Trying to read mysql database

[Rayj2019]

So I have a very small database right now.   It looks like this:

mysql> select * from broadcast;
+----+-------------------+
| id | streamID          |
+----+-------------------+
|  1 | lxdlp0u20D1QdLt6s |
+----+-------------------+
1 row in set (0.00 sec)

I can update the streamID just fine using another php file.

The problem I am having it accessing it and reading it out.

Here is the php/mysql I am using.  When I browse to this I never see the output!

<?php
$debug = TRUE;
ini_set( "log_errors", TRUE);
ini_set( "error_log", "/var/log/php_error.log");
ini_set( "display_errors", $debug );
error_reporting(E_ALL);

  //--------------------------------------------------------------------------
  // Example php script for fetching data from mysql database
  //--------------------------------------------------------------------------
  $servername = "localhost";
  $username = "root";
  $password = "1*General2019";
  $dbname = "WebRTCApp";
  $tableName = "broadcast";

  //--------------------------------------------------------------------------
  // 1) Connect to mysql database
  //--------------------------------------------------------------------------
  //
  mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
  $conn = mysqli_connect($servername,$username,$password, $dbname);
  if (mysqli_connect_error()) {
          $err = mysqli_connect_error();
          if( $debug ) exit( "Error: $err" );
          logerr( $err );
          exit( "Failed to connect to the database!");
  }

  echo "Connected Successfully";

  $sql = "SELECT streamID FROM broadcast WHERE id=1";
 // $sql = 'SELECT id FROM broadcast WHERE streamID = E0ArubgkL9q2SCrXL';
  $stream = mysqli_query($conn,$sql);

        if( !$stream) {
                $err = mysqli_error( $conn );
                logerr( $err );
                if( $debug ) echo "Error: $err<br>" };

  echo "<br>";
  echo "stream = " $stream;

  mysqli_close($conn);

?>

Any idea why I am not seeing the stream (mysql streamID)?

Appreciate your response.

Ray

[Barand]

You have to fetch the row from your result set that the query returned

https://www.php.net/manual/en/mysqli-result.fetch-assoc.php

[Rayj2019]

2 hours ago, Barand said:

You have to fetch the row from your result set that the query returned

https://www.php.net/manual/en/mysqli-result.fetch-assoc.php

Does not seem to be working?

 $sql = "SELECT streamID FROM broadcast WHERE id=1";
  if ($result = mysqli_query($conn, $sql)) {
          $row = $result->fetch_assoc();

 echo "stream = " $row; }

 

stream = 

???

[Barand]

Look more carefully at the examples on that manual page

Edited July 23, 2019 by Barand

[cyberRobot]

Note that fetch_assoc() returns an associative array, if there are matching results. So you'll need to modify how you are using $row.

Also note that fetch_assoc() returns NULL when there are no (or no more) rows in the result set. So you'll want to test the value of $row before assuming it contains a "streamID".

As Barand mentioned, the manual provides examples for using fetch_assoc(). Assuming that your query will always return 1 row, you don't need to use a loop to process the output from fetch_assoc(). You could use a plain if statement.

[MPM]

You can't just paste the example code in as they are using different variable names to you.  You have to learn the principles and then apply them.  The steps are:

1. connect (you do)
2. select the data from the table (you did int he first example)
3. extract a row from that data (as mentioned no need for a loop for the first test just get one row)
4. extract a single item from that array
5. echo that item

Or replace 4 and 5 with a print_r of the array.