Question

我正在尝试从数据库加载数据并将其显示在表中，如下所示：

http://img196.imageshack.us/img196/1857/cijene.jpg

在数据库中，我有两个表：

cities (id, name)
prices (id, city1_id, city2_id, price)

例如，4个城市的打印表格如下所示：

<table>
 <tr>
  <td>city1</td>
  <td> </td>
  <td> </td>
  <td> </td>
 </tr>

 <tr>
  <td> price_city1-city2</td>
  <td>city2</td>
  <td> </td>
  <td> </td>
 </tr>

 <tr>
  <td> price_city1-city3</td>
  <td> price_city2-city3</td>
  <td>city3</td>
  <td> </td>
 </tr>

 <tr>
  <td> price_city1-city4</td>
  <td> price_city2-city4</td>
  <td> price_city3-city4</td>
  <td>city4</td>
 </tr>
</table>

有人知道回复这种表的PHP语句是什么？

Answer 1

我使用了一种非常简单的方法将查询结果转换为HTML表格。我测试$ query_result为true并将结果作为关联数组获取...

$query_result = mysqli_query($db, $query);

if ($query_result) {

    while ($columns = mysqli_fetch_assoc($query_result)) {

        echo "<tr>\n";

        foreach ($columns as $name => $column) {
            echo "<td>";
            printf("%s",$columns[$name]);
            echo "</td>\n";
        }

        echo "</tr>\n";
    }
}

编辑：现在我已经能够看到你的桌子，我可以看到我没有真正回答你的问题。查询显然非常重要。这个问题是否成为“我如何创建一个返回结果的查询，这些结果可以通过我的简单方法转化为表格？”

我希望别人有一些想法。

Answer 2

这有点处于实用性的边缘，但我更倾向于将这个问题作为“正确行事”的练习。

我会对每个表做一个简单的选择并插入一个数组：

$query1 = "SELECT * FROM cities";
$query2 = "SELECT * FROM prices";

$results1 = mysql_query( $query1 );
$results2 = mysql_query( $query2 );

while( $rows = mysql_fetch_array( $results1 ) ) $cities[] = $row['name'];
while( $rows = mysql_fetch_array( $results2 ) ) $prices[] = array(
                                                               $row['city1_id'],
                                                               $row['city2_id'], 
                                                               $row['name']
                                                            );

然后我会用它来动态创建两个javascript列表：

$java_array1 = "var Cities = new Array;\n";
foreach( $cities as $key=>$city ) $java_array1 .= "Cities[$key] = \"$city\";\n";
$java_array2 = "var Prices = new Array;\n";
foreach( $cities as $key=>$price ) $java_array2 .= "Prices[$key] = new Array(
                                                                     \"{$price[0]}\",
                                                                     \"{$price[1]}\",
                                                                     \"{$price[2]}\",
                                                                  );\n";

我接下来会为每个单元格输出一个包含精心设计的ID的表格。我会使用与第一个答案非常相似的代码，但我会给每个单元格一个唯一的"cell-<row>-<col>" ID。

我要做的最后一件事就是掀起一些onload javascript，它会使用适当的配对填充表格。我的javascript很生锈，但它看起来像这样（注意：伪代码如下）：

n = 1;
for( var i in cities )
{
    // set city names on the diagonal
    document.getElementById( 'cell-'&i&'-'&i ).innerHTML = names[i];
    n = n + 1;
}

for( var j in prices )
{
    // use the city IDs in the prices array (see earlier code initializing
    // prices array) to place the prices
    document.getElementById( 'cell-'&prices[1]&'-'&prices[2] ).innerHTML = prices[0];
}

我几乎肯定搞砸了一些javascript;使用风险自负。

这比上面给出的更简单的答案要复杂得多，但这是我能想到做到这一点的唯一方法，这样才有意义。基本上，使用这种方法，你可以放置应该所在的所有内容，而且你不必担心奇数表和奇怪的选择。

Answer 3

// This single query gets all required data
// NOTE: this query assumes that your price data is entered
//   such that from_city always alphabetically precedes to_city!
$sql =
    "SELECT a.name AS from_city, b.name AS to_city, price ".
    "FROM prices ".
    "INNER JOIN cities AS a ON a.id=prices.city1_id ".
    "INNER JOIN cities AS b ON b.id=prices.city2_id";


// connect and do query
$conn = mysql_connect($host, $user, $pass);
mysql_select_db($db, $conn);
$q = mysql_query($sql, $conn);


// Stuff all data into an array for manipulation;
//   keep list of all cities mentioned
$price = array();
$cities = array();
while (($res = mysql_fetch_assoc($q)) !== false) {
    $from = $res['from_city'];
    $cities[ $from ] = true;

    $to = $res['to_city'];
    $cities[ $to ] = true;

    $price[$to][$from] = $res['price'];
}

// Invert to get alphabetical list of cities
$cities = array_keys($cities);
sort($cities);
$num = count($cities);


// some utility functions
function table($rows) {
    return '<table>'.join($rows).'</table>';
}
function row($cells) {
    return '<tr>'.join($cells).'</tr>';
}
function cell($text) {
    $text = (string) $text;
    if (strlen($text)==0)
        $text = '&nbsp;';
    return '<td>'.$text.'</td>';
}


// The data is now in the desired order;
// produce output as HTML table
$rows = array();
for ($to = 0; $to < $num; ++$to) {
    $t = $cities[$to];
    $cells = array();

    for ($from = 0; $from < $to; ++$from) {
        $f = $cities[$from];

        if isset($price[$t]) && isset($price[$t][$f])
            $text = $price[$t][$f];
        else
            $text = '';

        $cells[]= cell($text);
    }

    $cells[]= cell($t);    // add destination-label

    for ($from = $to+1; $from < $num; ++$from)   // pad to proper number of cells
        $cells[]= cell('');

    $rows[]= row($cells);
}
$html = table($rows);


// show results!
echo $html;

打印级联表

3 个答案: