在 MySQL UPDATE (PHP/MySQL) 中使用变量
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Using variables in MySQL UPDATE (PHP/MySQL)
提问by MU_FAM
I am using this code so I can update a record in database:
我正在使用此代码，以便可以更新数据库中的记录：
$query = mysql_query("UPDATE article
SET com_count = ". $comments_count
WHERE article_id = .$art_id ");
My question is: How can I use variables in a MySQL UPDATE statement.
我的问题是：如何在 MySQL UPDATE 语句中使用变量。
回答by morgar
$query = mysql_query("UPDATE article set com_count = $comments_count WHERE article_id = $art_id");
$query = mysql_query("UPDATE article set com_count = $comments_count WHERE article_id = $art_id");
You was messing up the quotes and concats.
你弄乱了引号和连接符。
You can use inline vars like the previous example or concat them like:
你可以像前面的例子一样使用内联变量，或者像这样连接它们：
$query = mysql_query("UPDATE article set com_count = " . $comments_count . " WHERE article_id = " . $art_id);
$query = mysql_query("UPDATE article set com_count = " . $comments_count . " WHERE article_id = " . $art_id);
回答by dee-see
You messed up on your " .pattern.
你搞砸了你的" .模式。
$query = mysql_query("UPDATE article set com_count = ". $comments_count . " WHERE article_id = " . $art_id . ");
回答by Fahim Faysal
Use apostrophes when using variables in a MySQL UPDATE statement:
在 MySQL UPDATE 语句中使用变量时使用撇号：
$query = mysql_query("UPDATE article
SET com_count = '$comments_count'
WHERE article_id = '$art_id'");
Be careful about space and apostrophes.
注意空格和撇号。
