php - Warning: mysql_fetch_array() expects parameter 1 to be resource sql -

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i'm having trouble inserting new post in database. ran on shared host , there no problem, on vps gives me following error:

 warning: mysql_fetch_array() expects parameter 1 resource, boolean given in /home/sam/public_html/admin/include/functions.php on line 37   warning: mysql_fetch_array() expects parameter 1 resource, boolean given in /home/sam/public_html/admin/include/functions.php on line 46   warning: mysql_fetch_array() expects parameter 1 resource, boolean given in /home/sam/public_html/admin/include/functions.php on line 55

here code (functions.php) :

<?php include("../../connections/confing.php");  function offset_time($time,$ofset) { $time_array = explode(":",$time); $am = "";  $hours   = $time_array[0]; $minutes = $time_array[1];  $new_hour = $hours +10;   if($new_hour>24){ $new_hour = $new_hour - 24; $am = "am"; }else{ $am = "pm"; }  if ($new_hour>12){ $new_hour = $new_hour -12; }   $offset_time = $new_hour . ":". $minutes ." ".$am ;  return $offset_time; }  //---------------------- cat id ------------------------------------------- function getcatid($newsid){  $minsql  = "select catid art_news newsid=".$newsid; $rsofget = mysql_query($minsql); $rsget   = mysql_fetch_array($rsofget);   $catid   = $rsget["catid"]; return $catid;   }  function getzoneid($newsid){  $minsql  = "select zoneid art_news newsid=".$newsid; $rsofget = mysql_query($minsql); $rsget   = mysql_fetch_array($rsofget); $zoneid  = $rsget["zoneid"]; return $zoneid;  }  function get_news_lang($newsid){  $minsql  = "select lang art_news newsid=".$newsid; $rsofget = mysql_query($minsql); $rsget   = mysql_fetch_array($rsofget); $lang  = $rsget["lang"]; return $lang;    }  function get_art_lang($artid){  $minsql  = "select lang art_articles artid=".$artid; $rsofget = mysql_query($minsql); $rsget   = mysql_fetch_array($rsofget); $lang  = $rsget["lang"]; return $lang;    }  function get_video_lang($albumid){  $minsql  = "select lang video_albums albumid=".$albumid; $rsofget = mysql_query($minsql); $rsget   = mysql_fetch_array($rsofget); $lang  = $rsget["lang"]; return $lang;    }  function get_gallery_lang($albumid){  $minsql  = "select lang glr_albums albumid=".$albumid; $rsofget = mysql_query($minsql); $rsget   = mysql_fetch_array($rsofget); $lang  = $rsget["lang"]; return $lang;    }   function getphoto_newsid($photoid){  $minsql  = "select newsid art_photos photoid=".$photoid; $rsofget = mysql_query($minsql); $rsget   = mysql_fetch_array($rsofget); $newsid  = $rsget["newsid"]; return $newsid;  }   function get_photo_albumid($photoid){  $minsql  = "select albumid glr_photos photoid=".$photoid; $rsofget = mysql_query($minsql); $rsget   = mysql_fetch_array($rsofget); $albumid = $rsget["albumid"]; return $albumid;     } ?>

you should provide valid sql results resource first parameter (mysql_query return false in case of error):

<?php mysql_connect("localhost", "user", "password") or     die("connect error: " . mysql_error()); mysql_select_db("mydb");  $sqlresource = mysql_query("select id, name mytable");  if ($sqlresource === false) // check if query return error     die("query error: " . mysql_error());  while ($row = mysql_fetch_array($sqlresource, mysql_num)) {     printf("id: %s  name: %s", $row[0], $row[1]);   }  mysql_free_result($sqlresource); ?>

ps: mysql* functions ares deprecated, use mysqli or better pdo extensions.


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