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我正在尝试从四个单独的文本字段中收集信息并在后端生成查询以显示结果。我已经尝试了许多已经在这里提出的解决方案，但我不断得到重复的结果。我让它处理一个查询：

<?php    
$query = "SELECT * FROM book WHERE booktitle LIKE '%" . mysql_real_escape_string($_POST['title']) . "%'";
$result = mysqli_query($con, $query) or die("Error in query");
while ($row = mysqli_fetch_array($result)) {
?>
<img src="<?php echo $row[12]; ?>" width="112px" height="150px" />
<?php
echo "<b>Title: </b>" . $row[1] . "<br />";
echo "<b>Author: </b>" . $row[2] . "<br />";
echo "<b>Price:</b> &pound;" . $row[9] . "<br />";
echo "<b>Description:</b> " . substr($row[3],0,320) . "...<br /><hr>";
}

但我还需要以某种方式合并其他 3 个结果？

$query = "SELECT * FROM book WHERE author LIKE '%" . mysql_real_escape_string($_POST['author']) . "%'";
$query = "SELECT * FROM book WHERE isbn LIKE '%" . mysql_real_escape_string($_POST['isbn']) . "%'";
$query = "SELECT * FROM book WHERE description LIKE '%" . mysql_real_escape_string($_POST['keyword']) . "%'";

任何帮助将不胜感激。提前致谢。

4

2 回答 2

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<?php
$mysqli = new mysqli("localhost", "username", "password", "database_name");

$book_title = mysql_real_escape_string($_POST['title']);
$author = mysql_real_escape_string($_POST['author']);
$isbn = mysql_real_escape_string($_POST['isbn']);
$keyword = mysql_real_escape_string($_POST['keyword']);

$query = "SELECT * FROM book WHERE booktitle LIKE '%{$book_title}' OR author LIKE'%" .     $author . "%'
OR isbn LIKE '%" . $isbn . "%' OR description LIKE '%" . $keyword . "%'";
$result = $mysqli->query($query);

 while($row = $result->fetch_array()):?>
<img src="<?=$row['image_column_name']; ?>" width="112px" height="150px" />
    <b>Title:</b> <?=$row['title_column_name']; ?> <br>
    <b>Author:</b> <?=$row['author_column_name']; ?> <br>
    <b>Price:</b> <?=$row['price_column_name']; ?> <br>
    <b>Description:</b> <?=substr($row['desc_column_name'],0,320)?>
<hr>
<?php endwhile; ?>
于 2013-03-22T00:04:46.140 回答
0

您可以使用带有 LIKE 关键字的 OR 语句同时搜索所有这些。

$query = "SELECT * FROM book WHERE booktitle LIKE '%" . $title . "%' OR author LIKE'%" . $author . "%' OR isbn LIKE '%" . $isbn . "%' OR description LIKE '%" . $keyword . "%';

如果变量为空，那么您将拥有WHERE booktitle LIKE '%%'，它返回所有行。您将需要根据输入构建查询。

// Determine the WHEREs to use
$where = array();
if ( ! empty($title))
    $where[] = "booktitle LIKE '%" . $title . "%'";
if ( ! empty($author))
    $where[] = "author LIKE '%" . $author . "%'";
if ( ! empty($isbn))
    $where[] = "isbn LIKE '%" . $isbn . "%'";
if ( ! empty($description))
    $where[] = "description LIKE '%" . $description . "%'";

// Build the query
$query = 'SELECT * FROM book';
if ( ! empty($where))
    $query .= ' WHERE ' . implode(' OR ', $where);

注意：为了便于阅读，我省略了转义输入。

于 2013-03-22T00:04:46.127 回答