如何更改MySQL查询的结果排序？

Question

我试图获取我的页面的上一个和下一个节点链接/缩略图，并根据它们的标题或文件URI或文件名排序结果...

代码查询数据库并输出上一个和下一个节点链接根据节点ID（nid，n.nid）。我想根据节点标题（标题，n.title），文件名（文件名，f.filename）或文件URI（uri，f.uri）来排序结果。

然而，当我改变这一行：

->orderBy('n.nid', $order) 

到：

->orderBy('n.title', $order) 

将无法​​正常工作。唯一的区别在于，如果您从一个图库中的最后一页移动到另一个图库，它会略微改变顺序，但是在图库内部的其他内容都是一样的。问题是如果你有一个画廊，并决定在一段时间后插入一个新的图像。节点ID现在完全不同于其他节点，并且此代码不会提取它。

我也尝试改变nid显示的其他部分，但它不起作用。 我认为对于了解MySQL和PHP的人来说这是微不足道的，但我坚持了几个小时，并希望得到任何帮助。

这里是整个代码（从维拉德Stratulat，最初发现here）：

的template.php

function dad_prev_next($nid = NULL, $op = 'p', $start = 0) { 
if ($op == 'p') { 
    $sql_op = '>'; 
    $order = 'ASC'; 
} 
elseif ($op == 'n') { 
    $sql_op = '<'; 
    $order = 'DESC'; 
} 
else { 
    return NULL; 
} 

$output = ''; 

// your node must have an image type field 
// let's say it's name is IMAGEFIELD 
// select from node table 
$query = db_select('node', 'n'); 
// join node table with image field table 
$query->leftJoin('field_data_field_IMAGEFIELD', 'i', 'i.entity_id = n.nid'); 
// join file managed table where all data about managed files stored 
$query->leftJoin('file_managed', 'f', 'f.fid = i.field_IMAGEFIELD_fid'); 
$query 
    // select nid and title from node 
    ->fields('n', array('nid', 'title')) 
    // select uri from file_managed (image path) 
    ->fields('f', array('uri')) 
    // select image alt and title 
    ->fields('i', array('field_IMAGEFIELD_alt', 'field_IMAGEFIELD_title')) 
    // where nid "greater than"/"lower than" our current node nid 
    ->condition('n.nid', $nid, $sql_op) 
    // where node type in array('your content types') 
    ->condition('n.type', array('PHOTOS'), 'IN') 
    // where node is published 
    ->condition('n.status', 1) 
    // where requested node has image to display (if you want thumbnail) 
    ->condition('f.uri', '', '!=') 
    // order by nid 
    ->orderBy('n.nid', $order) 
    // limit result to 1 
    ->range($start, 1);  

// make query 
$result = $query->execute()->fetchAll(); 

foreach ($result as $node) { 
    // theme your thumbnail image 
    $variables = array(
     // default image style name `thumbnail` 
     // you can use your own by following 
     // admin/config/media/image-styles on your site 
     'style_name' => 'thumbnail', 
     'path' => $node->uri, 
     'alt' => $node->field_IMAGEFIELD_alt, 
     'title' => $node->field_IMAGEFIELD_title 
    ); 
    $image = theme('image_style', $variables); 

    $options = array(
     'html' => TRUE, 
     'attributes' => array(
      'title' => $node->title 
     ) 
    ); 
    $output = l($image, "node/{$node->nid}", $options); 
} 

return $output; 
} 

Node.tpl.php

<?php print dad_prev_next($node->nid, 'p', 0); ?> 
<?php print dad_prev_next($node->nid, 'n', 0); ?> 

编辑2：

试图使用strcmp功能：

function dad_prev_next($title = NULL, $op = 'p', $start = 0) { 
if ($op == 'p') { 
    $strcmp = '1'; 
    $order = 'ASC'; 
} 
elseif ($op == 'n') { 
    $strcmp = '2'; 
    $order = 'DESC'; 
} 
else { 
    return NULL; 
} 

$output = ''; 

// your node must have an image type field 
// let's say it's name is IMAGEFIELD 
// select from node table 
$query = db_select('node', 'n'); 
// join node table with image field table 
$query->leftJoin('field_data_field_IMAGEFIELD', 'i', 'i.entity_id = n.nid'); 
// join file managed table where all data about managed files stored 
$query->leftJoin('file_managed', 'f', 'f.fid = i.field_IMAGEFIELD_fid'); 
$query 
    // select nid and title from node 
    ->fields('n', array('nid', 'title')) 
    // select uri from file_managed (image path) 
    ->fields('f', array('uri')) 
    // select image alt and title 
    ->fields('i', array('field_IMAGEFIELD_alt', 'field_IMAGEFIELD_title')) 
    // where node type in array('your content types') 
    ->condition('n.type', array('PHOTOS'), 'IN') 
    // where node is published 
    ->condition('n.status', 1) 
    // where requested node has image to display (if you want thumbnail) 
    ->condition('f.uri', '', '!=') 
    // order by nid 
    ->orderBy('n.title', $order) 
    // limit result to 1 
    ->range($start, 1);  

// make query 
$result = $query->execute()->fetchAll(); 

foreach ($result as $node) { 
    // theme your thumbnail image 
    $variables = array(
     // default image style name `thumbnail` 
     // you can use your own by following 
     // admin/config/media/image-styles on your site 
     'style_name' => 'thumbnail', 
     'path' => $node->uri, 
     'alt' => $node->field_IMAGEFIELD_alt, 
     'title' => $node->field_IMAGEFIELD_title 
    ); 
    $image = theme('image_style', $variables); 

    $options = array(
     'html' => TRUE, 
     'attributes' => array(
      'title' => $node->title 
     ) 
    ); 
    $output = l($image, "node/{$node->nid}", $options); 
} 

return $output; 
} 

现在没有错误记录，但总是显示相同的照片 - 第一个和第二个按字母顺序排列的列表中的2个。

Answer 1

您正在按n.title排序，但您的第一个条件是查找大于/小于$ nid的n.nid。这是没有意义的，并导致半随机选择。我建议按标题进行排序时，将条件改为n.title $ node-> title，它会起作用。