PHP PDO不會更新數據庫

Question

我嘗試使用PDO，但是我得到1,1,1的結果，但是數據庫不會更新。 我不知道我的代碼有什么問題。

try {$db = new PDO("mysql:dbname=$db_name;host=$db_host", $db_user, $db_pass );}
catch(PDOException $e){echo $e->getMessage();}

$title_insert = $db->prepare("UPDATE topics SET topic_title = ? WHERE id = ?");
$title_insert->bindParam(1, $id);
$title_insert->bindParam(2, $topic_title);

$tag_insert = $db->prepare("UPDATE topics SET topic_tags = ? WHERE id = ?");
$tag_insert->bindParam(1, $id);
$tag_insert->bindParam(2, $topic_tags);

$story_insert = $db->prepare("UPDATE topics SET topic_story = ? WHERE id = ?");
$story_insert->bindParam(1, $id);
$story_insert->bindParam(2, $topic_story);


$id = 1;
$topic_title = "title";
$topic_tags = "tags";
$topic_story = "story";

$result_title = $title_insert->execute();
echo $sonuc_title.',';

$sonuc_tag = $tag_insert->execute();
echo $sonuc_tag.',';

$result_story = $story_insert->execute();
echo $result_story;

感謝您的回答...

Answer 1

您是否嘗試過切換參數編號？

$title_insert = $db->prepare("UPDATE topics SET topic_title = ? WHERE id = ?");
$title_insert->bindParam(1, $topic_title);
$title_insert->bindParam(2, $id);

祝好運！

PDO綁定參數

Answer 2

您應該在bind語句之前影響變量：

$id = 1;
$topic_title = "title";
$topic_tags = "tags";
$topic_story = "story";

$title_insert = $db->prepare("UPDATE topics SET topic_title = ? WHERE id = ?");
$title_insert->bindParam(1, $id);
$title_insert->bindParam(2, $topic_title);

$tag_insert = $db->prepare("UPDATE topics SET topic_tags = ? WHERE id = ?");
$tag_insert->bindParam(1, $id);
$tag_insert->bindParam(2, $topic_tags);

$story_insert = $db->prepare("UPDATE topics SET topic_story = ? WHERE id = ?");
$story_insert->bindParam(1, $id);
$story_insert->bindParam(2, $topic_story);


$result_title = $title_insert->execute();
echo $sonuc_title.',';

$sonuc_tag = $tag_insert->execute();
echo $sonuc_tag.',';

$result_story = $story_insert->execute();
echo $result_story;