upsert 期间出现重复键错误 [解释]
我正在类中执行以下语句。这段代码来自
$query = array('_id' => $id, 'lock' => 0);
$update = array('$set' => array('lock' => 1));
$options = array('safe' => true, 'upsert' => true);
$result = $this->_mongo->update($query, $update, $options);
if ($result['ok'] == 1) {
return true;
}
但是我不明白如何得到重复的键错误。 有人可以解释一下我收到此错误的可能情况和可能性吗？
我一直在广泛研究这个问题，但在任何地方都找不到答案。因此，如果它在 SO 或任何其他网站上，请分享！
提前致谢。
I am executing the below statement in a class. This code is from
$query = array('_id' => $id, 'lock' => 0);
$update = array('$set' => array('lock' => 1));
$options = array('safe' => true, 'upsert' => true);
$result = $this->_mongo->update($query, $update, $options);
if ($result['ok'] == 1) {
return true;
}
However I do not understand how I would get a duplicate key error.
Can someone explain the possible scenarios and likelihood that I will receive this error?
I have been researching this extensively, cannot find my answer anywhere. So if it is on SO or any other website please share!
Thanks in advance.
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由于您正在进行更新插入并在查询中包含
_id，因此您不应该在该键上获得任何重复项。这让我认为您已经在lock上创建了一个唯一索引，该索引不适用于 2 个以上的文档，因为该字段只有 2 个值。如果您尚未锁定唯一索引，则您必须在此处未显示的字段上拥有唯一索引。这也不起作用，因为在插入时，您的 upsert 将仅设置
_id和lock，任何其他具有索引的字段将被插入为null。如果这些字段之一具有唯一索引，则只有单个文档在该字段中可以具有null。因此，当您尝试为该字段插入另一个null时，您将收到重复键错误。Since you're doing an upsert and including
_idin your query, you shouldn't be getting any duplicates on that key. This makes me think that you've created a unique index onlock, which isn't going to work for more than 2 documents because you only have 2 values for that field.If you haven't put a unique index on lock, then you must have a unique index on a field you aren't showing here. That won't work either because on an insert, your upsert is going to set
_idandlockonly, any other field with an index will be inserted asnull. If one of those fields has a unique index, then only a single document can have anullin that field. So when you try and insert anothernullfor that field, you'll get a duplicate key error.