mysqli_query not working in php despite correctly built query

Question

I am trying to get what should otherwise be a simple but of php to insert sample data into a table, but something just isn't having any of it.

Table Definition:

CREATE TABLE IF NOT EXISTS teams ( 
     token varchar(12) COLLATE utf8_unicode_ci NOT NULL, 
     tname varchar(48) COLLATE utf8_unicode_ci NOT NULL, 
     captain varchar(64) COLLATE utf8_unicode_ci NOT NULL, 
     email varchar(64) COLLATE utf8_unicode_ci NOT NULL, 
     phone varchar(14) COLLATE utf8_unicode_ci NOT NULL, 
          PRIMARY KEY (token), 
          UNIQUE KEY name (tname), 
          KEY id (token) 
     )  
     ENGINE=MyISAM DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

Code:

$con = mysqli_connect("localhost","username","password","database");

$tname = "Big Bang";
$cname = "Mike";
$cemail = "[email protected]";
$cphone = "123-456-7898";

$teamToken = strtoupper(bin2hex(mcrypt_create_iv(6, MCRYPT_DEV_URANDOM)));

$query = "INSERT INTO teams (token, tname, captain, email, phone) VALUES ('" . $teamToken . "', '" . $tname . "', '" . $cname . "', '" . $cemail . "', '" . $cphone . "')";

if (mysqli_query($con, $query))
{
    echo "Pass!";
}
else
{
    echo $query;
}

mysqli_close($con);

What's odd is the php echos the query, because the mysqli_query result is false, yet the echoed query, when copied and pasted right into phpMyAdmin's terminal, works fine.

I am at my qit's end.

FYI: I have even tried $query = "INSERT INTO teams (token, tname, captain, email, phone) VALUES ('$teamToken', '$tname', '$cname', '$cemail', '$cphone')"; It produces identical results. — Michael De Luca
– Michael De Luca, Commented Oct 8, 2015 at 9:03
You should print mysqli_error($con) when the query fails so you see the reason for the failure. — Barmar
– Barmar, Commented Oct 8, 2015 at 9:06
I find that hard to believe. If a query fails, mysqli_error always returns the reason for the failure. — Barmar
– Barmar, Commented Oct 8, 2015 at 9:11
I have replaced the original if/else block with mysqli_query($con, $query) or die(mysqli_error($con)); echo mysqli_error($con); Still a blank page. If this helps, the file has 0644 permissions. EDIT: I greatly appreciate your help, btw :) — Michael De Luca
– Michael De Luca, Commented Oct 8, 2015 at 9:13

CodeLove · Accepted Answer · 2015-10-08 09:38:11Z

2

Your Code:

$con = mysqli_connect("localhost","username","password","database");

Edited code:

$con = mysqli_connect("localhost","root","","database");
if(!$con):
die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());
endif;

Its worked fine on my localhost

edited Oct 8, 2015 at 9:38

answered Oct 8, 2015 at 9:34

CodeLove

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6 Comments

Michael De Luca Over a year ago

Connect Error (1045) Access denied for user 'mytestuser'@'localhost' (using password: NO) When I put the password in I get Connect Error (1045) Access denied for user 'xsoft16dbu'@'localhost' (using password: YES)

CodeLove Over a year ago

you should change root instead of username. Are you running your code on local?

Barmar Over a year ago

@MichaelDeLuca You need to add appropriate grants for mytestuser to the database.

Michael De Luca Over a year ago

I think i should add some background here. This is a Godaddy cPanel shared hosting. I have already went through the wizard to create both the table and the user, and grant the user select and insert access. Beyond that, unless Godaddy's cPanel is pulling me in for a nasty ride, I have no idea why a user I created and granted access still does not have access.

Michael De Luca Over a year ago

Ok, now I have simply reset the password to one that is only alphanumeric, and retried: Connect Error (1044) Access denied for user 'myuser'@'localhost' to database 'mydb'

|

Marcus · Accepted Answer · 2015-10-08 09:16:53Z

1

Maybe the datatypes of our columns make any problems.

I have once a similar "error" where my token field was to short. phpMyAdmin simply cut the long string to fit (or some thing this) and so it worked inside phpMyAdmin but not with my program.

Please post the CREATE statement of your table.

answered Oct 8, 2015 at 9:16

Marcus

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2 Comments

Michael De Luca Over a year ago

CREATE TABLE IF NOT EXISTS teams ( token varchar(12) COLLATE utf8_unicode_ci NOT NULL, tname varchar(48) COLLATE utf8_unicode_ci NOT NULL, captain varchar(64) COLLATE utf8_unicode_ci NOT NULL, email varchar(64) COLLATE utf8_unicode_ci NOT NULL, phone varchar(14) COLLATE utf8_unicode_ci NOT NULL, PRIMARY KEY (token), UNIQUE KEY name (tname), KEY id (token) ) ENGINE=MyISAM DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

Michael De Luca Over a year ago

I'm noticing an needed reference to an id key which no longer exists

Artemiy StagnantIce Alexeew · Accepted Answer · 2015-10-08 09:07:28Z

0

Try this

mysqli_query($con, $query) or die(mysqli_error($con));

answered Oct 8, 2015 at 9:07

Artemiy StagnantIce Alexeew

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2 Comments

Barmar Over a year ago

This will tell him the reason for the error, it won't solve the problem. It should be a comment.

Michael De Luca Over a year ago

I replaced the whole if/else block with: mysqli_query($con, $query) or die(mysqli_error($con)); The result was a blank page

Hrishikesh Waghmare · Accepted Answer · 2015-10-08 09:50:25Z

0

I will suggest go step by step step 1 : $con = mysqli_connect("localhost","username","password","database"); comment everything other than this statement. then below write

  if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

verify your username password and database name (by default:username=="root" password=="" )

step 2 : add the query statements and print it. also check that all values are within their bounds as specified id DB.

step 3 :if $con and $query are valid then you will get your o/p as you req.

If still you have an error please paste your error statement.

answered Oct 8, 2015 at 9:50

Hrishikesh Waghmare

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Comments

tchap · Accepted Answer · 2015-10-08 10:44:16Z

0

using following to connect:

username="root" password=""

Your code works fine in my system.

I used varchar as datatype. Maybe the length of the varchar which you have taken as 12 is less to store the $teamToken variable. Try increasing the size.

edited Oct 8, 2015 at 10:44

tchap

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answered Oct 8, 2015 at 10:32

Saurabh Katiyar

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