php - SQLSTATE[HY093] : Invalid parameter number: parameter was not defined : PHP PDO Object

Question

我知道这个错误之前已得到解答，但我无法解决我的代码中的问题。

这是我的代码:

//Basic SELECT query
$select = 'SELECT id, joketext';
$from = ' FROM joke';
$where = ' WHERE TRUE';

$placeholders = array();

if ($_GET['author'] != ' ') //author is selected
{
    $where .= 'AND authorid = :authorid';
    $placeholders[':authorid'] = $_GET['author'];
}

if ($_GET['category'] != ' ') //category is selected
{
    $from .= 'INNET SELECT jokecategory ON id = jokeid';
    $where .= 'AND categoryid = :categoryid';
    $placeholders[':categoryid'] = $_GET['category'];
}

if ($_GET['text'] != ' ') //text was specified
{
    $where .= 'AND joketext LIKE :joketext';
    $placeholders[':joketext'] = '%' . $_GET['text'] . '%';
}

try
{
    $sql = $select . $from . $where;
    $s = $pdo->prepare($sql);
    $s->execute($placeholders);
}
catch (PDOException $e)
{
    $error = 'Error fetching jokes.' . $e->getMessage();
    include 'error.php';
    exit();
}

如果有人向我解释为什么会出现此错误，我将不胜感激？谢谢!

Answer 1

您需要在 $where 和 $from 的每次添加之前添加空格。

if ($_GET['author'] != ' ') //author is selected
{
    $where .= ' AND authorid = :authorid';
    $placeholders[':authorid'] = $_GET['author'];
}

if ($_GET['category'] != ' ') //category is selected
{
    $from .= ' INNER JOIN jokecategory ON id = jokeid';
    $where .= ' AND categoryid = :categoryid';
    $placeholders[':categoryid'] = $_GET['category'];
}

if ($_GET['text'] != ' ') //text was specified
{
    $where .= ' AND joketext LIKE :joketext';
    $placeholders[':joketext'] = '%' . $_GET['text'] . '%';
}

如果您不这样做，您会得到如下所示的查询:

SELECT id, joketext FROM joke WHERE TRUEAND authorid = :authoridANDjoketext = :joketext