Mysql加入，PHP

Question

就在我把代碼行放在讓我解釋我想要實現的東西之前。

我想查詢特定頁面具有特定parent_id的表。一旦我有了所有的結果，我就會喜歡帶有這個特定parent_id的頁面的id，並從另一個表中獲取它的所有相關頁面部分。

然後，我只是想在準備打印數組此信息...

任何幫助將不勝感激......

$q = $CONN->query("SELECT cms_page.id, cms_page.title, cms_page.slug, cms_page_part.* FROM cms_page LEFT JOIN cms_page_part ON cms_page_part.page_id=cms_page.id WHERE cms_page.parent_id='8'"); 

Answer 1

您可以發佈您查詢的結果？據我可以看到你的選擇語句應該做的重複每個匹配部分的cms_page.id等字段的技巧。

又見http://www.w3schools.com/Sql/sql_join_left.asp

這（我希望）應該解決您的$行[ 'page_part']問題：

$rows = array(); 
$result = mysql_query($sql); 
$row = mysql_fetch_assoc($result); 
while ($row) { 
    $rows[$row['title']][] = $row; // or select only the fields that you need 
    $row = mysql_fetch_assoc($result); 
} 

foreach($rows as $portfolio => $parts) { 
    // adjust for actual html rendering 
    echo '<pre>'; 
    print_r($portfolio); 
    echo "\n"; 
    $count = 0; 
    foreach($parts as $part) { 
    echo ++$count . ' - ' . $part['cms_page_part_field1'] . ' - ' . $part['cms_page_part_field2'] . "\n"; 
    } 
    echo '</pre>'; 
} 

Answer 2

應該有與檢索多行給你當前的查詢沒有問題：

$sql = "SELECT cms_page.id, cms_page.title, cms_page.slug, cms_page_part.* 
FROM cms_page 
LEFT JOIN cms_page_part ON (cms_page_part.page_id = cms_page.id) 
WHERE cms_page.parent_id = '8'"; 

你想遍歷一個結果從mysql_query像這樣：

$result = mysql_query($sql); 
if ($result) { 
    // keep track of number of portfolio items 
    $i = 1; 
    // iterate over rows 
    while($row = mysql_fetch_assoc($result)) { 
     // output data to page for each row 
     echo 'Portfolio item ' . $i; 
     echo 'Title: ' . $row['title'] . '<br />'; 
     echo 'Slub: ' . $row['slug'] . '<br /><br />'; 
    } 
}