douhutongvm382381 2012-09-06 06:33
浏览 27

什么会导致传递给mysql_fetch_object()的资源无效? [关闭]

I got a warning in php as follows

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in E:\Program Files\VertrigoSer\www\b.php on line 16

My program is:

<?php
$connection=mysql_connect("localhost","root","vertrigo");
$db=mysql_select_db("mydb",$connection);

$query="INSERT INTO text VALUES(1003,'PHP','Mcgrowhill','Ivan')";
$r=mysql_query($query);

$query="INSERT INTO book VALUES(201,3,1003)";
$r=mysql_query($query);

$query1="select distinct title from text,book,course where text.isbn=book.isbn and book.cno=course.cno and course.dept='CS'";
$r=mysql_query($query1);

echo"row inserted";

while($row=mysql_fetch_object($r))
{
    echo $row->title;
}

mysql_close($connection);
?>

kindly give suggestions that how to solve this problem

  • 写回答

4条回答 默认 最新

  • dongluo9156 2012-09-06 06:35
    关注

    Here is how you can find out:

    $query = "INSERT INTO text VALUES(1003,'PHP','Mcgrowhill','Ivan')";
    $r = mysql_query($query);
    if (!$r) {
        echo mysql_error();
    }
    
    $query = "INSERT INTO book VALUES(201,3,1003)";
    $r = mysql_query($query);
    if (!$r) {
        echo mysql_error();
    }
    
    $query1 = "select distinct title from text, book, course where text.isbn = book.isbn and book.cno = course.cno and course.dept = 'CS'";
    $r = mysql_query($query1);
    if (!$r) {
        echo mysql_error();
    } else {
        echo"row inserted";
    
        while($row = mysql_fetch_object($r)) {
            echo $row->title;
        }
    }
    
    mysql_close($connection); ?>
    
    评论

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