php - 从 PHP/MYSQL 选择值填充下拉菜单。如何重新排列输出

Question

我将继续展示当前 HTML 输出的外观，下面的同一个小提琴是我希望输出 HTML 的方式：http: //jsfiddle.net/xQ95X/3/

一切都很好，我只需要这个 jQuery 插件 ( Superfish ) 的 HTML 结构即可运行。该下拉菜单不适用于 LI 之外的子 UL。它们必须嵌套在里面。

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>";

    ?>

我希望我提供了足够的信息，如果您需要其他信息，请告诉我。虽然 JSFiddle 应该让您很好地了解我希望如何输出 HTML。

PS - 我完全知道这不是最干净的代码，我的猜测是它需要大量的重新排列才能获得所需的输出。我愿意接受建议。

score 4 · Accepted Answer

我个人会在第一个循环中嵌套第二个 for 循环（我已更改为 while 循环）（这是非常粗略的代码），但请注意在父循环中已从部分标题中删除。：

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

从伪代码到接近您的代码的东西：

$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
}

score 2 · Accepted Answer

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子，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>';

score 2 · Accepted Answer

我在开发我们公司的网站导航时遇到了同样的问题，提供最佳可扩展解决方案的是所谓的“嵌套集模型”。

迈克·希勒 (Mike hiller) 的精彩帖子在这里：

http://mikehillyer.com/articles/managing-hierarchical-data-in-mysql/

解释了所有并给出了嵌套集模型的一个很好的总结。这应该是一个很好的起点。

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 干净。

基本上，它在数据库中设置了一个“类别”表，并使用一个左右值来相互关联并定义它们的层次结构和关系。

是的，只有一张桌子......没有“子类别”表格或任何东西......它们都是平等的并且在一张桌子上。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。

它将引导您了解如何构建 db 表以及如何布置站点的类别和子类别。这对我来说扩展得很好，每次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。

希望这对您和我一样有帮助！

score 2 · Accepted Answer

tl;博士获取所有数据然后构建菜单

有很多方法可以做到这一点，而且没有一个是错的；在我不得不这样做的任何地方，我们的 Web 服务器都比数据库服务器多，因此为了更好地扩展，我们尝试最大限度地减少数据库调用，并且为了让开发人员更轻松，我们将尽可能多的逻辑从数据库中提取出来并完成所有工作php/javascript 中的处理。

由于您在一开始就知道您想要什么数据（即您不必根据传入的某些逻辑或任何东西从数据库中提取不同的东西），您可以在开头提取您想要的所有数据页面，然后构建菜单，而不是在从数据库中提取时尝试这样做。这使您可以 1) 将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分开，以及 2) 将数据库调用减少到两个而不是 1 + 个部分。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，它会更容易阅读。

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>';
}

score 2 · Accepted Answer

你需要在里面输入你的第二个选择语句

  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>"; 
  }

那么你将不得不避免第二个 for 循环

 for($s=0;$s<count($sections);$s++)

因为你可以直接在里面获取section id

$sql3="SELECT * FROM articles WHERE id='$sectionid'";

所以在 if 里面写代码然后在 else 里面写代码是不好的。

我对您的代码进行了一些修改，希望对您有所帮助，尚未删除第二个 for 循环，但我认为不需要，请根据您的要求进行修改。

<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>";

?>

score 2 · Accepted Answer

我认为您需要将第二个 for 循环嵌套在第一个循环中，以便 ul 标记嵌套在 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>"; 
          }
        }

php - 从 PHP/MYSQL 选择值填充下拉菜单。如何重新排列输出

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