从PHP / MYSQL选择值填充下拉菜单。 如何重新排列输出
[英]Populating Drop Down Menu from PHP/MYSQL Select Values. How to rearrange output
问题描述
I'm going to go ahead and show how the current HTML output looks and in the same fiddle below is how I wish the HTML to be output: http://jsfiddle.net/xQ95X/3/ 
我将继续展示当前HTML输出的外观以及下面的同一个小提示我希望如何输出HTML： http ： //jsfiddle.net/xQ95X/3/
Everything works great, I just need this HTML structure for the jQuery plugin ( Superfish ) to be function. 一切都很好，我只需要这个jQuery插件（ Superfish ）的HTML结构就可以了。 The dropdown won't work with the child UL's outside of the LI's. 下拉列表不适用于儿童UL在LI之外。 They must be nested inside. 它们必须嵌套在里面。
The part which is stopping me from formatting the HTML in this manner is below: 阻止我以这种方式格式化HTML的部分如下：
<ul class="sf-menu">
<li><a href="#">HOME</a></li>
<?php
$sql="SELECT * FROM section ORDER BY orderID";
$result=mysql_query($sql) or die ("Error!! BAD SELECT SEARCH STATEMENT");
$nrows = mysql_num_rows($result);
// Top level Items
for($i=1;$i<=$nrows;$i++) {
$row = mysql_fetch_array($result);
extract($row);
$sections[] = $section;
$sectionids[] = $sectionid;
if ($section == "FORUM") {
echo"<li><a href='#' rel='dropmenu$i'>$section</a></li>";
} else {
echo"<li><a href='#h' rel='dropmenu$i'>$section</a></li>";
}
}
// Submenu Items
for($s=0;$s<count($sections);$s++) {
/*$sm1 = ($s - 1);*/
$sp1 = ($s + 1);
$sql3="SELECT * FROM category WHERE sectionid='$sectionids[$s]'";
$result3=mysql_query($sql3) or die ("Error!! BAD SELECT SEARCH STATEMENT");
$nrows3 = mysql_num_rows($result3);
if ($nrows3 > 0) {
echo"<ul>";
for($t=0;$t<$nrows3;$t++) {
$row3 = mysql_fetch_array($result3);
extract($row3);
if ($type=='Photo') {
$title20 = strtolower($title);
$title20 = str_replace(" ", "_", $title20);
$sec = strtolower($sections[$s]);
$sec = str_replace(" ", "_", $sec);
if ($sec=='photos') {
echo"<li><a href='" . dirname($_SERVER['PHP_SELF']) . "$sec/$catID-$title20/1.html'>$title</a></li>";
}
else {
echo"<li><a href='" . dirname($_SERVER['PHP_SELF']) . "$sec/photos/$catID-$title20/1.html'>$title</li></a>";
}
} elseif ($type=="Contact") {
$title20 = strtolower($title);
$title20 = str_replace(" ", "_", $title20);
echo "<li><a href='" . dirname($_SERVER['PHP_SELF']) . "contactus/contact_$title20.html'>$title</a></li>";
} elseif ($type=='Quote') {
$sql4="SELECT articleID FROM article WHERE category='49' AND Now() > publishingdate ORDER BY publishingdate DESC LIMIT 0, 1";
$result4=mysql_query($sql4) or die ("Error! bad select statement");
$row4=mysql_fetch_array($result4);
extract($row4);
echo "<li><a href='" . dirname($_SERVER['PHP_SELF']) . "everton_quotes.php?id=$articleID'>$title</a></li>";
}
elseif ($type=='LastMatch') {
$sql4="select r.matchID as matchID2, r.venue, r.versus as versus2, s.year FROM regmatch r, season s WHERE r.season=s.seasonID AND Now() > r.matchdate ORDER BY r.matchdate DESC LIMIT 0,1";
$result4=mysql_query($sql4) or die ("Error! bad select statement");
$nrows4=mysql_num_rows($result4);
if ($nrows4 > 0) {
$row4=mysql_fetch_array($result4);
extract($row4);
if ($venue=='Home') {
$teams = "Everton V $versus2";
$teams2 = "everton_vs_" . str_replace(" ", "_", strtolower($versus2));
}
else {
$teams = "$versus V Everton";
$teams2 = str_replace(" ", "_", strtolower($versus2)) . "_vs_everton";
}
$year = str_replace("/", "-", $year);
echo "<li><a href='" . dirname($_SERVER['PHP_SELF']) . "match/$year/$matchID2-$teams2/" . strtolower($venue) . "/matchreport.html'>$title</a></li>";
}
} elseif ($type=='NextMatch') {
