我正在寻找一种方法来做一些字符串替换.
在这里找到了这个片段,但它给了我错误MvcHtmlString is not defined:
@MvcHtmlString.Create(Html.Encode(comic.name).Replace(" ", "-"));
代码全部:
@{
var comicName = UrlData[0];
var db = Database.Open("PhotoGallery");
var comics = db.Query(@"SELECT * FROM Comics WHERE name = @0", comicName).ToList();
Page.Title = comicName;
}
<div><a href="@Href("~/")">Home</a> -> <a href="@Href("~/Comics")">Comics Categories</a> -> <strong>@comicName</strong></div>
<div style="clear: both;"> </div>
<div class="sidebar">
<center><img src="@Href("~/Images/Comics", comicName + ".jpg")" title="@comicName" width="320" height="498" /><br />
@comics.Count Issue(s)</center>
</div>
<div class="main">
<h1>@comicName</h1>
@foreach (var comic in comics) {
@MvcHtmlString.Create(Html.Encode(comic.name).Replace(" ", "-"));
<ul class="thumbnails gallery">
<li class="gallery">
<a …当此文件夹中有数千条记录时,此计数为onyl返回1.
// TV Shows
$dir = 'G:/TV';
if ($handle = opendir($dir)) {
/* This is the correct way to loop over the directory. */
while (false !== ($file = readdir($handle))) {
$check = mysql_query("SELECT * FROM tv_shows WHERE title = '$file'");
$checkcount = mysql_num_rows($check);
if ($checkcount == 0) {
mysql_query("INSERT INTO tv_shows (id, title) VALUES ('', '$file')");
}
}
echo count($dir)." Records Traversed!<br/>";
closedir($handle);
}
表结构:id,title
文件夹结构:主文件夹中的子文件夹
G:\TV
G:\TV\24
G:\TV\Family Guy
我为我的生活无法弄清楚为什么当我试图回应它时,这不是输出0.它适用于上一个查询,但不适用于最后一个查询.
// This one works and if there are no movies above 0 then it outpus 0 fine
$tag_movies_result = mysql_query("SELECT * FROM tags WHERE tagname='$n' AND movie > 0");
$total_times_used_movies = mysql_num_rows($tag_movies_result);
// Where as this query returns an error
$tag_shows_result = mysql_query("SELECT * FROM tags WHERE tagname='$n' AND show > 0"); <-- `show`
$total_times_used_shows = mysql_num_rows($tag_shows_result); <-- Line 12
// the error
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/a5410474/public_html/tags/tag.php on line 12 …我有一个简单的PHP变量,我需要继续javascript部分.
码:
<?php
if ($p == 'home') { $selected == '0'; }
if ($p == 'music') { $selected == '1'; }
if ($p == 'videos') { $selected == '2'; }
if ($p == 'search') { $selected == '3'; }
if ($p == 'about') { $selected == '4'; }
if ($p == 'contact') { $selected == '5'; }
?>
<script type="text/javascript">
//SYNTAX: tabdropdown.init("menu_id", [integer OR "auto"])
tabdropdown.init("colortab", $selected) <-- $selected is the variable I want to carry over
</script>
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？
我有这个简单的查询,并想知道是否有可能只拉信息arc只等于数字而不是文本.
码:
$sql = mysql_query("SELECT * FROM `comics` ORDER BY `id` DESC LIMIT 20");