I have been working on a program for a job interview coming up soon, and I was nearing the completion of the program which had everything running the way I needed it to... but then my computer crashed. When I opened up the files again, everything was the same, nothing changed, all my changes were saved before it crashed. Only thing is, now my MySQL table doesn't get the data sent to it from the INSERT code. Is there something I can do to make this work?
I have tried creating a new database, a different table, restarting my computers, restarting Chrome, everything.... I can't get it and I'm desperate at this point.
Please see the code below...
// Connect to the database.
$link2 = mysqli_connect("localhost", "cl60-booking", "XXXXXXX", "cl60-booking");
if (mysqli_connect_error()) {
die ("There was an error connecting to the database");
}
// Update the bookings DB with the user's hotel room.
$query = "INSERT INTO booking
(`beds`, `baths`, `booked`, `checkInDate`, `checkOutDate`)
VALUES('".mysqli_real_escape_string($link2,
$_POST['bedNumber'])."',
'".mysqli_real_escape_string($link2, $_POST['bathNumber'])."',
'Yes', '".mysqli_real_escape_string($link2,
$_POST['checkIn'])."',
'".mysqli_real_escape_string($link2,
$_POST['checkOut'])."')";
Use the following code segment :
mysqli_query($link2, $query);
Related
I am currently looking to run a basic insert query using PHP to submit HTML form data to MySQL database.
Unfortunately however the insert process isnt running.
In my Insert syntax I have tried including $_POST[fieldname], ive tried including variables as below, and ive even played around with different apostrphes but nothing seems to be working.
as a side dish, im also getting truck load of wamp deprication errors which is overwhelming, ive disabled in php.ini and php for apache.ini file and still coming up.
If anyone can advise what is wrong with my insert and anything else id be much thankful.
Ill keep this intro straightfoward.
Person logs in, if they try to get in without login they go back to login page to login.
I connect to database using external config file to save me updating in 50 places when hosting elsewhere.
Config file is working fine so not shown below.
database is called mydb.
Im storing the text field items into variables, then using the variables in the insert query.
unitID is an auto increment field so I leave that blank when running the insert.
Unfortunately nothing is going in to the mysql database.
Thanks in advance.
PS the text fieldnames are all correctly matched up
<?php
//Start the session
session_start();
//check the user is logged in
if (!(isset($_SESSION['Username']) )) {
header ("Location: LoginPage.php?i=1");
exit();
}
//Connect to the database
include 'config.php';
$UserName = $_SESSION['Username'];
$UserIdentification = $_SESSION['UserID'];
if(isset($_GET['i'])){
if($_GET['i'] == '1'){
$tblName="sightings";
//Form Values into store
$loco =$_POST['txtloco'];
$where =$_POST['txtwhere'];
$when =$_POST['txtdate'];
$time =$_POST['txttime'];
$origin =$_POST['txtorigin'];
$dest =$_POST['txtdest'];
$headcode =$_POST['txtheadcode'];
$sql= "INSERT INTO sightings (unitID, Class, Sighted, Date, Time, Origin, Destination, Headcode, UserID) VALUES ('','$loco', '$where', '$when', '$time', '$origin', '$dest', '$headcode', '$UserIdentification')";
mysql_select_db('mydb');
$result=mysql_query($sql, $db);
if($result){
$allocationsuccess = "Save Successful";
header ('Refresh: 2; url= create.php');
}
else {
$allocationsuccess = "The submission failed :(";
}
}
}
?>
"unitID is an auto increment field so I leave that blank when running
the insert"
That's not how it works. You have to omit it completely from the INSERT statement. The code thinks you're trying to set that field to a blank string, which is not allowed.
$sql= "INSERT INTO sightings (Class, Sighted, Date, Time, Origin, Destination, Headcode, UserID) VALUES ('$loco', '$where', '$when', '$time', '$origin', '$dest', '$headcode', '$UserIdentification')";
should fix that particular issue. MySQL will generate a value automatically for the field and insert it for you when it creates the row.
If your code had been logging the message produced by mysql_error() whenever mysql_query() returns false then you'd have seen an error being generated by your query, which might have given you a clue as to what was happening.
P.S. As mentioned in the comments, you need to re-write your code with a newer mysql code library and better techniques including parameterisation, to avoid the various vulnerabilities you're currently exposed to.
I am using the external authentication system. Therefore, there are a lot of user data, which is not available in Phorum.
I am using the last post module, although I want to get the information from the last post user, from my own user table (I have some data, like avatar, birth info etc). I want to show in my Phorum. How can I achieve this?
I've tried to simply connect via a: mysql_query(); but then I just get No database selected error.
I've searched for hours - I cannot find any documentation regarding getting custom data from your own user table.
I would recommend using mysqli, as mysql is deprecated. First make sure that your connection is correct. No database selected means you probably do not have your connection included at the top.
$con = mysqli_connect("localhost","username","password","database");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: ".mysqli_connect_error());
}
Make sure that your sql statement looks like so (Notice the $con in the mysqli_query()):
$sql = "select * from TableName";
if ($que = mysqli_query($con, $sql)) {
// Query has ran
}
This is the code that connects to my SQL database. I'm new with this stuff and it seems to be semi-working but certain features on my website still don't work.
