PHP - SQL query is not returning any results

Question

I must be missing something very obvious here, but I don't understand what I am doing wrong here. For some reason, $row is coming out empty, which should not happen. I am getting no error though.

$uname=mysqli_real_escape_string($conn,$_SESSION["username"]);
$query1="SELECT user_id FROM users WHERE username='$uname'";

$results=$conn->query($query1);
if($results->num_rows>0){
    while($row=$results->fetch_assoc()){
        $uid=mysqli_real_escape_string($row["user_id"]);
    }
}

I am sorry if this is a bad question, I will delete it right after, but I would appreciate any help.

You're not actually checking for errors, so how do you know? — aynber
– aynber, Commented Jun 22, 2017 at 19:27
echo your query try to run in PHPMyAdmin. is it returning some data? — Muhammad Usman
– Muhammad Usman, Commented Jun 22, 2017 at 19:29
You are wide open for SQL injection. Since you're using mysqli, take advantage of prepared statements and bind_param. And you'll never have to worry about pesky quoting issues again. — aynber
– aynber, Commented Jun 22, 2017 at 19:42
Actually, no error is being displayed in the log, so I thought I might not have errors. I am still learning. Thank you for the links you provided. @aynber — Archon
– Archon, Commented Jun 22, 2017 at 19:53

Flexi · Accepted Answer · 2017-06-22 19:35:50Z

2

No question is a bad question imo..we are all here to learn isnt it?I'd rather construct that code this way:

$uname=mysqli_real_escape_string($conn,$_SESSION["username"]);
$query1=mysqli_query($conn, "SELECT user_id FROM users WHERE username='$uname'");

if ($query1){
   while ($row = mysqli_fetch_assoc($query1){
      $user_id = $row['user_id'];
      echo $user_id;
   }
}

answered Jun 22, 2017 at 19:35

Flexi

1952 silver badges12 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

Archon Over a year ago

This is working. I did not know we could do it like this. Thank you! I need to read up more.

Archon Over a year ago

And about the bad question thing, it happened to me once before, it was a question like this and I had to delete it without ever getting an answer.

Flexi Over a year ago

It happened to me too severally, but the key is never to get deterred.

Archon Over a year ago

I try not to. Btw, I did understand what you did, but could you tell me what I was doing wrong?

Flexi Over a year ago

This line $uid=mysqli_real_escape_string($row["user_id"]); was unneccessary. I guess it broke ur code. Try with $uid = ($row['user_id']);

mayank · Accepted Answer · 2017-06-22 20:16:17Z

1

Try this

    $uname=mysqli_real_escape_string($conn,$_SESSION["username"]);
$query1="SELECT user_id FROM users WHERE username=
'$uname'";

$results=mysqli_query($conn, $query1);
if(mysqli_num_rows($results) >0){
    $row=$results->fetch_assoc();
    $uid=$row["user_id"];
}

edited Jun 22, 2017 at 20:16

answered Jun 22, 2017 at 19:32

mayank

4625 silver badges14 bronze badges

4 Comments

aynber Over a year ago

Bad concatenation there. You don't need the periods.

mayank Over a year ago

Echo ur query and check

mayank Over a year ago

@aynber try it now.

Archon Over a year ago

I tried this, but this is not working for some reason. Can you please explain what you have done?

Collectives™ on Stack Overflow

PHP - SQL query is not returning any results

2 Answers 2

5 Comments

4 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

5 Comments

4 Comments

Your Answer

Sign up or log in

Post as a guest

Related