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  1. #1
    Senior Member
    Join Date
    Mar 2013
    Posts
    118

    MySQL Query Not Working

    I'm trying to code a addon for Xenforo and the following code isn't working.

    Code:
    <?php
    
    mysql_connect("localhost", "root", "") or die(mysql_error());
    mysql_select_db("thepoker_db");
    
    $query = mysql_query("SELECT user.username
                    , user.bcp_lbp
                    , user.bcp_freeroll_money_won
                    , profile.field_value AS poker_user
                 FROM xf_user
               LEFT OUTER
                JOIN xf_user_field_value AS profile
                  ON profile.user_id = user.user_id
                 AND profile.field_id = '626c61636b5f636869705f706f6b65725f757365726e616d65'
               WHERE user.secondary_group_ids IN (2,4,5)");
    $n = 0;
    
    	 while($results = mysql_fetch_array($query))
                {
    			    $n++;
                    print "<table><tr><td>#." . $n++ . "</td>";
                    print "<td>" .$results['username'] . "</td>";
    				print "<td>" . $results['poker_user'] . "</td>";
                    print "<td align=\"center\">". $results['bcp_lbp'] ."</td>";
    				print "<td align=\"center\">$". $results['bcp_freeroll_money_won'] ."</td></tr></table>";
                }
    
    ?>
    I get the error:
    Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\pokertalk\leaderstest.php on line 19

    I'm using Wamp Server. Any idea how to fix it?

  2. #2
    Senior Member
    Join Date
    Jul 2013
    Location
    Bhubaneswar, Odisha, India
    Posts
    112

    Re: MySQL Query Not Working

    Quote Originally Posted by bauss View Post
    I'm trying to code a addon for Xenforo and the following code isn't working.

    Code:
    <?php
    
    mysql_connect("localhost", "root", "") or die(mysql_error());
    mysql_select_db("thepoker_db");
    
    $query = mysql_query("SELECT user.username
                    , user.bcp_lbp
                    , user.bcp_freeroll_money_won
                    , profile.field_value AS poker_user
                 FROM xf_user
               LEFT OUTER
                JOIN xf_user_field_value AS profile
                  ON profile.user_id = user.user_id
                 AND profile.field_id = '626c61636b5f636869705f706f6b65725f757365726e616d65'
               WHERE user.secondary_group_ids IN (2,4,5)");
    $n = 0;
    
    	 while($results = mysql_fetch_array($query))
                {
    			    $n++;
                    print "<table><tr><td>#." . $n++ . "</td>";
                    print "<td>" .$results['username'] . "</td>";
    				print "<td>" . $results['poker_user'] . "</td>";
                    print "<td align=\"center\">". $results['bcp_lbp'] ."</td>";
    				print "<td align=\"center\">$". $results['bcp_freeroll_money_won'] ."</td></tr></table>";
                }
    
    ?>
    I get the error:
    Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\pokertalk\leaderstest.php on line 19

    I'm using Wamp Server. Any idea how to fix it?
    You have missed querying the database. Your Query has not yet been executed by the time the
    Code:
    while($results = mysql_fetch_array($query))
    gets executed. The mysql_fetch_array - needs a recordset as a parameter and not the query you are passing to it.
    here should be the fixed code -
    Code:
    <?php
    
    mysql_connect("localhost", "root", "") or die(mysql_error());
    mysql_select_db("thepoker_db");
    
    $query = mysql_query("SELECT user.username
                    , user.bcp_lbp
                    , user.bcp_freeroll_money_won
                    , profile.field_value AS poker_user
                 FROM xf_user
               LEFT OUTER
                JOIN xf_user_field_value AS profile
                  ON profile.user_id = user.user_id
                 AND profile.field_id = '626c61636b5f636869705f706f6b65725f757365726e616d65'
               WHERE user.secondary_group_ids IN (2,4,5)");
    $n = 0;
    $result=mysql_query($query)                                  // Execute the Query to generate the Recordset
    	 while($results = mysql_fetch_array($result))    // Loop through the Recordest
                {
    			    $n++;
                    print "<table><tr><td>#." . $n++ . "</td>";
                    print "<td>" .$results['username'] . "</td>";
    				print "<td>" . $results['poker_user'] . "</td>";
                    print "<td align=\"center\">". $results['bcp_lbp'] ."</td>";
    				print "<td align=\"center\">$". $results['bcp_freeroll_money_won'] ."</td></tr></table>";
                }
    
    ?>

  3. #3
    Junior Member
    Join Date
    Jul 2014
    Posts
    12

    Re: MySQL Query Not Working

    Try this:
    <?php
    $con=mysqli_connect("localhost","my_user","my_pass word","my_db");
    // Check connection
    if (mysqli_connect_errno())
    {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $sql="SELECT Lastname,Age FROM Persons ORDER BY Lastname";
    $result=mysqli_query($con,$sql)

    // Numeric array
    $row=mysqli_fetch_array($result,MYSQLI_NUM);
    printf ("%s (%s)\n",$row[0],$row[1]);

    // Associative array
    $row=mysqli_fetch_array($result,MYSQLI_ASSOC);
    printf ("%s (%s)\n",$row["Lastname"],$row["Age"]);

    // Free result set
    mysqli_free_result($result);

    mysqli_close($con);
    ?>

 

 

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