您的位置: 首页  >  技术问答  >  mysql_fetch_array（）期望参数1是资源，在[重复]中给出布尔值

mysql_fetch_array（）期望参数1是资源，在[重复]中给出布尔值

分类: 技术问答 • 2022-03-11 20:31:02
问题描述：

I am not the first guy who is asking this question here but seriously, I tried possible duplicates before asking.

Here is my code: Updated 

 foreach ($pieces as $v) {
        $get_user ="SELECT * from demo_user WHERE user_name ='$v'";
        $result = mysql_query($get_user);
        if (!$result) {
             die(mysql_error());
        }
        while($row = mysql_fetch_array($result))
        {
            $GLOBAL_REST_URL = "http://192.168.0.100:9000/meeting/";
            $part_name = $row['user_name'];
            $participant = array("email"=>$row['primary_email'],
                    "contact_no"=>$row['primary_mobile'],
                    "password"=>$attendee_password,
                    );

            $LOCAL_REST_URL = $GLOBAL_REST_URL .$mtngid.'/participant/'.$row['user_name'] ;

           $json_part = array2json($participant);
           $curl_handle=curl_init();
           curl_setopt($curl_handle,CURLOPT_URL,$LOCAL_REST_URL);
           curl_setopt($curl_handle,CURLOPT_CONNECTTIMEOUT,20);

           curl_setopt($curl_handle, CURLOPT_CUSTOMREQUEST, "POST");
           curl_setopt($curl_handle, CURLOPT_POSTFIELDS,$json_part);

           curl_setopt($curl_handle, CURLOPT_HTTPHEADER, $headers);
           curl_setopt($curl_handle,CURLOPT_RETURNTRANSFER,1);
           $buffer = curl_exec($curl_handle);
           curl_close($curl_handle);
           $getit_part =  json_decode($buffer, true);
           if ($getit_part['code'] == 200){
            $key = $getit_part['data'];

            $join_url = "http://192.168.0.100/UI/user/joinuser.php?meetingid=".$mtngid."&key=".$getit_part['data']; 
            $pr_mobile = $row['primary_mobile'];
            $pr_email = $row['primary_email'];
            $sql = "INSERT INTO demo_participant(name, 
                             meeting_id_id, 
                             password, 
                             user_view__url, 
                             `key`, 
                             contact_no,
                             email)
                        SELECT '$part_name', 
                               id, 
                               '$attendee_password',
                               '$join_url',
                                {$getit_part['data']},
                               '$pr_mobile',
                               '$pr_email'
                        FROM demo_meeting
                        WHERE meetingID = '$mtngid'";

            $result = mysql_query($sql) or die (mysql_error());
           }
          // $message = 'Hi '.$_GET['meeting_name'].',and associated participant Updated successfully ';
        }
        }

When I execute the above code, I get the following warning:

mysql_fetch_array() expects parameter 1 to be resource, boolean given in

print_r($pieces) outputs:

Array
(
    [0] => Digvijay
    [1] => philip
    [2] => fake
)

Please let me know, what I might be doing wrong.

</div>

答

With your code, $result won't be a boolean.

if mysql_query return false, then the program should exit.

Check the error line once more.

Update:

After reading your updated code, the problem is the line below (at the end of the while loop):

$result = mysql_query($sql) or die (mysql_error());

You are overwriting the variable $result.

It is an insert query, so you only need to do:

mysql_query($sql) or die (mysql_error());

答

The query has failed, and you are passing the "false" to the fetch. That can't be because of the or die; You are not showing the real code?

  • Do not use the mysql_* function
  • Put error_reporting on
  • Build some error checking into your script.

相关推荐