php insert mysql not working

Question

.I don't know if it's syntax or what. I've tried a variety of ways this is the simplest I thought would work.

I send info to the userData.php using:

http://mydomain.com/adverts/userStats.php?name=001EC946C2F4&adNum=1&playClick=1

On the userData.php I have:

<?php 
    $db = mysql_connect('localhost', 'username', 'password') or die('Could not connect: ' . mysql_error()); 
   $db_selected =  mysql_select_db('databaseName', $db) or die('Could not select database');
if (!$db_selected)
  {
  die ("Can\'t use test_db : " . mysql_error());
  }

    $name = mysql_real_escape_string($_GET['name']); 
    $date = date("d/m/Y");
    $adClick = mysql_real_escape_string($_GET['adNum]);
    $playN = mysql_real_escape_string($_GET['playClick']);

$query = mysql_query("INSERT INTO playerData VALUES ('$name', '$date','$adClick','$playN')");
$result = mysql_query($query) or die('Query failed: ' . mysql_error())); 

mysql_close($db);
?>

I manually added 2 records to the table from phpMyAdmin, and I can display or update them just fine but adding a new record isn't working. I simply want to start a new record each time the link is called from another program, and store the mac address, date, adNum, and playClick.

EDIT2:: echo $query; for

http://simplehotkey.com/adverts/userStats.php?name=001EC946C2F4&adNum=1&playClick=1

outputs:
INSERT INTO playerData(mac,date,AdClick,PlayNum) VALUES ('001EC946C2F4', '26/07/2012','1','1')

Which is what I want it's just not adding it to the DB.

there could be the syntax error as @Daedalus mansion. Ans Daedalus and Swapnesh both of you let me tell you one thing you are right we have to pass DB handler, But its fine without DB handler it will work even he has not passed DB handler — Hiren Soni
– Hiren Soni, Commented Jul 27, 2012 at 4:46
@HirenSoni its better if we follow the procedures to minimize the error as we are not into his system, what if he is running php code without having php on it system as this might be a case from infinite error possibilities ;) — swapnesh
– swapnesh, Commented Jul 27, 2012 at 4:50
@swapnesh I am totally agreed with you, but as far as the error is concern that is not the mistake :) — Hiren Soni
– Hiren Soni, Commented Jul 27, 2012 at 4:54
I'm using bluehost. I can manually add records to the DB in phpMyAdmin through the control panel. I'm reading data out of the db and into my game, but trying to add records from my game to the mysql isn't working where would php errors be thrown in bluehost? — user1108224
– user1108224, Commented Jul 27, 2012 at 5:02
@user1108224 Check my new edited answer and let me know then — swapnesh
– swapnesh, Commented Jul 27, 2012 at 5:06

swapnesh · Accepted Answer · 2012-07-27 05:38:41Z

2

Correct syntax is --

mysql_select_db("databaseName", $db);

And its better if u use something like this for connection errors--

$db_selected= mysql_select_db("databaseName", $db);
if (!$db_selected)
  {
  die ("Can\'t use test_db : " . mysql_error());
  }

EDIT

You are writing all wrong :(

$query = mysql_query("INSERT INTO playerData VALUES ('$name', '$date','$adClick','$playN')");
$result = mysql_query($query)  <--------------WRONG

Try Something like this----

$query = "INSERT INTO playerData(CORRECT_COL_NAMES) VALUES ('$name', '$date','$adClick','$playN')";
$results = mysql_query($query, $connection);

NEW EDIT

AREA OF ERROR---- WRONG DATATYPE

','1','1' <--- this is passing as string while u have have this as an int in your db structure ..now run the same query as it is to figure out the error..also u can figure out using $result = mysql_query($query) or die(mysql_error());

edited Jul 27, 2012 at 5:38

answered Jul 27, 2012 at 4:39

swapnesh

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7 Comments

user1108224 Over a year ago

ugh I keep hitting enter to make a new line before I've typed my comment

swapnesh Over a year ago

@user1108224 echo the query and tell me then and what is the structure of playerData ???

user1108224 Over a year ago

problem is I'm not viewing it in the browser. I send the url from game. If I go to the URL of the php file nothing shows up after sending the URL from the game, in the game I can debug out that it sent the URL. I'm calling simplehotkey.com/adverts/… in the game where simplehotkey.com/adverts/userStats.php is the base script

user1108224 Over a year ago

I tried the $results = mysql_query($query, $db); still no go ugh

swapnesh Over a year ago

@user1108224 i think ','1','1' <--- this is passing as string while u have have this as an int in your db structure ..now run the same query as it is to figure out the error and let me know then...also u can figure out using $result = mysql_query($query) or die(mysql_error());

|

Daedalus · Accepted Answer · 2012-07-27 05:20:48Z

2

It's pretty easy to see what's wrong here, especially with syntax highlighting.

$adClick = mysql_real_escape_string($_GET['adNum]);

This line is missing a single quote mark; it should be:

$adClick = mysql_real_escape_string($_GET['adNum']);

This is a syntax error that ruins everything else.

Not to mention that your database selection is missing your database handler, ie:

mysql_select_db('databasename',$db);

As pointed out by @swapnesh, and as noted here.

Edit

I have been unable to reproduce your lack of an error, what I have gotten however, are errors. Firstly, you have an extra ) at line 12:

$result = mysql_query($query) or die('Query failed: ' . mysql_error()));

Should be:

$result = mysql_query($query) or die('Query failed: ' . mysql_error());

Lastly, you actually improperly execute your query twice, so the second time, the query is empty. What you have:

$query = mysql_query("INSERT INTO playerData VALUES ('$name', '$date','$adClick','$playN')");
$result = mysql_query($query) or die('Query failed: ' . mysql_error()));

Should instead be:

$query = "INSERT INTO playerData VALUES ('$name', '$date','$adClick','$playN')";
$result = mysql_query($query) or die('Query failed: ' . mysql_error());

edited Jul 27, 2012 at 5:20

answered Jul 27, 2012 at 4:39

Daedalus

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4 Comments

user1108224 Over a year ago

ah, I added the connection parameter from the answer above and added the missing ' but it still isn't working.

Daedalus Over a year ago

@user1108224 It isn't working isn't a problem description, unfortunately; please post the error as well.

user1108224 Over a year ago

where would I get an error? It's not throwing any so it just does nothing. I'm sending the link from a game. I'm retrieving the records from the table and displaying in game. When I submit a score, it does not show up in the database or in my game stats from display.php

user1108224 Over a year ago

my display.php: simplehotkey.com/adverts/display.php shows the 2 records that I manually added to the DB, I use that in my game to show the stats just fine, but insert isn't doing anything

Conrad Lotz · Accepted Answer · 2012-07-27 04:47:36Z

0

Instead of using the insert statement the way you do add the fields that will receive entries explicitly. The database table might have more fields and the insert statement does not explcitly state which fields will receive data.

$query = mysql_query("INSERT INTO playerData (Name,Date,AdClick,PlayN) VALUES ('$name', '$date','$adClick','$playN')");

answered Jul 27, 2012 at 4:47

Conrad Lotz

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1 Comment

user1108224 Over a year ago

I had it that way. I just swapped that in and it's still not adding anything...thanks though

Sumesh TG · Accepted Answer · 2012-07-27 05:01:05Z

0

You have the syntax error on this line

Wrong :

$adClick = mysql_real_escape_string($_GET['adNum]);

Correct :

$adClick = mysql_real_escape_string($_GET['adNum']);

answered Jul 27, 2012 at 5:01

Sumesh TG

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