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  1. #1
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    php/mysql script that is meant up update..

    ...some info in a table doesnt work..
    hmm you see the $id is from the url the rest is either in the script or the db.... its meant to get the current balance... and work out how much it should go up and then updat eit to the new baance.. it errors.. any one see why?
    Code:
    <?php
    $dbc = mysql_connect('localhost', '', '') or die("Couldn't connect to database"); 
     mysql_select_db('GAME') or die("Couldn't select database"); 
    
    $query ="SELECT * FROM founds WHERE id=$id";
    $query ="UPDATE founds SET balance = $income WHERE id =$id";
    
    $result = mysql_query($query) or die("There is an error with you account, please contact support and inform them of this"); 
     while ($r = mysql_fetch_array($result)) { 
     extract($r);
    
            $fincome = $extractors * 2000 ;
            $sincome = $extractors * $advanced ;
            $income = $fincome + $sincome ;
    
    echo("Account Balance Updated");
    
    ?>
    thanks a lot

  • #2
    Supreme Overlord Spookster's Avatar
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    You are only running the second query. A query is executed using this command:

    mysql_query($query)

    so in yours, you are only executing the second one:

    $query ="UPDATE founds SET balance = $income WHERE id =$id";

    because you overwrite the first one with the second one.
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  • #3
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    oh i see.
    thanks
    hmm is there anyway i can run to querys in the same sript in that case?
    I think i saw it somewhere befor... maybe I was wrong

  • #4
    Supreme Overlord Spookster's Avatar
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    $query ="SELECT * FROM founds WHERE id=$id";

    $result = mysql_query($query)

    //put your code here to extract results to get value for the second query

    $query ="UPDATE founds SET balance = $income WHERE id =$id";

    $result = mysql_query($query)
    Spookster
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  • #5
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    oh thanks Spook
    so like this yeh..
    Code:
    <?php
    $dbc = mysql_connect('localhost', '', '') or die("Couldn't connect to database"); 
     mysql_select_db('GAME') or die("Couldn't select database"); 
    
    $query ="SELECT * FROM founds WHERE id=$id"; 
    $result = mysql_query($query) 
     while ($r = mysql_fetch_array($result)) { 
     extract($r);
    
            $fincome = $extractors * 2000 ;
            $sincome = $extractors * $advanced ;
            $income = $fincome + $sincome ;
    
    $query ="UPDATE founds SET balance = $income WHERE id =$id"; 
    
    $result = mysql_query($query)
    
    echo("Account Balance Updated");
    
    ?>
    that what i think your tellin me to do.. done that tho.
    however error line 7 parse error... hmm any further ideas?
    thanks for all yoru time

  • #6
    Supreme Overlord Spookster's Avatar
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    Well other than a couple of missing semi-colons at the ends of two lines I don't see anything else wrong. I've never used the extract function before. I usually just pull the data out of the array like so:

    while ($row = mysql_fetch_array($result)) {
    $yaks = $row["tablecolumnname"];
    }
    Spookster
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  • #7
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    You're missing a curly bracket...

    Code:
    <?php
    $dbc = mysql_connect('localhost', '', '') or die("Couldn't connect to database"); 
    mysql_select_db('GAME') or die("Couldn't select database"); 
    
    $query ="SELECT * FROM founds WHERE id=$id"; 
    $result = mysql_query($query) 
    
    while ($r = mysql_fetch_array($result)) { 
      extract($r);
      $fincome = $extractors * 2000 ;
      $sincome = $extractors * $advanced ;
      $income = $fincome + $sincome ;
    
      $query ="UPDATE founds SET balance = $income WHERE id = $id"; 
      $result = mysql_query($query);
    }
    echo("Account Balance Updated");
    
    ?>

  • #8
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    oh thanks spooks, for all yer help.
    wich.. hm ok i put that code you typed in a file and ran it and got
    Parse error: parse error in c:\phpdev\www\public\game\hourlyfounds.php on line 8

    hmm doesnt make sense.. im lost arg

  • #9
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    Forgot another semicolon (on line 6)...

    Code:
    <?php
    $dbc = mysql_connect('localhost', '', '') or die("Couldn't connect to database"); 
    mysql_select_db('GAME') or die("Couldn't select database"); 
    
    $query ="SELECT * FROM founds WHERE id=$id"; 
    $result = mysql_query($query);
    
    while ($r = mysql_fetch_array($result)) { 
      extract($r);
      $fincome = $extractors * 2000 ;
      $sincome = $extractors * $advanced ;
      $income = $fincome + $sincome ;
    
      $query ="UPDATE founds SET balance = $income WHERE id = $id"; 
      $result = mysql_query($query);
    }
    echo("Account Balance Updated");
    
    ?>
    That should work

  • #10
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    oh yeh did nt see that thanks
    arg more trouble now.. all it puts in browser is this
    Warning: Supplied argument is not a valid MySQL result resource in c:\phpdev\www\public\game\hourlyfounds.php on line 8
    Account Balance Updated
    so many problems hehe

  • #11
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    That simply means that your query: "select * from founds where id=$id" didn't return any results, so either you have no data in you table matching id to $id, or $id has an inapropriate value or the table 'founds' doesn't exist...

    Try using this, that should give you some more info:

    Code:
    <?php
    $dbc = mysql_connect('localhost', '', '') or die("Couldn't connect to database");
    mysql_select_db('GAME') or die("Couldn't select database");
    
    $query ="SELECT * FROM founds WHERE id=$id";
    $result = mysql_query($query);
    
    if (!$result)
      die(mysql_error());
    
    while ($r = mysql_fetch_array($result)) {
      extract($r);
      $fincome = $extractors * 2000 ;
      $sincome = $extractors * $advanced ;
      $income = $fincome + $sincome ;
      
      $query ="UPDATE founds SET balance = $income WHERE id = $id";
      $result = mysql_query($query);
      if (!$result)
        die(mysql_error());
    }
    
    echo("Account Balance Updated");
    
    ?>


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