Question

我遇到了INT字段的问题。问题是当我在屏幕上打印值是OK但是当我更新数据库寄存器时再添加一个。

$today = date('Y-m-d H:i:s');
$query = "SELECT id_ad, ad_printed FROM ads WHERE (ad_type = \"990x90\" OR (ad_type = \"728x90\" OR ad_type = \"250x90\")) AND ad_date_start <= \"$today\" AND ad_date_finish >= \"$today\"";
$result = mysqli_query($link, $query);
while ($row = mysqli_fetch_array($result)) {
    $new_value = $row['ad_printed'] + 1;
    $curr_id = $row['id_ad'];
    $query_upd = "UPDATE ads SET ad_printed = '".$new_value."' WHERE id_ad = '".$curr_id."' LIMIT 1;";
    $upd = mysqli_query($link, $query_upd); 
}

有人知道会发生什么吗？ I.E.如果原始值为26，则新值必须为27. $new_value为27，但它注册为28 ... :(

Answer 1

如果你想在sql中增加一个值，请不要将列值包装成引号，否则sql会将列值作为字符串处理。请确保您的列类型正确，例如integer而不是varchar

UPDATE ads SET ad_printed = (ad_printed + 1) WHERE id_ad = '".$curr_id."' LIMIT 1;

//修改

如果直接将变量传递给sql，请查看mysqli_real_escape_string函数以防止sql注入。

http://de2.php.net/manual/de/mysqli.real-escape-string.php

sql将一个添加到字段值

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