SQL: mysqli_query, no output

Question

session_start();
$user=$_SESSION['username'];
$con=mysqli_connect("$host","$username","$password","$db_name");
$sql1="SELECT UserID FROM User WHERE username='$user'";
$query=mysqli_query($con,$sql1);
echo $query;

Im not sure what's the problem with this, I've tried searching around here and tried different ways of writing the syntax and checking for errors in parameter but still no output for the command. What am i missing here?

UPDATE

$sql1="SELECT UserID FROM User WHERE username='$user'";
$query=mysqli_query($sql1,$con);
$row=mysqli_fetch_array($query);
$uid=$row[0];
echo $uid;

As answered by the others, I tried doing this. There is still no output or an output of 0 was given. Where if I tried the query in sql, it shows the right one.

UPDATE 2

It tried the var_dump() command and it returns NULL. Does that mean it can't read the database?

@Sadikhasan oh sorry, forgot to indicate in the post. I did use echo. — user3799541
– user3799541, Commented Jul 3, 2014 at 6:58
It's required to fetch data once query is executed successfully. — Rikesh
– Rikesh, Commented Jul 3, 2014 at 6:59

Yannici · Accepted Answer · 2017-06-14 09:15:57Z

1

mysqli_query gives no output. A successfully query returns a resource. You have to handle the resource with mysqli_fetch_object, mysqli_fetch_assoc or mysqli_fetch_array.

Also you gave $con parameter before $sql1 which is wrong. You have to put the connection first, then the query variable.

Example:

<?php

session_start();
$user = $_SESSION['username'];
$con = mysqli_connect("$host", "$username", "$password", "$db_name");
$sql1 = "SELECT UserID FROM User WHERE username='$user'";
$query = mysqli_query($con, $sql1);

$fetch = mysqli_fetch_assoc($query);

var_dump($fetch);

?>

edited Jun 14, 2017 at 9:15

answered Jul 3, 2014 at 7:00

Yannici

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1 Comment

user3799541 Over a year ago

Please see my updated post, I tried what you said but still no output was given.

Zander · Accepted Answer · 2014-07-03 07:01:00Z

0

$query will be a "Resource". You can use for example $result = mysqli_fetch_array($query); and then echo $result[0];May there's another way, but that's a simple one!

answered Jul 3, 2014 at 7:01

Zander

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6 Comments

user3799541 Over a year ago

i tried using that, but still there's no output. Is there a mistake with my query?

Zander Over a year ago

Try echoing your query and see if it's printing what you want. May the variable inside the simplequotes are doing something. Take a try.

user3799541 Over a year ago

This is what I got from echoing it SELECT UserID FROM User WHERE username='1234' Is there a mistake in the query?

Zander Over a year ago

Query seems to be OK, if the filed UserID is exactly typed like that and the table is User (all beign case sensitive)

user3799541 Over a year ago

I typed them all correctly, what would it mean if var_dump() gets null?

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user3799541 · Accepted Answer · 2014-07-10 07:00:47Z

0

I was able to do some workaround with my problem using mysql syntax. Still not able to determine the problem with mysqli but able to finished my task. Thanks for answering guys.

answered Jul 10, 2014 at 7:00

user3799541

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