【发布时间】:2017-10-21 11:18:40
【问题描述】:
我正在 postgresql 中开发一个简单的问题标签数据库关联。一个问题可以有多个标签,一个标签可以关联多个问题。这就是为什么我有一个 M-M 表,问题标签。 使用 php 我想创建一个问题并验证给定的标签是否已经存在于标签表中。如果它们确实存在:
- 不要将我的标签添加到标签表中
- 将预先存在的标签与我的 questiontags 表中的问题相关联
如果标签不存在,我想将它添加到标签表中,然后创建一个关联。
为此,我正在尝试这样的事情:
function update_tags($questionid, $tags) {
global $conn;
//Check if question already exists. If Yes, delete it from the array -> //EDIT PROPOSES
$questiontags = get_all_tags();
$existant_tags = [];
foreach ($questiontags as $tag_idx => $tag) {
if(in_array($tag['name'], $tags)){
$key = array_search($tag['name'], $tags);
unset($tags[$key]);
$existant_tags[] = $tag['tagid'];
associate_only_tag($tag['tagid'], $questionid);
}
$questiontags[$tag_idx] = $tag['tagid'];
}
foreach ($tags as $tag) {
associate_tag($tag, $questionid);
}
$tags_to_delete = array_diff($questiontags, $existant_tags);
foreach ($tags_to_delete as $tagid) {
delete_tag_from_question($tagid, $questionid);
}
}
function get_all_tags() {
global $conn;
$query=$conn->prepare("SELECT tags.tagid, tags.name FROM tags ");
$query->execute();
return $query->fetchAll();
}
function get_tags_from_question($questionid) {
global $conn;
$query=$conn->prepare("SELECT tags.tagid, tags.name FROM tags
INNER JOIN questiontags
ON tags.tagid = questiontags.tagid
WHERE questiontags.questionid = :question
");
$query->execute(['question' => $questionid]);
return $query->fetchAll();
}
function insert_tag($tag)
{
global $conn;
$stmt = $conn->prepare("INSERT INTO tags (name) VALUES(:tag)");
$stmt->execute([$tag]);
return (int)$conn->lastInsertId();
}
function associate_tag($tag, $questionid)
{
global $conn;
$tagid = insert_tag($tag);
$stmt = $conn->prepare("INSERT INTO questiontags (questionid, tagid) VALUES(:question, :tag)");
$stmt->execute(['question' => $questionid, 'tag' => $tagid]);
}
function associate_only_tag($tagid, $questionid)
{
global $conn;
$stmt = $conn->prepare("INSERT INTO questiontags (questionid, tagid) VALUES(:question, :tag)");
$stmt->execute(['question' => $questionid, 'tag' => $tagid]);
}
function delete_tag_from_question($tagid, $questionid) {
global $conn;
$query = $conn->prepare("DELETE FROM questiontags WHERE questionid = :question AND tagid = :tag");
$query->execute(['question' => $questionid, 'tag' => $tagid]);
}
问题是这仅适用于新问题,而不适用于我更新问题时。当我做associate_only_tag 时,我需要检查问题是否存在然后更新而不是尝试创建新行questiontags。经过一番努力,我无法弄清楚。
有没有办法做到这一点?
【问题讨论】:
标签: php postgresql common-table-expression upsert