mysqli_fetch_assoc() 期望参数 1 是 mysqli_result，while 循环中给出的字符串

Question

I currently have this code but it gives me the error mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, string given on line 22

The code is this:

1   $servername = "localhost";
2   $user = "root";
3   $pass = "";
4   $db = "mafioso";
5
6   $con = mysqli_connect($servername, $user, $pass, $db);
7
8   $cash_utbetaling[0] = 50000000;
9   $cash_utbetaling[1] = 40000000;
10  $cash_utbetaling[2] = 30000000;
11  $cash_utbetaling[3] = 20000000;
12  $cash_utbetaling[4] = 10000000;
13
14  $kuler_utbetaling[0] = 25;
15  $kuler_utbetaling[1] = 20;
16  $kuler_utbetaling[2] = 15;
17  $kuler_utbetaling[3] = 10;
18  $kuler_utbetaling[4] = 5;
19
20  $i = 0;
21  $result = mysqli_query($con, "SELECT * FROM daily_exp ORDER BY exp DESC LIMIT 5");
22  while($row_best = mysqli_fetch_assoc($result)) {
23    
24    $acc_id = $row_best['acc_id'];
25
26    $sql = "SELECT * FROM accounts WHERE ID='".$acc_id."'";
27    $query = mysqli_query($con, $sql) or die (mysqli_error());
28    $row_top5 = mysqli_fetch_assoc($query);
29    
30    $result = "UPDATE accounts SET money = (money + ".$cash_utbetaling[$i]."), 
      bullets = (bullets + ".$kuler_utbetaling[$i].")  WHERE ID = ".$acc_id."";
31    mysqli_query($con, $result) or die("Bad query: $result");
32
33    $i++;
34  }

I can't seem to find the error, I am running the same code in another file and there is no issues.

Answer 1

You are overwritting $result

// This is supposed to be
//  a mysqli_result object ----------v-----v
while($row_best = mysqli_fetch_assoc($result)) 
{
    // some code
    // $result is now a string. Next iteration will raises the warning
    $result = "UPDATE accounts SET ...";
}

So you need to give your variables distinct names. Naming a query $result is not the best choice.

---

Side note.

Your inner queries are vulnerable to SQL injections. One should use prepared statements instead of concatening strings.

In example :

// prepare the query
$query = "SELECT * FROM accounts WHERE ID=?";
if ($stmt = mysqli_prepare($con, $query)) {

    // bind the param
    mysqli_stmt_bind_param($stmt, "s", $acc_id);

    // execute the query
    mysqli_stmt_execute($stmt);

    // get the result. Of course, avoir using the same variable name, again :)
    $result = mysqli_stmt_get_result($stmt);
}

For more informations about prepared statements with mysqli, read the official documentation or the tutorials written by @YourCommonSense

Answer 2

Your main problem here is that you called the SQL query variable the same name as the result you used in the while loop. You overwrite it with a string on line 30:

30    $result = "UPDATE accounts SET money = (money + ".$cash_utbetaling[$i]."), 
      bullets = (bullets + ".$kuler_utbetaling[$i].")  WHERE ID = ".$acc_id."";

The easy answer is it was a typo, but you made many more mistakes.

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$con = mysqli_connect("localhost", "root", "", "mafioso");
$con->set_charset('utf8mb4');


$cash_utbetaling[0] = 50000000;
$cash_utbetaling[1] = 40000000;
$cash_utbetaling[2] = 30000000;
$cash_utbetaling[3] = 20000000;
$cash_utbetaling[4] = 10000000;
$kuler_utbetaling[0] = 25;
$kuler_utbetaling[1] = 20;
$kuler_utbetaling[2] = 15;
$kuler_utbetaling[3] = 10;
$kuler_utbetaling[4] = 5;
$i = 0;

$result = mysqli_query($con, "SELECT * FROM daily_exp ORDER BY exp DESC LIMIT 5");
$daily_exp = $result->fetch_all(MYSQLI_ASSOC);

// prepare update
$stmt = $con->prepare('UPDATE accounts SET money = (money + ? ), bullets = (bullets + ? ) WHERE ID=?');

foreach ($daily_exp as $row_best) {
    $stmt->bind_param('sss', $cash_utbetaling[$i], $kuler_utbetaling[$i], $row_best['acc_id']);
    $stmt->execute();
    $i++;
}

1. You need to enable error reporting instead of using or die (mysqli_error()) , which would never work, because of missing argument.
2. Don't use while loop to go through the results. It is much better to fetch all the records at once or loop on the result using foreach . If you used my suggestion, then most likely you would avoid your typo at all.
3. You must use prepared statements. In your code the values are integers and constant, but I assume you would use variable input at some point, which means you need to use placeholders and parameters.
4. The SELECT query inside the loop wasn't doing anything, so I removed it from my answer.
5. You should strive to avoid N+1 queries problem. Try doing the same in one query only if possible.