这是错误代码：
Cannot send session cookie - headers already sent by (output started at
C:\xampp\htdocs\log\New folder (4)\New folder (2)\iskono\forum.php:5) in
C:\xampp\htdocs\log\New folder (4)\New folder (2)\iskono\forum.php on line 244
Warning: session_start() [function.session-start]: Cannot send session cache limiter -
headers already sent (output started at C:\xampp\htdocs\log\New folder (4)\New folder
(2)\iskono\forum.php:5) in C:\xampp\htdocs\log\New folder (4)\New folder
(2)\iskono\forum.php on line 244
第244行是需要的代码
<? require("forum/index.php"); ?>
如何解决这个错误？
我有我的工作,我有一个表仅本地项目id,title以及price多个领域.
示例信息:
ID || Title || Price
1 - Title 1 - 8.00
2 - Title 2 - 75.00
3 - Title 3 - 70.00
当我尝试ORDER BY price它回来像这样:
8.00
75.00
70.00
声明:
$query = mysql_query("Select * From table ORDER BY price DESC");
我究竟做错了什么？
现在,我将以下代码放入header.php.
我认为解决方案不是很优雅.
如何将这个CSS代码添加functions.php到我的标题中(该代码将如何显示)？
wp_head();
?>
<style>
.jimgMenu ul li.landscapes a {
background: url(<?php bloginfo('template_directory'); ?>/images/<?php echo get_option(THEME_PREFIX . 'intro_image'); ?>) repeat scroll 0%;
}
.jimgMenu ul li.people a {
background: url(<?php bloginfo('template_directory'); ?>/images/<?php echo get_option(THEME_PREFIX . 'slider_image'); ?>) repeat scroll 0%;
}
.jimgMenu ul li.nature a {
background: url(<?php bloginfo('template_directory'); ?>/images/nature.jpg) repeat scroll 0%;
}
.jimgMenu ul li.abstract a {
background: url(<?php bloginfo('template_directory'); ?>/images/abstract.jpg) repeat scroll 0%;
}
.jimgMenu ul li.urban a {
background: url(<?php bloginfo('template_directory'); …我正在使用Dispatch Queue加载图像,但如果我不希望它在某些时候加载所有图像,我不知道如何停止队列.
- (void) loadImageInBackgroundArr1:(NSMutableArray *)urlAndTagReference {
myQueue1 = dispatch_queue_create("com.razeware.imagegrabber.bgqueue", NULL);
for (int seriesCount = 0; seriesCount <[seriesIDArr count]; seriesCount++) {
for (int i = 0; i < [urlAndTagReference count]; i++) {
dispatch_async(myQueue1, ^(void) {
NSData *data0 = [[NSData alloc]initWithContentsOfURL:[NSURL URLWithString:[urlAndTagReference objectAtIndex:i]]];
UIImage *image = [[UIImage alloc]initWithData:data0];
dispatch_sync(dispatch_get_main_queue(), ^(void) {
for (UIView * view in [OD_ImagesScrollView subviews]) {
if ([view isKindOfClass:[UIView class]]) {
for (id btn in [view subviews]) {
if ([btn isKindOfClass:[UIButton class]]) {
UIButton *imagesBtn = (UIButton *)btn; …我已经读到使用Akka时的一个重要规则是避免任何阻塞输入/输出操作,轮询,忙等待,睡眠等等.但是,如果我真的需要一些流量控制呢？
我正在使用Akka演员向我们的客户发送邮件,并且对邮件服务器友好,每5秒发送一封邮件.我的计划是使用调度程序actor执行流控制,使用发送方actor执行邮件发送工作.
class Dispatcher extends Actor {
def receive = {
case ResetPassword(to, data) =>
senderActor ! Mail("resetPassword", to, data)
Thread.sleep(5000)
...
}
}
class Sender extends Actor {
def receive = {
case Mail(to, data) => // send the mail immediately
...
}
}
这是正确的方法吗？如果没有,我该如何进行流量控制？