Question

我的查询中有一些不好的sytax，我正在尝试创建'prev_id'以进入$ row结果链接。有关更好的方法的任何想法吗？

编辑：错误是“检查与您的MySQL服务器版本相对应的手册，以便在'prev_id'附近使用正确的语法。”

function traversePhoto($the_selected_id) {
global $pdo;

$id = $the_selected_id;

$stmt_a = $pdo->prepare("
    (SELECT * FROM images WHERE id < '.intval($id).' ORDER BY id DESC LIMIT 1)
     UNION (SELECT * FROM images WHERE id = (SELECT MAX(id) FROM images)) LIMIT 1 prev_id");
$stmt_b = $pdo->prepare("
    (SELECT * FROM images WHERE id > '.intval($id).' ORDER BY id ASC LIMIT 1)
     UNION (SELECT * FROM images WHERE id = (SELECT MIN(id) FROM images)) LIMIT 1 next_id");

// $vars = array(':id' => $id);
$prev = $stmt_a->execute();
$next = $stmt_b->execute();

if ($prev) {
    while($row = $stmt_a->fetchObject()) {
        echo '<a href="' . $row['prev_id'] . '">Previous</a>';
    } 
} else {
    echo 'no previous';
}

if ($next) {
    while($row = $stmt_b->fetchObject()) {
        echo '<a href="' . $row['next_id'] . '">Next</a>';
    }
} else {
    echo 'no next';
}

}

Answer 1

尝试这样

<?php function traversePhoto($the_selected_id) {
global $pdo;

$id = $the_selected_id;

$stmt_a = $pdo->prepare("
(SELECT * FROM images WHERE id < ? ORDER BY id DESC LIMIT 1)
 UNION (SELECT * FROM images WHERE id = (SELECT MAX(id) FROM images)) LIMIT 1");
$stmt_b = $pdo->prepare("
(SELECT * FROM images WHERE id > ? ORDER BY id ASC LIMIT 1)
 UNION (SELECT * FROM images WHERE id = (SELECT MIN(id) FROM images)) LIMIT 1");

// $vars = array(':id' => $id);
$prev = $stmt_a->execute(array( (int)$id ));
$next = $stmt_b->execute(array( (int)$id ));

if ($stmt_a->rowCount() > 0) {
    while($row = $stmt_a->fetch(PDO::FETCH_ASSOC)) {
        echo '<a href="' . $row['prev_id'] . '">Previous</a>';
    } 
 } else {
    echo 'no previous';
}

if ($stmt_b->rowCount() > 0) {
   while($row = $stmt_b->fetch(PDO::FETCH_ASSOC)) {
      echo '<a href="' . $row['next_id'] . '">Next</a>';
  }
} else {
echo 'no next';
}

使用PDO获取上一行的下一行来创建链接

1 个答案: