CODEWITHSUNDEEP
php
raj
raj

Reputation: 129

data not getting stored properly in database

I am using this code for storing values in database ,but when i select multiple check-box only one eid value is getting stored in database ( which should be no of checkbox i have selected) and also rid value is stored as 0 where is the mistake? this is php code

<?php //insert selected candidate in table                  
                    //echo "checkcandi".$_POST['check_candi']; 
if(isset($_POST['select_candidate']) && logged_in())
{
  $checkcandi_sel = $_POST['check_candi'];
  if(empty($checkcandi_sel)) 
  {
    echo("You didn't select any Candidates.");
  } 
  else
  {
    $N = count($checkcandi_sel);

    echo("You selected $N Candidate(s): ");
    for($i=0; $i < $N; $i++)
    {
      echo($checkcandi_sel[$i] . " ");
      $eid=$checkcandi_sel[$i];
       echo "candi id".$eid;
    }
  }

//$rid=$value[1];
if(logged_in())
    {
    $emplyrid=$_SESSION['uid'];
    echo "emplyrid".$emplyrid;
    }
//$rid=$_GET['rid'];
//var_dump($emplyrid);
$query = "INSERT INTO selected_candidate (
                            eid,rid
                        ) VALUES (
                             '{$eid}','{$emplyrid}'
                        )";
            $result = mysql_query($query, $connection);
            if ($result) {
                $message = "<p style=\"color:green\">The Candidate selected and saved to your my seleted.</p>";
            } else {
                $message = "Sorry job was not applied.";
                $message .= "<br />" . mysql_error();
            }
}


?>

this is code for check-box

echo "<td><input id=\"select_candi{$i}\" onclick=\"javascript:func(this.id,{$_SESSION['uid']})\" type=\"checkbox\" name=\"check_candi[]\" value=\"{$data_set['eid']}\"/></td>";

Upvotes: 0

Views: 146

Answers (3)

Prasanth Bendra
Prasanth Bendra

Reputation: 32740

You can not add array to database 

"INSERT INTO selected_candidate (
                            eid,rid
                        ) VALUES (
                             '{$eid}','{$emplyrid}'
                        )"

In this query $eid will always contain last value of the array $checkcandi_sel, you convert the array $checkcandi_sel to string and add to data base

OR

Add it as multiple entries by looping the sql query.

Upvotes: 0

Sahal
Sahal

Reputation: 4136

Put the sql query inside the foreach...

if(count($checkcandi_sel) > 0) {
  for($i=0; $i < count($checkcandi_sel); $i++) {
    $eid=$checkcandi_sel[$i];
    $query = "INSERT INTO selected_candidate ( eid,rid) VALUES ('{$eid}','{$emplyrid}' )";
    $result = mysql_query($query, $connection);
  }
}

Upvotes: 1

Edwin Alex
Edwin Alex

Reputation: 5108

Pull your for loop near insert query and try this,

$N = count($checkcandi_sel);

if($N > 0) {
  for($i=0; $i < $N; $i++)
  {
    $eid=$checkcandi_sel[$i];
    $query = "INSERT INTO selected_candidate (
                        eid,rid
                    ) VALUES (
                         '{$eid}','{$emplyrid}'
                    )";
    $result = mysql_query($query, $connection);

  }
}

Upvotes: 1

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