警告:第82行/opt/lampp/htdocs/linom/index.php中的字符串偏移量'slimg‘非法
<?php
$slider = "SELECT * FROM post ORDER BY id ASC LIMIT 4";
$slider1 = mysqli_query($connect,$slider);
$sliderV = mysqli_fetch_array($slider1);
do {
printf ("<div class='slider'>
我正在尝试有一个登录脚本burt我有这个错误未定义的变量:_session请参阅下面的页面
//checklogin.php
<?php
ob_start();
$host="localhost"; // Host name
$username="xxxxx"; // Mysql username
$password="xxxx"; // Mysql password
$db_name="test"; // Database name
$tbl_name="members"; // Table na
如何在PHP中访问单独类中的静态变量?作用域解析操作符是不是不适合这项工作?示例:
class DB {
static $conn = 'Connection';
}
class User {
function __construct() {
DB::conn; //throws "Undefined class constant 'conn' error.
}
}
尝试填充列表框中的字段时,第23行出现错误未定义的变量$alltext。已经检查了数据库行,它们是匹配的。请帮帮忙。提前谢谢。
<?php
include 'db.inc.php';
$sql = "SELECT supplierID,suppliername,street,town,county,telephone,balance FROM Suppliers";
if(!$result = mysql_query($sql, $con))
{
die("Error in querying the database".
我很好奇如何获得包含空格或特殊字符的变量。
有办法吗?
<?php
$first_day = "monday";
$$first_day = "first day of the week";
echo $monday." <br />";
$days = 'saturday sunday';
$$days = 'days of the weekend';
// how to echo $days of the weekend ???
print_r(get_defined_vars());