String given in mysql_fetch_array

Question

$con = mysql_connect("localhost","root","");    
if (!$con) {
    die('Could not connect: ' . mysql_error());
}

mysql_select_db("vinzq", $con);
session_start();
$confirmation = $_SESSION['confirm'];

$result = mysql_query("SELECT * FROM reservation WHERE confirmation = '$confirmation'");

while ($row = mysql_fetch_array($result)) {

    $arrival = $row['arrival'];
    $departure = $row['departure'];
    $adults = $row['adults'];
    $child = $row['child']; 
    $nroom = $row['no_room'];
    $result = $row['result'];
    $name = $row['firstname'];
    $last = $row['lastname'];
    $city = $row['city'];
    $zip = $row['zip'];
    $country = $row['country'];
    $password = $row['password'];
    $email = $row['email'];
    $cnumber = $row['contact'];
    $stat= 'Active';
}

I have my codes here written in php. At the bottom after this code snippet, I have used html to display out the different variables. The values of the variables are correct, however the error that

mysql_fetch_array() expects parameter 1 to be resource, string given,

still pops up even tho it's functional. The error seems to belong to the while statement line.

possible duplicate of mysql_fetch_array() expects parameter 1 to be resource, string given — Matheno
– Matheno, Commented Apr 13, 2015 at 15:18
I am not using oop, I'm embedding html in my php, for a website. @MHakvoort — Jeany
– Jeany, Commented Apr 13, 2015 at 15:19
I have printed $confirmation out, and it is the same as the session variable being stored. @BeingSunny I think it is correct also, because the data it prints out like all the $arrival, $departure etc are all correct. I can get what I want, but it just shows that error over there. — Jeany
– Jeany, Commented Apr 13, 2015 at 15:22

Barmar · Accepted Answer · 2015-04-13 19:43:06Z

1

The problem is this line inside the while loop:

$result = $row['result'];

The first time through the loop, $result contains the result of mysql_query(). The second time, it contains the value of this field from the first row of results, so mysql_fetch_array($result) tries to use this string instead of the original query result.

Use different names for these two variables.

edited Apr 13, 2015 at 19:43

answered Apr 13, 2015 at 15:23

Barmar

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4 Comments

Jeany Over a year ago

I don't have a line that reads: $result = $row['result']; are you talking about the data nested within the while loop? Like $arrival = $row['arrival'];? Sorry haven't been touching php for some time @Barmar

Barmar Over a year ago

Yes, that's the line I'm talking about.

Jeany Over a year ago

That was the error, duplicate variable name, because the original $result was within the while loop of the query $result execution... Thank you! @Barmar

Barmar Over a year ago

Thank you for confirming the obvious.

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