php - "number of bound variables does not match number of tokens"——但确实如此

Question

我见过很多人收到错误 SQLSTATE[HY093]: Invalid parameter number: number of bound variables does not match number of tokens，但据我所知，他们都没有问题对我有帮助或适用。我确信这对我来说真的很愚蠢......

当我回显 $ps 和 $ps_val(在我的数据库类和页面上)时，它们具有相同数量的值。我什至会回显 SQL 并计数并仔细检查占位符的拼写，所有这些都会加起来。我什至进入了 phpMyAdmin，复制粘贴了我的 SQL，并复制粘贴了所有占位符及其值，这很有效，所以我知道这不是 SQL 语法。

$ps[] = ":category";
$ps_val[] = $_GET['cat'];

if (isset($_GET['color'])){
    // resort_color converts hex color into comma delimited text colors
    // e.g. resort_color("#000000") returns "black, black with white trim, etc."
    $colors = resort_color($_GET['color']);

    if (strpos($colors, ",") >= 0){
        $colors = explode(",", $colors);

        $i = 0;
        foreach ($colors as $color){
            $color = trim($color);
            if ($i == 0){
                $sql_color = " AND `Color` LIKE \"%:color{$i}%\"";
                $ps[] = ":color{$i}";
                $ps_val[] = $color;
            } else {
                $sql_color = $sql_color . " OR `Color` LIKE \"%:color{$i}%\"";
                $ps[] = ":color{$i}";
                $ps_val[] = $color;
            }
            $i++;
        }
    } else {
        $sql_color = 'AND `Color` LIKE \"%:color%\"';
        $ps[] = ":color";
        $ps_val[] = $colors;
    }
} else {
    $sql_color = "";
}

$sql[] = 'SELECT * FROM `bullet_catalog` WHERE `Category` = :category ' . $sql_color;
$sql[] = 'SELECT * FROM `leeds_catalog` WHERE `Category` = :category ' .  $sql_color;
$sql[] = 'SELECT * FROM `trimark_catalog` WHERE `Category` = :category ' .  $sql_color;

if ($db->query_all($sql, $ps, $ps_val)){
    $products = $db->query_all($sql, $ps, $ps_val);
} else {
    $products = "";
}

还有我的query_all 函数:

public function query_all($sql, $ps="", $ps_val=""){
    if (!$sql){
        return false;
    } else {
        foreach ($sql as $ind_sql){
            $query = $this->pdo->prepare($ind_sql);
            if ($ps){
                if (is_array($ps)){
                    $i = 0;
                    foreach ($ps as $pss){
                        $query->bindValue($pss, $ps_val[$i]);
                        $i++;
                    }
                } else {
                    $query->bindValue($ps, $ps_val);
                }
            }

            try {
                $query->execute();

                $result = $query->fetchAll(PDO::FETCH_ASSOC);
                foreach ($result as $res){
                    $results[] = $res;
                }   
            } catch (PDOException $error){
                echo "<div><pre>" . print_r($error->getMessage(), true) . "</pre></div>";
            }
        }
        if (isset($results)){
            return $results;
        } else {
            return false;
        }
    }
}

PDOStatement::debugDumpParams 返回

SQL: [295] SELECT * 
FROM `table` 
WHERE `Category` = ":category" 
AND `Color` LIKE "%:color0%" 
OR `Color` LIKE "%:color1%" 
OR `Color` LIKE "%:color2%" 
OR `Color` LIKE "%:color3%" 
OR `Color` LIKE "%:color4%" 
OR `Color` LIKE "%:color5%" 
OR `Color` LIKE "%:color6%" 

Params: 8 

Key: Name: [9] :category 
paramno=-1 
name=[9] ":category" 
is_param=1 param_type=2 

Key: Name: [7] :color0 
paramno=-1 
name=[7] ":color0" 
is_param=1 param_type=2 

Key: Name: [7] :color1 
paramno=-1 
name=[7] ":color1" 
is_param=1 param_type=2 

Key: Name: [7] :color2 
paramno=-1 
name=[7] ":color2" 
is_param=1 param_type=2 

Key: Name: [7] :color3 
paramno=-1 
name=[7] ":color3" 
is_param=1 param_type=2 

Key: Name: [7] :color4 
paramno=-1 
name=[7] ":color4" 
is_param=1 param_type=2 

Key: Name: [7] :color5 
paramno=-1 
name=[7] ":color5" 
is_param=1 param_type=2 

Key: Name: [7] :color6 
paramno=-1 
name=[7] ":color6" 
is_param=1 param_type=2 

Answer 1

如果你使用PDO bindValue要使用 LIKE 条件进行搜索，您不能将百分比和引号放在参数占位符 '%:color0%' 中。

这是错误的:

SELECT * FROM `bullet_catalog` WHERE `color` LIKE '%:color0%'

正确的解决方案是像这样只使用占位符:

SELECT * FROM `bullet_catalog` WHERE `color` LIKE :color0

然后将百分比添加到存储关键字的 php 变量中:

$query->bindValue(':color0', "%" . $color0 . "%");

另外 % 和 _ 在 $color0 中没有转义，所以你应该自己转义它们。

也有解释

• How do I create a PDO parameterized query with a LIKE statement?
• implement LIKE query in PDO