php pdo 查询不起作用

Question

我一直在尝试使用prepare()、bindParam() 和execute() 函数来实现PDO，以允许根据用户输入的数据构造查询。

我想显示书籍列表，然后允许用户过滤列表，然后查看新列表和完整列表。

当我在表单中输入条件进行搜索时，什么也没有发生。我在忽略什么？

这是代码

<?php
        $pageTitle = "Book List";
        $pageHeading = "Book List";
        include_once ('header.php');
        include_once('databaseConnection.php');


        if(isset($_POST['txtSearchBookTitle'])) {
            $db = new DatabaseConnection();
            $db = $db->db_connection;
            $searchTitle = ($_POST['txtSearchBookTitle']);
            $sql = $db->prepare("SELECT title FROM tblBook WHERE title LIKE ('%:searchTitle%') ORDER BY title");
            $sql->bindParam(':searchTitle', $searchTitle);
            $sql->execute();
            $result = $sql->fetchAll();

            print_r($result);

            foreach ($result as $row) {
                echo  "<li>" .  " " . $row["title"]. " " . "</li>";
              } 
        }

    ?>
        <form name="searchBookTitle" method="post" action="<?php echo htmlentities($_SERVER['PHP_SELF']);?>" >
            <fieldset>
                <legend>Search Books</legend>
                <label for="txtSearchBookTitle">Search by Book Title</label>
                <input type="text" name="txtSearchBookTitle" id="txtSearchBookTitle">
                <input type="submit" value="Submit">
            </fieldset>
        </form>

    <?php 
            include_once('getBooks.php'); 
            getBooks(); 
        include 'footer.php';
    ?>

Answer 1

您需要以这种方式准备输入:

$searchTitle = $_POST['txtSearchBookTitle'];
$sql = $db->prepare("SELECT title FROM tblBook WHERE title LIKE :searchTitle ORDER BY title");
$sql->execute(array(':searchTitle' => '%' . $searchTitle . '%'));

或者像这样:

$searchTitle = $_POST['txtSearchBookTitle'];
$sql->bindParam(':searchTitle', "%{$searchTitle}%");