$sql5="select r.matchID as matchID3, r.venue, r.versus as versus2, s.year FROM regmatch r, season s WHERE r.season=s.seasonID AND r.matchdate > Now() ORDER BY r.matchdate ASC LIMIT 0,1";
$result5=mysql_query($sql5) or die ("Error! bad select statement");
$nrows5=mysql_num_rows($result5);
if ($nrows5 > 0) {
$row5=mysql_fetch_array($result5);
extract($row5);
if ($venue=='Home') {
$teams = "Everton V $versus2";
$teams2 = "everton_vs_" . str_replace(" ", "_", strtolower($versus2));
}
else {
$teams = "$versus2 V Everton";
$teams2 = str_replace(" ", "_", strtolower($versus2)) . "_vs_everton";
}
$year = str_replace("/", "-", $year);
echo "<li><a href='" . dirname($_SERVER['PHP_SELF']) . "match/$year/$matchID3-$teams2/" . strtolower($venue) . "/teamnews.html'>$title</a></li>";
}
} else {
// Overides for missing pages
if($title == "Gwladys Street Preacher") {
echo"<li><a href='/fans/121-gwladys_street_preacher/index.html'>$title</a></li>";
}
elseif($title == "The Secret Fan") {
echo"<li><a href='/news/124-secret-fan/index.html'>$title</a></li>";
}
elseif($title == "Toffee Girl") {
echo"<li><a href='/fans/125-toffee_girl/index.html'>$title</a></li>";
}
else {
echo"<li><a href='" . dirname($_SERVER['PHP_SELF']) . $link . "'>$title</a></li>";
}
}
}
echo"</ul>";
}
}
echo "</ul>";
?>
I hope I have provided enough information, let me know if you require anything else. 我希望我提供了足够的信息，如果您需要其他任何信息，请告诉我。 Though the JSFiddle should give you a good idea of how I would like the HTML to be output. 虽然JSFiddle应该让你很好地了解我希望如何输出HTML。
PS - I'm fully aware this isn't the cleanest code around, my guess is that it will need a fair amount of rearranging to get the desired output. PS - 我完全清楚这不是最干净的代码，我的猜测是它需要大量的重新排列来获得所需的输出。 I'm open to suggestions. 我愿意接受建议。
6 个解决方案
解决方案1
4 已采纳 2013-10-23 15:42:42
I would personally nest the second for loop (which I have changed to while loops) in the firsts (and this is very much rough code) but notice in the parent loop the has been removed from the section title.: 我个人将第二个for循环（我已经更改为while循环）嵌套在第一（这是非常粗略的代码）但是在父循环中的注意事项已从部分标题中删除：
Some pseudo code: 一些伪代码：
print parent unordered list start
get section data
loop through section results
print parent list item start
print section title
get page data
print unordered child list start
loop through page results
print child list item start
print page info
print child list item end
print child unordered list end
print parent list item end
print parent unordered list end
And from pseudo code to something close to your code: 从伪代码到接近代码的东西：
$sql="SELECT * FROM section ORDER BY orderID";
$result=mysql_query($sql) or die ("Error!! BAD SELECT SEARCH STATEMENT");
// Top level Items
while($row = mysql_fetch_array($result){
extract($row);
if ($section == "FORUM") echo"<li><a href='#' rel='dropmenu$i'>$section</a>";
else echo"<li><a href='#h' rel='dropmenu$i'>$section</a>";
$sql3="SELECT * FROM category WHERE sectionid='$sectionid'";
$pageResult=mysql_query($sql3) or die ("Error!! BAD SELECT SEARCH STATEMENT");
echo "<ul>";
while($page = mysql_fetch_array($pageResult){
extract($page);
//process and print each child <li>...</li> here
//(this is a cut down example) where you would add your if/elseif section
echo "<li>$title</li>";
}
echo "</ul>"; //close child UL
echo "</li>"; //close parent LI
}
解决方案2
2 2013-11-18 20:14:32
I think you need to nest the second for loop within the first one so that the ul tags are nested within the li: A general guideline would be: 我认为你需要在第一个中嵌套第二个for循环，以便ul标签嵌套在li中：一般准则是：
for($i=1;$i<=$nrows;$i++) {
$row = mysql_fetch_array($result);
