SESSION AS $key=>$value)
{
session_unset($_SESSION[$key]);
}
session_destroy();
}
echo "<script language='javaScript' type='text/javascript'>
window.location.href='".$destinationPath."';
</script>";
}
function validation_check($checkingVariable, $destinationPath)
{
if($checkingVariable == '')
{
echo "<script language='javaScript' type='text/javascript'>
window.location.href='".$destinationPath."';
</script>";
}
}
function realStrip($input)
{
return mysql_real_escape_string(stripslashes(trim($input)));
}
function no_of_record($table, $cond)
{
$sql = "SELECT COUNT(*) AS CNT FROM ".$table." WHERE ".$cond;
$qry = mysql_query($sql);
$rec = mysql_fetch_assoc($qry);
$count = $rec['CNT'];
return $count;
}
//drop down
function drop_down($required=null, $text_field, $table_name, $id, $name, $cond, $selected_id=null)
{
$qry = mysql_query("SELECT $id, $name FROM $table_name WHERE $cond ORDER BY $name ASC");
$var = '';
if(mysql_num_rows($qry)>0)
{
$var = '<select id="'.$text_field.'" name="'.$text_field.'" '.$required.'>';
$var .='<option value="">--Choose--</option>';
while($r = mysql_fetch_assoc($qry))
{
$selected = '';
if($selected_id==$r[$id]){
$selected = 'selected="selected"';
}
$var .='<option value="'.$r[$id].'" '.$selected.'>'.$r[$name].'</option>';
}
$var .='</select>';
}
echo $var;
}
function uploadResume($title,$uploaddoc,$txtpropimg)
{
$upload= $uploaddoc;
$filename=$_FILES[$txtpropimg]['name'];
$fileextension=strchr($filename,".");
$photoid=rand();
$newfilename=$title.$photoid.$fileextension;
move_uploaded_file($_FILES[$txtpropimg]['tmp_name'],$upload.$newfilename);
return $newfilename;
}
function fRecord($field, $table, $cond)
{
$fr = mysql_fetch_assoc(mysql_query("SELECT $field FROM $table WHERE $cond"));
return $fr[$field];
}
function get_values_for_keys($mapping, $keys) {
$output_arr = '';
$karr = explode(',',$keys);
foreach($karr as $key) {
$output_arr .= $mapping[$key].', ';
}
$output_arr = rtrim($output_arr, ', ');
return $output_arr;
}
function getBaseURL() {
$isHttps = ((array_key_exists('HTTPS', $_SERVER)
&& $_SERVER['HTTPS']) ||
(array_key_exists('HTTP_X_FORWARDED_PROTO', $_SERVER)
&& $_SERVER['HTTP_X_FORWARDED_PROTO'] == 'https')
);
return 'http' . ($isHttps ? 's' : '') .'://' . $_SERVER['SERVER_NAME'];
}
function request_uri()
{
if ($_SERVER['REQUEST_URI'])
return $_SERVER['REQUEST_URI'];
// IIS with ISAPI_REWRITE
if ($_SERVER['HTTP_X_REWRITE_URL'])
return $_SERVER['HTTP_X_REWRITE_URL'];
$p = $_SERVER['SCRIPT_NAME'];
if ($_SERVER['QUERY_STRING'])
$p .= '?'.$_SERVER['QUERY_STRING'];
return $p;
}
preg_match ('`/'.FOLDER_NAME.'(.*)(.*)$`', request_uri(), $matches);
$tableType = (!empty ($matches[1]) ? ($matches[1]) : '');
$url_array=explode('/',$tableType);
?>
此外，我通过单词和时间创建了用户ID，如LH1450429882，列是verture类型。我认为这对登录没有影响。
我认为主要错误来自function.php很抱歉有很长的代码，但我试图涵盖编码的所有部分。
我正在努力使用这个代码一周。在此先感谢您的帮助。
答案 0 :(得分：0)
通过实现代码ini_set（＆＃39; display_errors＆＃39;，1）; error_reporting（E_ERROR | E_WARNING | E_PARSE）;我在登录php的第6行遇到了标题ploblem的错误我删除了吗？＆gt;和
现在我在login.php中的工作代码是
<?php
include_once("../init.php");
$msg='';
if(isset($_POST['click']))
{
$user = trim($_POST['user']);
$pass = trim($_POST['pass']);
if(($user =='' )|| ($pass=='')){
$msg ='Please enter username & password';
}else{
$npass = ($pass);
$qry = mysql_query("select * from user where user ='$user'");
if(mysql_num_rows($qry)==0) {
$msg ='Invalid UserName';
} else {
$res = mysql_fetch_array($qry);
if($res['pass']==$npass) {
$_SESSION['USE_USER'] = $res['user'];
$_SESSION['SID'] = $res['id'];
$_SESSION['USE_NAME'] = $res['fname'];
$_SESSION['USE_SPONSOR'] = $res['sponsor'];
$_SESSION['PACKAGE_AMT'] = $res['package_amt'];
$_SESSION['ADDRESS'] = $res['address'];
$_SESSION['PHONE'] = $res['phone'];
$_SESSION['JOIN_DATE'] = $res['join_date'];
header('location: main.php');
} else {
$msg ='Invalid Password';
}
}
}
}
?>
答案 1 :(得分：0)
这可能是error_reporting将展示的错误。始终在开发模式下使用它，以捕获一些疏忽错误并确保代码的清晰度。
ini_set('display_errors',1);
error_reporting(E_ERROR | E_WARNING | E_PARSE);