我无法将两个avi视频合并在一起.谷歌充满了以下例子:
cat file1.avi file2.avi file3.avi > video_draft.avi
after appending the data together using cat above, you need to re-index the draft movie like this:
mencoder video_draft.avi -o video_final.avi -forceidx -ovc copy -oac copy
Now you're video_final.avi file will be right to go.
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但它对我不起作用,第一个视频被转换,就是这样.
今天我将我的图像从一台服务器迁移到另一台服务器,并面临一个奇怪的许可问题
[Mon Mar 25 08:42:23.676315 2013] [core:crit] [pid 15182] (13)Permission denied: [client 24.14.2.22:48113] AH00529: /files/domain.com/public_html/images/.htaccess pcfg_openfile: unable to check htaccess file, ensure it is readable and that '/files/domain.com/public_html/images/' is executable, referer: http://domain.com.com/file.php
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我试过的:
chmod -R 777 root_dir
restorecon -r
disable selinux
Chown -R root:root root_dir
Run Code Online (Sandbox Code Playgroud) 我想告诉谷歌不要索引页面的某些部分,在yandex(俄语se)中有一个非常有用的标签叫做<noindex>.怎么用谷歌呢?
我正在构建菜单,显示一个字母列表,然后根据用户选择的内容加载正确的页面.我能够加载一个字母,但我怎么能使用"字母"作为id,所以我不需要为每个字母复制相同的jquery代码.
<div id="idshow"><a id="hidelinks" href="#">Hide</a> <br /><br /></div>
<div id="letter_a"><a id="success_a" href="#">A Show</a></div>
<div id="letter_b"><a id="success_b" href="#">B Show</a></div>
<div id="letter_c"><a id="success_c" href="#">C Show</a></div>
<div id="loadpage"></div>
<script>
$("#idshow").hide();
$("#success_a").click(function () {
$("#idshow").show();
$("#letter").hide();
$("#loadpage").load("letter_a.html", function(response, status, xhr) {
if (status == "error") {var msg = "Sorry but there was an error: ";$("#error").html(msg + xhr.status + " " + xhr.statusText);}});
});
$('#hidelinks').click(function() {
$('#letter_content').slideUp();
$("#idshow").hide();
$("#letter").show();
});
</script>
Run Code Online (Sandbox Code Playgroud) 我正在处理一个显示最近10个帖子的块,当用户点击"加载更多"按钮时,我想显示10个旧帖子.
如果我已经在使用DESC LIMIT,我如何选择最后10行?
mysql_query("SELECT title,id,alt_name FROM dle_post WHERE approve='1' AND date >= '$monthagodate'
AND date < '$curdate' + INTERVAL 1 day ORDER BY date DESC LIMIT $more;");
Run Code Online (Sandbox Code Playgroud) 如何判断php数组是否为空?我尝试了不同的方法,但它永远不会echo "no content";
while($row = mysql_fetch_array($result))
{
if(count($row['link']))
{
echo '<a id="link_' . $row['Id'] . '" href="' . $row['link'] . '" data="/short_info.php?id=' . $row['Id'] . '/">' . $row['title'] . '</a><div class="in...
}
else
{
echo "no content";
}
}
Run Code Online (Sandbox Code Playgroud) 我有三个变量,
$title_1 $title_2 $title_3
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我如何在for循环中打印它们?
我试过的:
$number = 3;
for($i=0;$i<$number;$i++){
echo "$title_($i+1)";
}
Run Code Online (Sandbox Code Playgroud) 如何将变量值添加到新变量中?我正在修改新闻滑块,我想在字幕中插入"日期"
我试过的:
(function( $ ){
$.fn.accessNews = function(settings){
$now = new Date();
var defaults = {
// title for the display
title: "news:",
// subtitle for the display
subtitle: "$now",
Run Code Online (Sandbox Code Playgroud) 如何从投票数和评分中发现大拇指和大拇指的数量.
例如我有rating of +203和301 votes.每次投票都是-1或+1
我试过的
$NEGATIVE_VOTES = ( $row['rating'] - $row['vote_num'] );
$POSITIVE_VOTES = ( $row['vote_num'] - $NEGATIVE_VOTES );
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