错误更新记录你在你的SQL语法错误

问题描述 投票:0回答:2

我尝试更新我的表,但是我给这个错误,请帮助我

function updatePost($id , $title, $content, $date, $groups)
    {
        $connection = mysqli_connect(DataBaseManager::HOST, DataBaseManager::USER, DataBaseManager::PASSWORD, DataBaseManager::DATABASENAME);
        $sqlCommand = "UPDATE posts
          SET title = '$title', content = '$content', date = '$date' , groups = '$groups' 
          WHERE id == 1";
        if ($connection->query($sqlCommand) === TRUE) {
            echo "Record updated successfully";
        } else {
            echo "Error updating record: " . $connection->error;
        }

        $connection->close();
    }
php json
2个回答
0
投票

where条款有2个等号。将其更改为一个等号......还,日期是保留字,所以你需要将它包装在反引号。

$sqlCommand = "UPDATE posts
    SET title = '$title'     , 
         content = '$content',
        `date`   = '$date'   , 
         groups  = '$groups' 
    WHERE id = 1";

0
投票

你需要把你的SQL查询像串并相应CONCAT您的变量:

"UPDATE posts SET title = '".$title."', content = '".$content."', date = '".$date."' , likes = '".$likes."', groups = '".$groups."' WHERE id = '".$id."'";
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