Question

我正在尝试为我的论坛创建一个函数，它会将我的用户的“帖子”属性增加1.无论出于何种原因，以下PHP都不起作用。

function postCountIncrease($username) {
    //get the connection variable
    global $con;
    //change to the users database (this function works correctly)
    sqlconnect_users();
    //get current post number (this is also working)
    $getCurrentPosts = "SELECT Posts\n"
    . "FROM users\n"
    . "WHERE Username='".$username."'";
    $query1 = mysqli_query($con, $getCurrentPosts) or die(mysqli_error($con));
    $currentPosts = mysqli_fetch_array($query1);
    //here is the problematic post.  Assume that $username is a valid value, and that I've already done mysqli_real_escape_string() on it
    $incrementPostsQuery = "UPDATE users.users SET Posts=". $currentPosts[0]+1 ." WHERE Username='". $username ."'";
    $query2 = mysqli_query($con, $incrementPostsQuery) or die(mysqli_error($con));
    //return the result
    $result = mysqli_fetch_array($query2);
    return $result;
}

老实说，我看不出我做错了什么，因为SQL运行正常。如果我在控制台中使用UPDATE users.users SET Posts=1 WHERE Username='Lampitosgames'，则无错误。帮助很多。此外，这是它向我投掷的错误：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1 WHERE Username='Lampitosgames''

Answer 1

你无法以"toto ".$var+1的方式连接，你必须用括号"toto ".($var+1)

包围

在您的情况下，这是var $incrementPostsQuery的声明失败

Answer 2

查看您的错误，您的语法已关闭

$getCurrentPosts = "SELECT Posts 
                    FROM users 
                    WHERE Username='$username'";

Answer 3

错误在于构建查询。

$incrementPostsQuery = "UPDATE users.users SET Posts=". $currentPosts[0]+1 ." WHERE Username='". $username ."'";

我会建议您创建这样的查询提示：

“update table set field = value”; //你可以直接写这个值
“update table set field =”。 $值; // easy
“update table set field =”。（$一个+ $ b）中; // ...
“update table set field = {$ value}”; //你可以添加带花括号的变量
“update table set field = {$ va [3]}”; //更无比的方式
“更新表格设置字段= {$ a-＆gt; b}”; //一个对象字段

我的PHP SQL查询抛出错误，即使它在SQL控制台中有效

3 个答案: