FETCH num rows with array

Question

I am using pdo to fetch data if email and password is same but it giving me an error and not fetching data below is the code .........

if(isset($_POST['submit'])){
  $email      = $_POST['email'];
  $password   = $_POST['password'];
  //Encrypt Password
  $encrypted  = md5($password);

  //Include Database Connection
  require 'assets/_conn.php';
  $result = dbConnect()->prepare("SELECT email, password FROM users WHERE email= :email AND password= :password");
  $result->bindParam(':email', $email);
  $result->bindParam(':password', $encrypted);
  $result->execute();

Here is the problem data in not fetching ....

$rows = $result->fetch(PDO::FETCH_NUM);
if(count($rows) > 0) {
    foreach($rows as $row){
       $image  = $row['image'];
       $pic    = array('1.jpg',$image);
    }
}

it giving me an error What error?

il_raffa
– il_raffa

2016-01-04 06:55:40 +00:00
Commented Jan 4, 2016 at 6:55 — il_raffa
– il_raffa, Commented Jan 4, 2016 at 6:55
Undefined variable: query

Stack Overflow
– Stack Overflow

2016-01-04 06:58:19 +00:00
Commented Jan 4, 2016 at 6:58 — Stack Overflow
– Stack Overflow, Commented Jan 4, 2016 at 6:58

Saty · Accepted Answer · 2016-01-04 07:10:19Z

2

PDO::FETCH_NUM

returns an array indexed by column number as returned in your result set, starting at column 0

So instead of using column name use indexing

$image  = $row[2];// your indexing

And you need to add image column in your query . Your query would be

$result = dbConnect()->prepare("SELECT email, password, image FROM users WHERE email= :email AND password= :password");

edited Jan 4, 2016 at 7:10

answered Jan 4, 2016 at 6:54

Saty

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1 Comment

Saty Over a year ago

You need to add image column in your query

Renish Khunt · Accepted Answer · 2016-01-04 07:16:25Z

1

for fetching data you should used to $result->fetch(PDO::FETCH_OBJ) like below

and as mention @Saty please add image column into your query.

$rows = $result->fetch(PDO::FETCH_NUM);
$pic = array();
if(count($rows) > 0) {
    /* foreach($rows as $row){
       $image  = $row['image'];
       $pic  = array('1.jpg',$image);
    } */
    while($row=$result->fetch(PDO::FETCH_OBJ)) {
        /*its getting data in line.And its an object*/
        $image  = $row->image;
        $pic[]    = array('1.jpg',$image);
    }
}

i hope this is working for you.

Thanks

edited Jan 4, 2016 at 7:16

answered Jan 4, 2016 at 6:53

Renish Khunt

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4 Comments

Stack Overflow Over a year ago

thanks reniskhunt but its not working it not fetching data error coming is Undefined variable: pic

Renish Khunt Over a year ago

it's not possible men just check your code. are you used $pic variable outside of while loop?

Renish Khunt Over a year ago

@StackOverflow if you used $pic variable outside f while loop then you should define the variable before while loop.

Stack Overflow Over a year ago

CHeck this code .... $rows = $result->fetch(PDO::FETCH_NUM); if(count($rows) > 0) { while($row=$result->fetch(PDO::FETCH_OBJ)) { $image = $row->image; $pic = array('1.jpg',$image); } } its in loop and print this $pic outside the loop <?php for($i=0;$i<2;$i++) echo"<option data-img-src=\"$pic[$i]\" value=\"$pic[$i]\" ></option>"; ?>

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