not going inside while loop in 'while($row = mysqli_fetch_array($stmt))'

Question

I tried to execute the following php program in my friend's laptop, but the nothing is getting executed inside the while loop.

<html>
<body>

<?php

    $db_hostname = "localhost";
    $db_username = "user";
    $db_password = "password";
    $db_name     = "resulta";

$link=mysqli_connect($db_hostname,$db_username,$db_password,$db_name);

// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$htno = $_POST['htno'];

$query="select * from marks where hallno=?";


if ($stmt = mysqli_prepare($link, $query)) {

mysqli_stmt_bind_param($stmt, 's', $htno);
/*http://php.net/manual/en/mysqli-stmt.bind-param.php*/

    /* execute query */
    mysqli_stmt_execute($stmt);

    /* store result */
    mysqli_stmt_store_result($stmt);

    printf("Number of rows: %d.\n", mysqli_stmt_num_rows($stmt));



if(mysqli_stmt_num_rows($stmt)>0)
{
echo "thank you";
echo "recieved";
echo "<table>";
echo "<tr><th>Hall Ticket No.</th><th>Sub. Code</th><th>Sub. Name</th><th>Internal Marks</th><th>External Marks</th><th>Total Marks</th><th>Credits</th></tr>";
//$stmt = mysqli_query($link,$query);

while($row = mysqli_fetch_array($stmt)) {
  echo "inside yeah!!!";
  echo "<tr><td>" . $row['hallno'] . "</td><td>" . $row['subcode'] . "</td><td>" . $row['subname'] . "</td><td>" . $row['intemarks'] . "</td><td>" . $row['extmarks'] . "</td><td>" . $row['totalmarks'] . "</td><td>" . $row['credits'] . "</td></tr>";

}
echo "</table>";
echo "over";
mysqli_stmt_close($stmt);
mysqli_close($link);

}

else
{
    /* close statement */
    mysqli_stmt_close($stmt);
echo "Roll No. not found";
 }   
}
?>

</body>
</html>

So, why is the code inside the loop not getting executed while all the code outside the while loop (also which prints the no. of rows in the result) is executing properly, I've searched in Google but and tried solutions but nothing is getting displayed if I use

while($row = mysqli_fetch_array($stmt, MYSQLI_ASSOC)) or 
while($row = mysqli_fetch_array($stmt, MYSQLI_NUM)) //with $row[0]...

I tried a similar code which is working, in case if it helps:

    $con=mysqli_connect($db_hostname,$db_username,$db_password,$db_name);

if (mysqli_connect_errno()) {
   echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

    // select everything from the news table
$st=0;
$result = 'False';
$result = mysqli_query($con,"SELECT * FROM notif ORDER BY dat desc LIMIT $st, 6");

echo "<ul>";
  while($row = mysqli_fetch_array($result)) {
  echo "<li style='font-family: 'Open Sans',sans-serif;' ><b><u>[" . $row['subject'] . "]</u></b> " . $row['note'] . " <i>by <b>" . $row['user'] . "</b> on <b>" . $row['dat'] . "       </b></i></li>";
  echo "<br>";
}
echo "</ul>";
    // disconnect from the database

mysqli_close($con);

To check if it was problem with the computer I tried the same code in other computer as well, but it wasn't executing, don't know why some code is executing and some are not, even when I'm not able to fetch the result but the function printf("Number of rows: %d.\n", mysqli_stmt_num_rows($stmt)); is giving the correct output, with correct no. of rows.

I'm running ubuntu 14.04 I also tried the code on my friend's laptop who is running unbuntu 14.04, can anyone please check both the codes if they are running, still can't understand why only that code is not working, both the codes(that work and that don't work) look same to me, but don't know why when I write code for different purposes, sometimes they work and sometimes they don't.

Can anyone please try the code?

Yes, that code given below is working, but the code given above is not. — user41965, Aug 02 '14 at 18:53
No, Manfred's code is not working the third code block I gave in the question is working, but the first and second block are not working(second block of code is only a small change in first block of code) — user41965, Aug 02 '14 at 21:28
also try this code for debugging what is in the $row, etc. variables: var_dump($row); //shows output of variable for debug purposes — amanda fouts, Aug 02 '14 at 23:09
do a dump of this var. as well: $htno = $_POST['htno']; as in var_dump($htno) to see if the post variable is pulling a null or invalid value. If this is not the issue, refer to this page: http://php.net/manual/en/mysqli.prepare.php and prepared statement example. — amanda fouts, Aug 03 '14 at 03:14
@amandafouts the post variable contains the correct variable, even the ` printf("Number of rows: %d.\n", mysqli_stmt_num_rows($stmt));` is also printing the correct no. of rows, only(according to what I observed) only the statement `$row = mysqli_fetch_array($result)` is not working and therefore the control is not going inside the while loop. Just can't understand why it's not working, can anyone please try the code — user41965, Aug 08 '14 at 16:24

score 1 · Answer 1 · answered Aug 02 '14 at 14:49

1

Your $stmt is of type mysqli_stmt, thats why it works with mysqli_stmt_num_rows

mysqli_fetch_array however expects a mysqli_result which you can get by storing the result of mysqli_stmt_store_result($stmt);

answered Aug 02 '14 at 14:49

Manfred Radlwimmer

13,257
13
53
62

But I've already stored using `mysqli_stmt_store_result($stmt);` in the code. – user41965 Aug 02 '14 at 15:19
yes, but you are passing $stmt to the method, what you need to do (at least I think so) is `$returnvalue = mysqli_stmt_store_result($stmt); mysqli_fetch_array($returnvalue);` – Manfred Radlwimmer Aug 02 '14 at 16:56
Thanks for responding but that's not working. I tried a similar code in my laptop and it was working without any errors – user41965 Aug 02 '14 at 17:18

score -3 · Answer 2 · answered Aug 02 '14 at 14:50

-3

Make the equal sign double, like this:

while($row == mysqli_fetch_array($stmt))

answered Aug 02 '14 at 14:50

Daiver

109
1
7

3

This shouldn't be a comparison, it's meant to be an assignment.... it's assigning the value returned by a call to `mysqli_fetch_array()` to the variable `$row` – Mark Baker Aug 02 '14 at 14:51

not going inside while loop in 'while($row = mysqli_fetch_array($stmt))'

2 Answers2