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Php MYSQL和状态不工作_Php_Mysql - Fatal编程技术网

Php MYSQL和状态不工作

Php MYSQL和状态不工作,php,mysql,Php,Mysql,我已经编写了获取mysql_num_rows()的代码,但是输出总是为零 $myc= "SELECT * FROM scores WHERE user_id=$userid AND quiz_id=$myqid"; $completed=mysql_query($myc,$kpsctuts->connect) or die('MySql Error! ' . mysql_error()); echo "<br/> row count: ".mysql_num_rows($

我已经编写了获取mysql_num_rows()的代码,但是输出总是为零

$myc= "SELECT * FROM scores WHERE user_id=$userid AND quiz_id=$myqid";
$completed=mysql_query($myc,$kpsctuts->connect) or  die('MySql Error!  ' . mysql_error());
echo "<br/> row count:  ".mysql_num_rows($completed);
if (mysql_num_rows($completed) > 0) {
exit;
}
$myc=“从用户id=$userid和测验id=$myqid的分数中选择*”;
$completed=mysql\u query($myc,$kpsctuts->connect)或die('mysql Error!'.mysql\u Error());
echo“
行数:“.mysql\u num\u行($completed); 如果(mysql_num_rows($completed)>0){ 出口 }
尝试使用此方法从mysql服务器获取数据

$connect=@mysql_connect("localhost","root","password");
mysql_select_db("db_name",$connect);

$myc= "SELECT * FROM scores WHERE user_id=$userid AND quiz_id=$myqid";
$completed=mysql_query($myc);
echo mysql_error();

echo "<br/> row count:  ".mysql_num_rows($completed);
if (mysql_num_rows($completed) > 0) {
exit;
}
$connect=@mysql\u connect(“localhost”、“root”、“password”);
mysql_select_db(“db_name”,$connect);
$myc=“从用户id=$userid和测验id=$myqid的分数中选择*”;
$completed=mysql\u查询($myc);
echo mysql_error();
echo“
行数:“.mysql\u num\u行($completed); 如果(mysql_num_rows($completed)>0){ 出口 }

这仅基于mysql\u查询,它有助于在执行查询之前建立连接

mysql\u查询($myc,$kpsctuts->connect)
?是否使用OOP类型的连接进行程序化的mysql\u查询?意味着您可能正在混合使用
mysql.*
mysqli.*
libraries@CD001不一定意味着OP混合了2个驱动程序。他本可以写自己的类,尽管我对此深表怀疑。@Daan-没错,也可能是OOP风格的PDO。。。我只是瞄准了最有可能的罪犯;)而且评论还足够新,可以编辑,所以从类调用Pits时,数据库连接正在工作fine@Gunasegar请注意,mysql库非常不推荐使用!您应该使用mysqli或PDO。看一看它的位置,让每个人都能看到@欢迎你。拉斐尔·穆勒将尝试调查PDO或迈斯克利。