extract($row);
$sections[] = $section;
$sectionids[] = $sectionid;
if ($section == "FORUM") {
echo"<li><a href='#' rel='dropmenu$i'>$section</a>";
for($s=0;$s<count($sections);$s++) { //Filter $sections for FORUM submenu items only
//submenu items
}
echo "</li>";
} else {
echo"<li><a href='#h' rel='dropmenu$i'>$section</a>";
for($s=0;$s<count($sections);$s++) {
//submenu items
}
echo </li>";
}
}
解决方案3
2 2013-11-19 04:32:34
I ran into this same problem developing our company's site navigation, what provided the best scalable solution was what's called 'The Nested Set Model.' 我遇到了同样的问题，开发我们公司的网站导航，提供最好的可扩展解决方案的是所谓的'嵌套集模型'。
Mike hiller's excellent post here: 迈克希勒在这里的优秀帖子：
http://mikehillyer.com/articles/managing-hierarchical-data-in-mysql/ http://mikehillyer.com/articles/managing-hierarchical-data-in-mysql/
explains all and gives a great summary of the nested set model. 解释了所有并给出了嵌套集模型的一个很好的总结。 It should be a great place to start. 它应该是一个很好的起点。
It allows you to put all of the complexity into your database queries and keep both your database tables and your php clean. 它允许您将所有复杂性放入数据库查询中，并保持数据库表和PHP清洁。
Basically, it sets up a, say, 'categories' table in the database and uses a value, left and right to relate to each other and define their hierarchy and relationship. 基本上，它在数据库中设置一个“类别”表，并使用左，右相互关联的值来定义它们的层次结构和关系。
And yes, just one table... No 'subcategories' table or anything....they are all equal and on one table. 是的，只有一个表...没有'子类别'表或任何东西......它们都是相同的并且在一张桌子上。 Again, what defines their placement and hierarchy is their left (lft) and right (rgt) value. 同样，定义它们的位置和层次结构的是它们的左（lft）和右（rgt）值。
It walks you through how to structure the db table and how your site's categories and sub-categories are laid out. 它将指导您如何构建db表以及如何布置站点的类别和子类别。 This has scaled very well for me and everytime I find myself having to add a category, all I do is add it to the categories table and change the left and right values to fit the new category in. 这对我来说非常好，每当我发现自己必须添加一个类别时，我所做的就是将它添加到类别表中并更改左右值以适应新类别。
Hope this is as much of a help to you as it was to me! 希望这对你有所帮助，就像对我一样！
解决方案4
2 2013-11-19 05:28:17
You need to enter your second select statement inside 您需要在里面输入第二个select语句
if ($section == "FORUM") {
echo"<li><a href='#' rel='dropmenu$i'>$section</a>";
//the ul should start here
echo "</li>";
} else {
echo"<li><a href='#h' rel='dropmenu$i'>$section</a>";
//the ul should start here
echo "</li>";
}
So then you will have to avoid the second for loop 那么你将不得不避免第二个for循环
for($s=0;$s<count($sections);$s++)
because you can get the section id directly inside 因为你可以直接在里面获得部分ID
$sql3="SELECT * FROM articles WHERE id='$sectionid'";
so writing the code inside if and then also inside else, is not good. 因此，如果然后在其他内部编写代码，则不是很好。
I have modified your code a bit, hope it helps, have not removed the second for loop yet but I think its not needed, modify according to your requirements. 我已经修改了你的代码，希望它有所帮助，还没有删除第二个for循环，但我认为它不需要，根据你的要求修改。
<ul class="sf-menu">
<li><a href="#">HOME</a></li>
<?php
$sql="SELECT * FROM section ORDER BY orderID";
$result=mysql_query($sql) or die ("Error!! BAD SELECT SEARCH STATEMENT");
$nrows = mysql_num_rows($result);
// Top level Items
for($i=1;$i<=$nrows;$i++) {
$row = mysql_fetch_array($result);
extract($row);
$sections[] = $section;
$sectionids[] = $sectionid;
echo "<li><a href='";
if ($section == "FORUM")
echo "#";
else
echo "#h";
echo "' rel='dropmenu$i'>$section</a>";
for($s=0;$s<count($sections);$s++) {
/*$sm1 = ($s - 1);*/
$sp1 = ($s + 1);
$sql3="SELECT * FROM category WHERE sectionid='$sectionids[$s]'";
$result3=mysql_query($sql3) or die ("Error!! BAD SELECT SEARCH STATEMENT");
$nrows3 = mysql_num_rows($result3);
if ($nrows3 > 0) {
echo"<ul>";
for($t=0;$t<$nrows3;$t++) {
$row3 = mysql_fetch_array($result3);
extract($row3);
if ($type=='Photo') {
$title20 = strtolower($title);
$title20 = str_replace(" ", "_", $title20);
$sec = strtolower($sections[$s]);
$sec = str_replace(" ", "_", $sec);
if ($sec=='photos') {
echo"<li><a href='" . dirname($_SERVER['PHP_SELF']) . "$sec/$catID-$title20/1.html'>$title</a></li>";
}
else {
echo"<li><a href='" . dirname($_SERVER['PHP_SELF']) . "$sec/photos/$catID-$title20/1.html'>$title</li></a>";
}
} elseif ($type=="Contact") {
$title20 = strtolower($title);
$title20 = str_replace(" ", "_", $title20);
echo "<li><a href='" . dirname($_SERVER['PHP_SELF']) . "contactus/contact_$title20.html'>$title</a></li>";
} elseif ($type=='Quote') {
$sql4="SELECT articleID FROM article WHERE category='49' AND Now() > publishingdate ORDER BY publishingdate DESC LIMIT 0, 1";
$result4=mysql_query($sql4) or die ("Error! bad select statement");
$row4=mysql_fetch_array($result4);
extract($row4);
echo "<li><a href='" . dirname($_SERVER['PHP_SELF']) . "everton_quotes.php?id=$articleID'>$title</a></li>";
}
elseif ($type=='LastMatch') {
$sql4="select r.matchID as matchID2, r.venue, r.versus as versus2, s.year FROM regmatch r, season s WHERE r.season=s.seasonID AND Now() > r.matchdate ORDER BY r.matchdate DESC LIMIT 0,1";
$result4=mysql_query($sql4) or die ("Error! bad select statement");
$nrows4=mysql_num_rows($result4);
if ($nrows4 > 0) {
$row4=mysql_fetch_array($result4);
extract($row4);
if ($venue=='Home') {
$teams = "Everton V $versus2";
$teams2 = "everton_vs_" . str_replace(" ", "_", strtolower($versus2));
}
else {
$teams = "$versus V Everton";
$teams2 = str_replace(" ", "_", strtolower($versus2)) . "_vs_everton";
}
$year = str_replace("/", "-", $year);
echo "<li><a href='" . dirname($_SERVER['PHP_SELF']) . "match/$year/$matchID2-$teams2/" . strtolower($venue) . "/matchreport.html'>$title</a></li>";
}
} elseif ($type=='NextMatch') {
$sql5="select r.matchID as matchID3, r.venue, r.versus as versus2, s.year FROM regmatch r, season s WHERE r.season=s.seasonID AND r.matchdate > Now() ORDER BY r.matchdate ASC LIMIT 0,1";
$result5=mysql_query($sql5) or die ("Error! bad select statement");
$nrows5=mysql_num_rows($result5);
if ($nrows5 > 0) {
$row5=mysql_fetch_array($result5);
extract($row5);
if ($venue=='Home') {
$teams = "Everton V $versus2";
$teams2 = "everton_vs_" . str_replace(" ", "_", strtolower($versus2));
}
else {
$teams = "$versus2 V Everton";
$teams2 = str_replace(" ", "_", strtolower($versus2)) . "_vs_everton";
}
$year = str_replace("/", "-", $year);
echo "<li><a href='" . dirname($_SERVER['PHP_SELF']) . "match/$year/$matchID3-$teams2/" . strtolower($venue) . "/teamnews.html'>$title</a></li>";
}
} else {
// Overides for missing pages
if($title == "Gwladys Street Preacher") {
echo"<li><a href='/fans/121-gwladys_street_preacher/index.html'>$title</a></li>";
}
elseif($title == "The Secret Fan") {
echo"<li><a href='/news/124-secret-fan/index.html'>$title</a></li>";
}
elseif($title == "Toffee Girl") {
echo"<li><a href='/fans/125-toffee_girl/index.html'>$title</a></li>";
}
else {
echo"<li><a href='" . dirname($_SERVER['PHP_SELF']) . $link . "'>$title</a></li>";
}
}
}
echo"</ul>";
}
}//end of for
echo "</li>";
}
// Submenu Items
echo "</ul>";
?>
解决方案5
2 2013-11-19 13:24:48
