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SQL 查询有什么问题?

[英]What's wrong within SQL query?

I'm using this query to get range of rows for example 0-5 , But I get an error message.我正在使用此查询来获取行范围,例如 0-5 ,但我收到一条错误消息。

Here is the code:这是代码:

if(isset($_GET['page']) && isset($_GET['per'])){        
    $per = $_GET['per'];
    $pageNumber = $_GET['page'];
    $from = ($pageNumber * $per) - $per;
    $results = $pdo->prepare('SELECT * FROM users LIMIT :sfrom , :sto');
    $results->execute(array('sfrom' => $from , 'sto' => $per));
}

I get the following error :我收到以下错误:

Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax;致命错误:未捕获的异常 'PDOException' 带有消息 'SQLSTATE[42000]:语法错误或访问冲突:1064 您的 SQL 语法有错误; check the manual that corresponds to your MariaDB server version for the right syntax to use near ''5' , '5''检查与您的 MariaDB 服务器版本相对应的手册,了解在 ''5' , '5'' 附近使用的正确语法

I think that's because the $from and $per are inserted as strings or something I tried to convert them to int using (int)$from , (int)$per But still having the same error我认为这是因为$from$per是作为字符串插入的,或者我试图使用(int)$from , (int)$per将它们转换为 int 但仍然有相同的错误

The values passed must be integers but casting them is not enought.传递的值必须是整数,但转换它们是不够的。

Instruct PDO the binded parameters' type is integer by using PDOStatement:: bindParam使用PDOStatement:: bindParam指示PDO绑定参数的类型为整数

Then call PDOStatement:: execute as last step (without parameters).然后调用PDOStatement:: execute作为最后一步(不带参数)。

$results = $pdo->prepare('SELECT * FROM users LIMIT :sfrom , :sto');
$results->bindParam(':sfrom', (int)$from, PDO::PARAM_INT );
$results->bindParam(':sto',   (int)$to,   PDO::PARAM_INT );
$results->execute();

See http://php.net/manual/en/pdostatement.bindparam.phphttp://php.net/manual/en/pdostatement.bindparam.php

Setting $per and $from as numeric strings is totally OK.$per$from设置为数字字符串是完全可以的。 Make sure you are not passing "'5'" as a value.确保您没有将"'5'"作为值传递。 Casting "'5'" to int results in 0 ."'5'"int结果为0 Try尝试

// remove " and ' then cast to int
$yourNumber = (int) str_replace(['"', "'"], '', $yourInput);

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