MySQLi 更新的增量值-php问题

Increment value on MySQLi Update(MySQLi 更新的增量值)

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问题描述

只是尝试将点添加到名为点"的列中的现有值.我已经阅读了一些建议下面的文章，但它对我不起作用.也许是因为我使用的是 mysqli 而不是 mysql?有什么想法吗?if 语句工作正常.

Simply trying to add points to an existing value in a column named 'Points'. I've read a few articles that suggest the beneath, but it's not working for me. Perhaps because I'm using mysqli rather than mysql? Any thoughts? The if statement works fine.

if (!empty($m1A) && ($r1A == 0)) {
    // Rewards pts
    $query = "UPDATE users SET Points=Points+3 WHERE 1A='$m1A'";
    $result = mysqli_query($conn, $query);

    // Record reward of pts
    $query1 = "UPDATE rounds SET 1A = 1";
    $result2 = mysqli_query($conn, $query1);
}

推荐答案

如果您的列名以数字开头，则必须用反引号将其引用:

If your column name starts with a number, you have to quote it in backticks:

$query = "UPDATE users SET Points=Points+3 WHERE `1A`=$m1A";

和:

$query1 = "UPDATE rounds SET `1A` = 1";

我建议使用带有绑定参数的准备好的语句来避免 sql 注入问题.

And I would recommend using a prepared statement with bound parameters to avoid sql injection problems.

如果您的 1A 列不是整数列并且值是字符串，则需要将它们引用起来.

If your 1A column is not an integer column and the values are strings, you need to quote them.

$query = "UPDATE users SET Points=Points+3 WHERE `1A`='$m1A'";
                                                      ^    ^

虽然这个问题会通过准备好的语句自动解决...

Although that problem would be solved automatically with a prepared statement...

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