See this, NOTE: i added an index called 'has_sub' to check if the section has sub, then create a sub menu for it 看到这个，注意：我添加了一个名为'has_sub'的索引来检查该部分是否有sub，然后为它创建一个子菜单
echo '<ul>';
$array = array();
while($row = mysql_fetch_array($query_main_data)) {
if($row['has_sub'] === 'yes') {
echo '<li><a>'.$row['section_name'].'</a></li>';
echo '<ul>';
while($row2 = mysql_fetch_array($query_sub_data)) {
echo '<li><a>'.$row['section_name'].'</a></li>';
}
echo '</ul>';
}
}
echo '</ul>';
解决方案6
2 2013-11-23 17:43:01
tl;dr get all of your data then build the menu tl; dr获取所有数据，然后构建菜单
There are a lot of ways to do this and none of them are wrong; 有很多方法可以做到这一点，但没有一个是错的; everywhere I've had to do this, we had more web servers than database servers and so to scale better we tried to minimize database calls, and to keep things easy on developers we pulled as much logic as possible outside of the database and did all of the processing in php/javascript. 在我必须这样做的任何地方，我们拥有比数据库服务器更多的Web服务器，因此为了更好地扩展我们尝试最小化数据库调用，并且为了让开发人员保持简单，我们尽可能多地在数据库之外提取逻辑并完成所有操作在php / javascript中处理。
Since you know what data you want at the beginning (ie you don't have to pull different things from the database depending on some logic that's passed in or anything), you can pull all of the data that you want at the beginning of the page and then build the menu, instead of trying to do it as you pull from the database. 既然您在开始时就知道自己想要什么数据（也就是说，根据传入的逻辑或任何内容，您不必从数据库中提取不同内容），您可以在开头提取所需的所有数据。页面然后构建菜单，而不是在从数据库中提取时尝试执行此操作。 This lets you 1) separate the code that makes database calls and the code that holds the logic for building the menus and 2) cut down your database calls to just two instead of 1 + number of sections. 这使您可以1）分离进行数据库调用的代码和保存用于构建菜单的逻辑的代码，以及2）将数据库调用减少到只有两个而不是1 +多个部分。 Depending on your needs, #2 may not be a big factor, but it makes it easier to read if you separate everything at the beginning. 根据您的需要，＃2可能不是一个重要因素，但如果您在开始时将所有内容分开，则可以更容易阅读。
Something like this (could be cleaner): 像这样的东西（可能更干净）：
<li><a href="#">HOME</a></li>
<?php
$sql="SELECT * FROM section ORDER BY orderID";
$result=mysql_query($sql) or die ("Error!! BAD SELECT SEARCH STATEMENT");
$nrows = mysql_num_rows($result);
// Top level Items
for($i=1;$i<=$nrows;$i++) {
$row = mysql_fetch_array($result);
extract($row);
$sections[$sectionid] = $section; //we want to reference the section by the id later
$sectionids[] = $sectionid;
}
// Here we can pull all of the subsections we need
$all_section_ids = implode(',', $sectionids); // join all of the ids with a comma
// SQL IN lets you select all rows where the sectionid matches any of a list
$sql3="SELECT * FROM category WHERE sectionid IN ($all_section_ids)";
$result=mysql_query($sql) or die ("Error!! BAD SELECT SEARCH STATEMENT");
$nrows = mysql_num_rows($result);
for($i=1;$i<=$nrows;$i++) {
$row = mysql_fetch_array($result);
extract($row);
//Whatever values that you need from the category table, depends on what your data looks like
$subsections[$sectionid][] = $subsection;
$types[$sectionid][] = $type; //each sectionid gets a separate array
$some_array[$sectionid][] = $whatever_variable;
}
//Now we can loop through and create everything
foreach ($sections as $id => $section) {
//now $section is the name and $id is the sectionid for that section
echo '<li><a href="whatever link"></a>';
echo '<ul>';
//since we separated the subsections by sectionid, we can do this
foreach ($subsections[$id] as $subsection) {
echo '<li>';
//whatever logic you need to to build the inner list
echo '</li>';
}
echo '</ul>';
}
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