Php postgresql error, prepared statement already exists -

am getting error of prepared statement "my_query7" exists, call function each time user tries update table leader_info in database, have gone through documentation pg_prepare , don't understand meant should run once. code snippets of help. thanks.

 function add_leader_country($user_id,$l_country)  {    global $connection;    $query = pg_prepare($connection,"my_query7","update leader_info set l_country = $1 user_id = $2 , status < 9");  $result = pg_execute($connection,"my_query7",array($l_country,$user_id)); if(!$result) {   echo pg_last_error($connection); }  else  {   echo "records created successfully\n"; } $row = pg_affected_rows($result); return $row;     } 

prepare execute not permit duplicate naming, error. query should prepared once, example, in cycle preparation state must set out of , execution in for.

 $result=$pg_prepare($connection,"my_query7",$query);     for($id=1;$id<3;$id++){       $result=pg_execute($connection,"my_query7",array($l_country,$user_id));        ...     } 

in case using functio use prepare , execute multiple times it's problem.

what trying accomplish function dispatches more code calling function. way might able you.

if want use functions use method

exemple https://secure.php.net

<?php function requesttodb($connection,$request){     if(!$result=pg_query($connection,$request)){         return false;     }     $combined=array();     while ($row = pg_fetch_assoc($result)) {         $combined[]=$row;     }     return $combined; } ?>  <?php $conn = pg_pconnect("dbname=mydatabase");  $results=requesttodb($connect,"select * mytable");  //you can access "cell" of table this: $rownumber=0; $columname="mycolumn";  $mycell=$results[$rownumber][$columname]; var_dump($mycell);  

if whant use preaper , execute functions try create function creates preparations once in session. not forget give different names same error not occur. tried find of genre , did not find. if find form presented here others learn. if in meantime find way present it.