PHP PDO 准备&执行 MySQL UPDATE 语句不起作用 - 难住了！

发布于 2024-11-09 02:33:24 字数 516 浏览 4 评论 0原文

我正在为我的应用程序开发一个超级简单的计数器。我可以使用“$date”和“$c”插入行，但更新不起作用。

这有效：

$c = 8;
$today = date('Y-m-d');
$insert_count = $db->prepare("INSERT INTO COUNTER (COUNTER.date, clicks) VALUES ('" . $today . "', ?)");
$insert_count->execute(array($c));

这不

$c = 8;
$today = date('Y-m-d');
$update_count = $db->prepare("UPDATE COUNTER SET clicks = clicks + ? WHERE COUNTER.date = '" . $today . "'");
$update_count->execute(array($c));

没有错误，什么也没有。

原文

I'm working on a super-simple counter for my application. I am able to insert rows using '$date' and '$c' just fine, but updating does not work.

This works:

$c = 8;
$today = date('Y-m-d');
$insert_count = $db->prepare("INSERT INTO COUNTER (COUNTER.date, clicks) VALUES ('" . $today . "', ?)");
$insert_count->execute(array($c));

This does not

$c = 8;
$today = date('Y-m-d');
$update_count = $db->prepare("UPDATE COUNTER SET clicks = clicks + ? WHERE COUNTER.date = '" . $today . "'");
$update_count->execute(array($c));

No errors, nothing.

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杀お生予夺 2024-11-16 02:33:24

我不认为？绑定占位符可用于计算。您是否尝试过使用命名绑定参数？

$update_count = $db->prepare("UPDATE COUNTER SET clicks = clicks + :increment WHERE COUNTER.date = '" . $today . "'");
$update_count->execute(array("increment"=>$c));

当你说没有错误时，这个命令执行后没有输出？

print_r($db->errorInfo());

I don't think the ? binding placeholder works with calculations. Have you tried using named binding parameters instead?

$update_count = $db->prepare("UPDATE COUNTER SET clicks = clicks + :increment WHERE COUNTER.date = '" . $today . "'");
$update_count->execute(array("increment"=>$c));

When you say there are no errors, there is no output for this command after execute?