Question

这是给我这个错误的代码。我已经四处搜索，发现了相关问题，但我无法应用实施。这有些不对劲，超出了我的理解......

function add_user_to_db()
{
    $dbhost = "111.111.111.111";
    $dbuser = "Bob";
    $dbpass = "password";
    $connection = mysql_connect($dbhost, $dbuser, $dbpass);

    if(!$connection){die('Could not connect: '.mysql_error());}
    echo 'Connected to Vikings Game DB!';

    mysql_select_db('vgDB');

    $tb_result = mysql_query("SHOW TABLES LIKE 'Players'");
    $table_exists = mysql_num_rows($tb_result) > 0;
    if($table_exists)
    {
        if( isset($_POST["ID"]) && isset($_POST["NAME"]) && isset($_POST["COMMENT"]) )
        {
            $id          = $_POST['ID'];
            $name        = $_POST["NAME"];
            $comment     = $_POST["COMMENT"];

            //Check if a row exists
            $row_result = mysql_query("SELECT `".$id."` FROM Players");
            if($row_result == FALSE)
            {
                $add_user_query = mysql_query( "INSERT INTO Players( ID , NAME , COMMENT )VALUES('123','Bob','Bob's comment')" );
                $retval = mysql_query($add_user_query, $connection);
                if(!$retval) die("Could not insert data: ".mysql_error());
                echo "User '".$id."' was added successfully!";
            }
            else
            { 
               $name_read = mysql_query("SELECT NAME FROM Players WHERE ID = `".$id."`");
               $comment_read = mysql_query("SELECT COMMENT FROM Players WHERE ID = `".$id."`");
               echo "Reading data of user (`".$id."`): Name = `".$name_read."`; Comment: `".$comment_read."`";
            }
        }
    }
    mysql_close($connection);
}

Answer 1

我认为你的问题在以下几行：

$add_user_query = mysql_query( "INSERT INTO Players( ID , NAME , COMMENT )VALUES('123','Bob','Bob's comment')" );

尝试修改（仅用于测试目的），如下所示：

$add_user_query = mysql_query( "INSERT INTO Players( ID , NAME , COMMENT )VALUES('123','Bob','Bob comment')" );

如果你使用它，就意味着'鲍勃的评论'是问题，鲍勃引起它之后的引用。

PHP错误：查询为空

1 个答案: