帶連接表的2D數組

Question

我正在使用連接從兩個表中查詢數據。對於每個a_id，我需要獲取關聯的image_id，過濾這些image_id結果，以便只保留第一個結果，最後將結果輸出到li中。我認爲我的查詢很好，但是我在包裝我的頭部時遇到了一些麻煩，要如何爲每個a_id獲取image_id並將其輸出到我的li。此代碼返回一些結果，但它們不是我正在尋找的。

<?php 
echo '<ul>'; 
$result = mysql_query ("SELECT * from artists left join images on artists.a_id = images.image_id where artists.display_works = '1' and artists.active = '1' order by artists.project_year desc, artists.fullname desc"); 

while ($row = mysql_fetch_array($result)){ 
    $data['a_id']['image_id']=$row->a_id; 

    foreach($data as $id=>$images) { 
      $totalimages=1; 
      $addstyle = ""; 
      $art_id = $data['a_id']; 
      $img_id = $data['image_id']; 

      foreach($images as $val){ 

        if($totalimages > 1){ $addstyle = 'style="display:none;"'; } 
        else { 
        $myimagename = "http://artists/$art_id/images/$img_id" . "_large.jpg"; 
        list($width, $height, $type, $attr) = getimagesize("$myimagename"); 
        $myimagename = "http://artists/resize.php/$art_id/images/$img_id" . "_large.jpg?resize(157x2000)"; 

        if($row["layout"] == "vert"){$pl = "_vertical";}else if($row["layout"] == "website"){$pl = "-s";}else if($row["layout"] == "video"){$pl = "_video";}else{$pl = "_horizontal";} 
        echo "<li class='thumbnail_container' $addstyle> <a class='thumbnail' href=\"../works$pl.php?a_id=" . $row["a_id"] . "\"><span><img src=\"$myimagename\" /></span>\n</a></li>\n"; 
        } 

        $totalimages++; 
     } 
    } 
} 
echo '</ul>'; 
?> 

嗯，我已經修改了代碼位和它的工作，但由於某種原因，我沒有圖像的URL或鏈接URL的第一圖像後獲得一個額外的縮略圖。我認爲這可能與我的方法檢查重複的a_ids有關：

<?php 
echo '<ul>'; 
$result = mysql_query ("SELECT * from artists left join images on artists.a_id = images.a_id where artists.display_works = '1' and artists.active = '1' order by artists.project_year desc, artists.fullname desc, images.position asc"); 

while ($row = mysql_fetch_array($result)){ 
    $check = $row['a_id']; 
    if (in_array($check, $a_ids)) {end;} 
    else { 
    $a_id=$row['a_id']; 
    $a_ids[] = $a_id; 
    $image_id=$row['image_id']; 

    $myimagename = "http://artists/$a_id/images/$image_id" . "_large.jpg"; 
    list($width, $height, $type, $attr) = getimagesize("$myimagename"); 
    $myimagename = "http://artists/resize.php/$a_id/images/$image_id" . "_large.jpg?resize(157x2000)"; 

    if($row["layout"] == "vert"){$pl = "_vertical";}else if($row["layout"] == "website"){$pl = "-s";}else if($row["layout"] == "video"){$pl = "_video";}else{$pl = "_horizontal";} 
     echo "<li class='thumbnail_container' $addstyle> <a class='thumbnail' href=\"../works$pl.php?a_id=" . $row["a_id"] . "\"><span><img src=\"$myimagename\" /></span>\n</a></li>\n"; 
    } 
} 
echo '</ul>'; 
?> 

Answer 1

這是我最終使用的解決方案。真正讓我感到困難的是，我沒有理解兩個概念。

第一次加入。我不明白它是如何合併這兩張桌子的。在閱讀了一些關於它之後，我現在知道在加入不同名稱的列時ON更合適，但我真正想要的是通過它們的a_id列加入兩個表。儘管我已經離開了ON並且使兩個列名都相同，但我應該使用關鍵字USING進行研究，因爲它專門用於具有相同名稱的列。

我很難理解的第二個概念是如何從新的連接表中獲取所需的所有信息。我不知道在while循環中使用mysql_fetch_array會遍歷每一行並獲取每列的所有數據。一旦我理解了這一點，雖然很容易通過每一行並獲得image_id和a_id。

我的最終代碼：

<?php 
echo '<ul>'; 
$result = mysql_query ("SELECT * from artists left join images on artists.a_id = images.a_id where artists.display_works = '1' and artists.active = '1' order by artists.project_year desc, artists.fullname desc, images.position asc"); 

while ($row = mysql_fetch_array($result)){ 
    $check = $row['a_id']; 
    if (!in_array($check, $a_ids) && $check !='') { 
    $a_id = $row['a_id']; 
    $a_ids[] = $a_id; 
    $image_id = $row['image_id']; 

    $myimagename = "http://artists/$a_id/images/$image_id" . "_large.jpg"; 
    list($width, $height, $type, $attr) = getimagesize("$myimagename"); 
    $myimagename = "http://artists/resize.php/$a_id/images/$image_id" . "_large.jpg?resize(157x2000)"; 

    if($row["layout"] == "vert"){$pl = "_vertical";}else if($row["layout"] == "website"){$pl = "-s";}else if($row["layout"] == "video"){$pl = "_video";}else{$pl = "_horizontal";} 
     echo "<li class='thumbnail_container' $addstyle> <a class='thumbnail' href=\"../works$pl.php?a_id=" . $row["a_id"] . "\"><span><img src=\"$myimagename\" /></span>\n</a></li>\n"; 
    } 
} 
echo '</ul>'; 
?>