MySQL data ouput in a printable PHP variable

Question

I know I am having a syntax error here but I can't figure this out.

I'm trying to get the result of a single specific record in my database to display in a PHP variable. Seems simple enough but apparently not for me. Here is the code I'm trying to run:

    // Create connection
    $con=mysqli_connect($mysqlserver,$mysqluser,$mysqlpass,$mysqldb);

    // Check connection
    if (mysqli_connect_errno()) {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $SQL = "SELECT * FROM players_brixx_gastonia WHERE email='catheygreaney@example.com'"; 
    $result = mysqli_query($SQL,$con);
    $result1 = mysqli_fetch_assoc($result);
    $prevpoints1 = $result1['points'];
    echo $prevpoints1;

    mysqli_close($con);

My page is outputting: "Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in"

Which normally means my query is failing however when running the query by itself in phpmyadmin, it completes just fine displaying the row I need.

Whats up with this?

UPDATE:

KaNch solved my issue. Thanks!

The connection is fine. I'm running other database commands in this same connection, so the $con is working properly. — Michael White, Oct 26 '14 at 14:46
OK so I fixed my code in the original question. This is what I have and it's throwing the error. — Michael White, Oct 26 '14 at 14:54

PrakashSharma · Accepted Answer · 2014-10-26T15:08:56.397

2

 // Create connection
$con=mysqli_connect($mysqlserver,$mysqluser,$mysqlpass,$mysqldb);

// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$SQL = "SELECT * FROM players_brixx_gastonia WHERE email='catheygreaney@example.com'"; 
$result = mysqli_query($con,$SQL);
$result1 = mysqli_fetch_assoc($result);
$prevpoints1 = $result1['points'];
echo $prevpoints1;

mysqli_close($con);

edited Oct 26 '14 at 15:08

answered Oct 26 '14 at 14:49

PrakashSharma

421
5
15

Hey i hope this works – PrakashSharma Oct 26 '14 at 15:09
Hey I think I about got it. I used your exact code and it's not throwing the errors but it should return a number and isn't. – Michael White Oct 26 '14 at 15:14
NO I WAS WRONG! IT IS WORKING! – Michael White Oct 26 '14 at 15:16
Thank you so much KaNcH I'm not even sure what it was but it's working fine now! Thanks. – Michael White Oct 26 '14 at 15:16
Good .. every time try using mysqli ... – PrakashSharma Oct 26 '14 at 15:17

Royden Rego · Answer 2 · 2014-10-26T15:18:02.677

0

Use this, you haven't created a connection to the database. First create a connection to the database as follows

$con = mysqli_connect("localhost","usrname","password","db_name");
// Check connection
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

Execute your query as below

$result = mysqli_query($con, "SELECT * FROM players_brixx_gastonia WHERE email='catheygreaney@example.com'");
while($row = mysqli_fetch_array($result))
{
    echo $row['points'];
}

edited Oct 26 '14 at 15:18

answered Oct 26 '14 at 14:47

Royden Rego

132
2
13

This returns: "Fatal error: Cannot use object of type mysqli_result as array in" – Michael White Oct 26 '14 at 14:50
Or else try using mysqli_fetch_assoc() instead of mysqli_fetch_array() – Royden Rego Oct 26 '14 at 14:52
$con should be second parameter.. first in string type Read : http://php.net/manual/en/function.mysql-query.php – PrakashSharma Oct 26 '14 at 14:53
I updated my code in the original question and it's still throwing errors. – Michael White Oct 26 '14 at 14:56
Note I'm using mysqli not mysql. – Royden Rego Oct 26 '14 at 14:56
I now have two errors "Warning: mysql_query() expects parameter 2 to be resource, object given in C:\xampp\htdocs\beerhandpoker\includes\brixx_gastonia_newgame.php on line 123 Warning: mysql_fetch_assoc() expects parameter 1 to be resource, null given in C:\xampp\htdocs\beerhandpoker\includes\brixx_gastonia_newgame.php on line 124" – Michael White Oct 26 '14 at 14:57
changed it to mysqli and same thing. – Michael White Oct 26 '14 at 14:58
you are for using mysql for mysql_fetch_assoc(), use mysqli_fetch_assoc(). – Royden Rego Oct 26 '14 at 14:59
Sorry Roy, I tried your exact code and it's still not working. – Michael White Oct 26 '14 at 14:59
Fatal error: Cannot use object of type mysqli_result as array in – Michael White Oct 26 '14 at 15:00
@roydenrego Sorry Bro – PrakashSharma Oct 26 '14 at 15:01
@MichaelWhite either it should be whole mysql or mysqli you cannot use both.. Please post your connect() code. – PrakashSharma Oct 26 '14 at 15:03
I put it in the original question. – Michael White Oct 26 '14 at 15:03
Hey .. your are using mysqli_connect()... Change everything to mysqli then... – PrakashSharma Oct 26 '14 at 15:06
Are you absolutely sure your query is returning at least one result? – Royden Rego Oct 26 '14 at 15:08
@KaNcH it is all mysqli I don't have any mysql commands left. – Michael White Oct 26 '14 at 15:09
@roydenrego Yes I am I have a screenshot if it would let me upload it lol. – Michael White Oct 26 '14 at 15:12
@roydenrego KaNcH figured it out! Thanks. – Michael White Oct 26 '14 at 15:17
@MichaelWhite Good to know. Well I did make a mistake, I used $result['points'] instead of $row['points']. I have updated my answer. – Royden Rego Oct 26 '14 at 15:20

score 0 · Answer 3 · edited Nov 01 '14 at 19:19

0

mysql_query returna a resource if the query is executed successfully, or FALSE if there is an error. Since mysql_fetch_assoc complains about the argument being a boolean, maybe you have an error in your query.

edited Nov 01 '14 at 19:19

Machavity

29,816
26
86
96

answered Oct 26 '14 at 15:03

riss

1
1

I thought that too but when I run it in the plain SQL it returns fine. – Michael White Oct 26 '14 at 15:06

MySQL data ouput in a printable PHP variable

3 Answers3