Question

在PDO和PHP中，我研究了一些我认为是查询语句函数的工作解决方案的研究。现在，它与语句

错误

PHP警告：PDOStatement :: execute（）：SQLSTATE [HY093]：参数号无效：绑定变量的数量与/var/www/html/php/php_tools/private/functions.php中的令牌数量不匹配第39行

这是我的临时文件的第15行，以使这更容易阅读而不是通过多个文件，它仍然工作相同，我检查。

 <?php

try {
  $db = new PDO("mysql:host=$host;dbname=$base;", $user, $pass);
}catch(Exception $e){
     echo $error = $e->getMessage();
}

function execute_query($con, $query, $statements) {

$query_string = $con->prepare($query);

foreach ($statements as $statement) {

    $query_string->bindValue(1, $statement['string'], PDO::PARAM_STR);
    $query_string->bindValue(2, $statement['value'], PDO::PARAM_STR);
    $query_string->bindValue(3, $statement['type'], PDO::PARAM_STR);// I believe this is the culprit?

    }

$query_string->execute();


return $query_string->fetchAll();
 }

   $multi_item_query = "SELECT t.t_id as id, t.item_code as code, 
 t.item_name as name,
 t.retail_price as retail, t.sale_price as price,
 t.item_pieces as  pieces, t.qty as quantity,
 t.sold as sold, t.description as description,
 b.brand as brand, c.category as category,
 tt.tool_type as sections, i.image as image
 FROM Tools as t
 INNER JOIN Brands as b on t.b_id = b.b_id
 INNER JOIN Categories as c ON t.c_id = c.c_id
 INNER JOIN Images AS i ON t.t_id = i.t_id
 LEFT OUTER JOIN Types AS tt ON t.tt_id = tt.tt_id
 WHERE tt.tool_type = :tool";

if ( isset($_GET['cat']) ) {
     if ( $_GET['cat'] == 'wrenches') {
        $page_name = 'Wrenches';
        $section = 'wrenches';
        $param = 'wrenches';
    } elseif ( $_GET['cat'] == 'blades') {
         $page_name = 'Blades';
        $section = 'blades';
        $param = 'blades';
    } else {
        $page_name = 'Full Catalog';
         $section = null;
    }
}
 $con = $db;
 $statement = array(); // Prepare a statement array.
$id = array(
     'string' => ':tool',
     'value' =>  $param,
     'type' => PDO::PARAM_STR
 );

 $statement[] = $id;

 ?>

<?php  $items = execute_query($con, $multi_item_query, $statement); ?>

正好在＆＃34; bindValue＆＃34;它是foreach循环中现在破坏的地方。随着我的学习更多做更多的研究我相信我的底部＆＃34; bindValue＆＃34;是罪魁祸首？但我不确定我会用什么作为PDO ::？项目宣布为..？任何帮助或指导都是不完美的..因为我可能会离开..还是新的，还有任何反馈请吗？

Answer 1

你使你的功能过于复杂，以至于你自己没有使用它。让它更简单，像这样：

function execute_query($con, $query, $variables) {
    $stmt = $con->prepare($query);
    $stmt->execute($variables)
    return $stmt;
}

所以你能够以这种方式运行它

$con = $db;
$variables['tool'] = $param;
$items = execute_query($con, $multi_item_query, $variables)->fetchAll();

Answer 2

肯定会是

foreach ($statements as $statement) {
    $query_string->bindValue($statement['string'], $statement['value'],$statement['type'] );
}

循环中的execute方法

foreach ($statements as $statement) {
    $query_string->bindValue($statement['string'], $statement['value'],$statement['type'] );
    $query_string->execute();
}

Answer 3

您尝试传递绑定的所有部分，但尝试单独绑定它们。您需要将语句的所有部分传递给一个绑定值：

foreach ($statements as $statement) {

    $query_string->bindValue($statement['string'], $statement['value'], $statement['type']);

}

PDO查询执行invaid，问题与bindValue不匹配

3 个答案: