How can I pass my query values from mysqli_fetch_array?

Question

I am trying to insert values into a table in my database. The first param is a non null variable, the next two are the two columns I want to pass in as well. What is wrong with my logic here.

$query  = "SELECT cnum, cname FROM course WHERE specialization = '0'";  
 $result = mysqli_query($conn,$query);

 if (!$result) die ("Database access failed: " . $conn->error);

 $rows = $result->num_rows;    
for ($j =0; $j<$rows;++$j) {

     $row = mysqli_fetch_array($result);
     $query = "INSERT INTO student_schedule VALUES ('$studentID', '$row[0]', '$row[1]', '0')";

     $result = $conn->query($query);
     if (!$result) die ("Database access failed: " . $conn->error);

 }

Yeah I get the expected $result to be parameter but boolean instead passed to mysqli_fetch_array — Murph_Fish
– Murph_Fish, Commented Apr 28, 2016 at 3:34

Sergey Mitroshin · Accepted Answer · 2016-04-28 15:27:21Z

1

Your solution

<?php
$query  = "SELECT cnum, cname FROM course WHERE specialization = '0'";  
$result = mysqli_query($conn,$query);
if (!$result) die ("Database access failed: " . $conn->error);

while ($row = mysqli_fetch_array($result)) {
    $insertQuery = "INSERT INTO student_schedule VALUES ('" . $conn->real_escape_string($studentID) . "', '" . $conn->real_escape_string($row[0]) . "', '" . $conn->real_escape_string($row[1]) . "', '0')");
    $insert = $conn->query($insertQuery);
    if (!$result) die ("Database access failed: " . $conn->error);
}
?>

Also, as a general rule, I suggest you don't mix MySQLi Procedural code with Object-Oriented code. Lastly, I also suggest you remove error outputting $conn->error, instead, capture the error and print out a custom error message instead. This reduces injection attacks.

edited Apr 28, 2016 at 15:27

Sergey Mitroshin

1137 bronze badges

answered Apr 28, 2016 at 3:23

user3785693

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10 Comments

Murph_Fish Over a year ago

after changing to your solution I get Notice: Undefined offset: 0 in C:\xampp\htdocs\journalism.php on line 140 Notice: Undefined offset: 1 in C:\xampp\htdocs\journalism.php on line 140 Notice: Use of undefined constant result - assumed 'result' in C:\xampp\htdocs\journalism.php on line 142

VipindasKS Over a year ago

@Murph_Fish change $insert = $conn->query($insertQuery); to $result = $conn->query($insertQuery); incase you are using the solution by @LFlare. also $ missing in your if (!result) die ("Database access failed: " . $conn->error);

Murph_Fish Over a year ago

@VipindasKS after those fixes I am getting undefined offset 0 and undefined offset 1 pointing to $insertquery any ideas?

VipindasKS Over a year ago

Did you change if (!result) to if (!$result)? I mean did you add dollar sign to the result variable?

VipindasKS Over a year ago

@Murph_Fish can you please post the error? And, which line causing error.

|

Sergey Mitroshin · Accepted Answer · 2016-04-28 03:43:23Z

0

Your code is vulnerable to SQL Injections, that might to be the reason why it doesn't work properly.

You should escape the data before including it into an SQL query:

for ($j =0; $j<$rows;++$j) {

    $row = mysqli_fetch_array($result);
    $query = $conn->prepare("INSERT INTO student_schedule VALUES (?, ?, ?, '0')";
    $query->bind_param('iss', $studentID, $row[0], $row[1]);
    $result = $query->execute();
    if (!$result) die ("Database access failed: " . $conn->error);

}

You can find more information on the bind_param() function in the PHP manual.

edited Apr 28, 2016 at 3:43

answered Apr 28, 2016 at 3:26

Sergey Mitroshin

1137 bronze badges

2 Comments

Murph_Fish Over a year ago

after plugging this in I get Fatal error: Cannot pass parameter 5 by reference in C:\......... on line 145

Sergey Mitroshin Over a year ago

@Murph_Fish yes, there was an issue. The answer is edited now, you can try again.

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