令人难以置信的是,我有一个确切的脚本,其中有一个不同于“additional”的变量。但是,即使有一个额外的MySQL列,我无法得到它的更新。我甚至尝试改变其他东西,但它无法与此专栏一起工作。什么是冲突?Mysql更新命令失败
<?php
echo "yup";
$lt2 = "yup";
$blah = "yup";
$username = "yup";
$lt1 = "yup";
$dbh = new PDO(censored);
$sql = "UPDATE purchases SET additional = ? WHERE username = ? AND blah = ? AND kusername = ?";
$q = $dbh->prepare($sql);
$params = array($lt1, $lt2, $username, $blah);
$q->execute($params);
$doc = new DOMDocument();
$r = $doc->createElement("location");
$doc->appendChild($r);
foreach ($q->fetchAll() as $row) {
$e = $doc->createElement("location");
$e->setAttribute('name', $row['additional']);
$r->appendChild($e);
}
print $doc->saveXML();
echo "work";
?>
感谢,但没有奏效 – user1803649
比任何你没有连接到数据库或条件查询不匹配 –
他们这样做。连接起作用。我对此100%肯定。第二个建议是我也想过的,所以我复制并粘贴了该名称,但仍然无效。 – user1803649