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我有这个 PHP 代码：

第一次查找 query-1 工作正常。但第二个 query2 工作正常，但 mysql_affected_rows() 没有告诉我记录是否已更新，为什么？

谢谢。

$uemail  = 'driggg2@gmail.com';
$authkey = '2f4071bffda44aa30064364055687d13';

require_once ('./domainsiteconnect.php'); // Connect to the database.

if ( $uemail && $authkey)
{

echo "<br>--uemail: [$uemail] -- && -- [$authkey] --- ";

$query1 = "SELECT * FROM member WHERE email='$uemail' AND actkey='$authkey' AND status='pending' ";

    $result1 = mysql_query ($query1);
    $num1 = mysql_num_rows ($result1);

    if ( $num1 == 1 )  // Exactly 1 rec found - Update member to verified.
    {
       echo '<br>+++ query1: successful +++>br>';

        $status = 'verified';

$query2 = " UPDATE member SET status='$status' WHERE email='$uemail'   ";    //AND actkey='$authkey'  ";
    //- $query2 = " UPDATE member SET status='$status' WHERE email='$uemail' AND actkey='$authkey'  ";

    $result2 = mysql_query($query2);
    $num2 = mysql_affected_rows($result2);


        if ( $num2  )  //  updated? - inform user
        {

            echo "<br>--$uemail updated to verified --";
        }
        else
            echo "<br>--$uemail  verfication [update] failed --";

            $val1=mysql_errno();
        $val2=mysql_error();

        //-- 
echo '<br>['.$query2.']--<br>--['.$result2.']<br>val-1:'.$val1.'<br> val-2'.$val2;


            }

    }   //-- eof. if auth/email flags ok ---

=========================

结果输出是：

 --uemail: [driggg2@gmail.com] -- && -- [2f4071bffda44aa30064364055687d13] ---

 +++ query1: successful +++>br>

 Warning: mysql_affected_rows() expects parameter 1 to be resource, boolean given in /home/arif/public_html/amy/public/domainsite/verifyfirst.php on line 125

 --driggg2@gmail.com verfication [update] failed --
[ UPDATE member SET status='verified' WHERE email='driggg2@gmail.com'; ]--
--[1]
val-1:0
val-2

我认为，警告是因为 query2 没有返回 mysql_affected_rows() 所需的资源 ID。

  • db 和 table 都在 utf8 中。我究竟做错了什么 ？
4

2 回答 2

2

Basically what is states is that mysql_affected_rows() does not take any parameters. Within the context of the update is will output the affected rows. Thus, you need not pass the $result2.

Instead of

$num2 = mysql_affected_rows($result2);

use

$num2 = mysql_affected_rows();

于 2013-03-02T17:28:28.957 回答
0

Indeed, put the result of mysql_connect in mysql_affected_rows to eliminate the error message and get the actual affected number of rows.

于 2013-03-02T17:27:54.903 回答