mysql_query() expects parameter 1 to be string

Question

I know this question had been asked and answered by many, but I don't seem to be able to get this work on my side with the references I had found.

I understand that mysql is phasing out, but I just want to get this work. My idea is to update the selected row with the ID I get from the previous page. Error and code are shown below.

Error:

Warning: mysql_query() expects parameter 1 to be string, resource given in C:\xampp\htdocs\file\doAttendance.php on line 8 doAttendance.php:

<?php 
include "dbFunctions.php";

$id = $_REQUEST['id'];
$presence = 'Yes';

$selectQuery = mysql_query("SELECT * FROM guests WHERE id = '$id'");
$selected = mysql_query($selectQuery, $connect) or die(mysql_error($connect));

if (mysql_num_rows($selected) > 0) {   
} else {
    $sql = mysql_query ("UPDATE guests SET presence='$presence' WHERE id = '$id'");
    $updated = mysql_query($sql, $connect) or die(mysql_error($connect));
}
?>

What version of PHP are you running? If it's PHP 7, there just ain't no way nohow you're going to get any **mysql** calls to work. — Darwin von Corax, Feb 21 '16 at 03:33
Possible duplicate of [MySQLi not found error](http://stackoverflow.com/questions/21864663/mysqli-not-found-error) — Panda, Feb 21 '16 at 03:33
mysql_* functions are removed from php7, use mysqli_* instead. — Sanzeeb Aryal, Feb 21 '16 at 03:34
the error is not for the code you posted here. you aren't using mysqli_ — baao, Feb 21 '16 at 03:36
@SanzeebAryal tried to convert everything to mysqli, but Fatal error: Class 'mysqli_query' not found though. i had followed this: http://stackoverflow.com/questions/21864663/mysqli-not-found-error but no luck. — Penelope, Feb 21 '16 at 04:35
@mmm sorry for the mistake. this supposed to be the question instead of the other. — Penelope, Feb 21 '16 at 05:47
Did you alredy try this? `$selectQuery = "SELECT * FROM guests WHERE id = '$id'";` — Thiago Ururay, Feb 21 '16 at 05:55
`mysql_query` This extension was deprecated in `PHP 5.5.0`, and it was removed in `PHP 7.0.0`. Instead, the `MySQLi` or `PDO_MySQL` extension should be used — Murad Hasan, Feb 21 '16 at 05:56
@Paige you're welcome. You were giving the mysql_query() result to mysql_query() again. It expects a string and you pass a resource. — Thiago Ururay, Feb 21 '16 at 06:16
i don't see the comment of @ThiagoUruray, if i see it before i make an answer, i will not. After i comment i re-check the warning of the question and i see `Paige`, make a mistake using mysql_query double time. — Murad Hasan, Feb 21 '16 at 06:27
@ThiagoUruray i see.. weird, it doesn't seem to pass the query anymore. — Penelope, Feb 21 '16 at 06:31
@FrayneKonok thank you for the code. no error now, but $sql is not being passed to update the database. please advice. — Penelope, Feb 21 '16 at 06:44
check your condition, if any rows found then you did't do anything but when not found anything you do the thing, so how can the update works?? see my answer, i will update it right now. — Murad Hasan, Feb 21 '16 at 06:59
@FrayneKonok i just checked $selected. the output was Resource id #5. so i tried http://stackoverflow.com/questions/9754320/figuring-out-why-i-am-getting-a-resource-id-5-error it did pass the query, but only once. — Penelope, Feb 21 '16 at 07:10
I think @FrayneKonok's code will help you. So I'll let here just a suggestion: after getting this working, look for PDO: https://phpdelusions.net/pdo — Thiago Ururay, Feb 21 '16 at 07:16
@FrayneKonok oh, will it be a possibility that i left the IF statement blank, that's why there is nothing to be passed through? — Penelope, Feb 21 '16 at 07:18

Murad Hasan · Accepted Answer · 2016-02-21T07:00:22.193

i don't see the comment of @ThiagoUruray, if i see it before i make an answer, i will not. After i comment i re-check the warning of the question and i see @Paige, make a mistake using mysql_query double time.

Here is the solution: you miss use of mysql_query.

you have to pass the query string into the mysql_query not the resource. i just remove the mysql_query from the $selectQuery and $sql variable.

<?php 
include "dbFunctions.php";

$id = $_REQUEST['id'];
$presence = 'Yes';

$selectQuery = "SELECT * FROM guests WHERE id = '$id'";
$selected = mysql_query($selectQuery, $connect) or die(mysql_error($connect));
//check if any rows found then update the rows.
if (mysql_num_rows($selected) > 0) {
    $sql = "UPDATE guests SET presence='$presence' WHERE id = '$id'";
    $updated = mysql_query($sql, $connect) or die(mysql_error($connect));
}
?>

This is the only mistake you did in this code, so your result says:

Warning: mysql_query() expects parameter 1 to be string, resource given in C:\xampp\htdocs\file\doAttendance.php on line 8 doAttendance.php:

now it will be fine, let me know.

mysql_query() expects parameter 1 to be string

1 Answers1