<?php
$con = mysql_connect("localhost","username","password");
$select_db = mysql_select_db('database1',$con);
/*$con = mysql_connect("localhost","username2","password2");
$select_db = mysql_select_db('database2',$con);*/
?>
This is the site in question: http://tmatube.com keep in mind the credentials above are filled in with what the programmer used for testing on his own server... ;) unfortunately I don't have access to him for support anymore.
Anyway, here's my thoughts on how this code needs to be edited maybe someone can chime in and let me know if I'm correct in my assumptions:
<?php
$con = mysql_connect("localhost","username1","password1"); -------------<<< leave this line
$select_db = mysql_select_db('DATABASE_NAME_HERE',$con);
/*$con = mysql_connect("localhost","DB_USERNAME_HERE","DB_PASSWORD_HERE");
$select_db = mysql_select_db('DATABASE_NAME_HERE',$con);*/
?>
Ok - now on to a few problems I noticed...
What does this do? /* code here */? It doesn't work at all if I leave that bit in.
Why is it connecting to database twice? and is it two separate databases?
$select_db = mysql_select_db('DATABASE_NAME_HERE',$con); <<<---- single '
When I tried to see if that line was correct the examples I saw had quotes like this
$select_db = mysql_select_db("DATABASE_NAME_HERE",$con); <<<---- double "
Which one is right?
He didn't leave it out. What he did was leave the database to be connected using the root, which has no password. The other connection (which is commented out) is using another user, rajvivya_video, with a password defined.
In testing it MIGHT be okay to connect to root and leave it without password, but even that is not recommended, since its so easy to work with a user and password defined (besides root).
Here is php mysql connect with mysqli:
<?php
$link = mysqli_connect("myhost","myuser","mypassw","mybd");
?>
No difference here with ' or ". (Anyway use mysqli and you can the wanted db as 4th parameter.) php quotes
/* comment */ is a commented out so the php does not care what is inside so only 2 first rows of are affecting (they are same mysql database on the local machine and 2 different user + password combinations). Comment in general are used to explain the code or removing part of the code with out erasing it. php commenting
I'm trying to increment +1 impression every time an ad is displayed on my site, however the variable increments +2 to +3 arbitrarily. I've removed everything that's working correctly and I made a page with only this code in it:
<?php
require "connect_to_mydb.php";
echo 'Hello***** '.$testVariable=$testVariable+1;
mysql_query("UPDATE `imageAds` SET `test`=`test`+1 WHERE `id`='1'");
?>
Every time the page is refreshed the, test increments arbitrarily either +2 or +3 and my page displays Hello***** 1 (Just to show its not looping). Access is restricted to this page so it's not other users refreshing the page.
Also, id and test are int(11) in the DB.
My DB required connection has nothing in it that would interfere.
Edit
Here is an updated code:
<?php
require "connect_to_mydb.php";
mysql_query("UPDATE `imageAds` SET `test`=`test`+1 WHERE `id`='1'");
$sql = mysql_query("SELECT * FROM imageAds WHERE id='1' LIMIT 1");
$check = mysql_num_rows($sql);
if($check > 0){
$row = mysql_fetch_array($sql);
echo $row['test'];
}
?>
Increments by +2 everytime
Edit
This is whats in connect_to_mydb.php
<?php
$db_host = "*************************";
$db_username = "*********";
$db_pass = "**********";
$db_name = "**************";
mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql");
mysql_select_db("$db_name") or die ("no database");
?>
Either there's a bug in MySQL's implementation of UPDATE, or you're doing something wrong in some code you haven't posted.
Hint: It's very unlikely to be a bug in MySQL. Other people would have noticed it.
From what you've shown, it looks like your page is being loaded multiple times.
This attempt to prove that the code is only being called once doesn't prove anything:
echo 'Hello***** '.$testVariable=$testVariable+1;
This will always print the same thing (Hello***** 1) even if you open this page multiple times because the value of $testVariable is not preserved across seperate requests.
This +2/+3 error is occurring only with Chrome and my Mobile Android browser and the code is solid. I looked to see if there is any issue with Chrome sending more than one http request (thx user1058351) and there is which is documented here:
http://code.google.com/p/chromium/issues/detail?id=39402
So since this way was unreliable I just completed a work around that is solid. Instead of including a PHP file that updates the amount of ad impressions on reload, I now have it so when the page loads, an AJAX request is sent to a separate PHP file which updates the ad stats and returns the appropriate data. The key I think is to send it through the JS code so only one http request can be sent to increment the data.
Thank you to all who responded especially user1058351 and Mark Byers (not a bug in MYSQL but possibly appears to be a bug in Chrome).
I am new to Joomla and new to php (wish I was so new in age too). I have installed joomla on a local apache webserver. I am trying to use some php code in a joomla article in order to fetch some data from a Sybase ASE 12.5 database. I installed sourcerer and started to try an ODBC connection using a system DSN (which I verified it is working):
{source}
<?php
echo 'This text is placed through <b>PHP</b>!';
echo '<p>';
echo '</p>';
$conn = odbc_connect('myDSN', 'sa', 'myPassword') or die('Could not connect !');
echo 'Connected successfully';
$sql = 'SELECT day, name FROM my_table where month = 1';
odbc_close($conn);
?>
{/source}
The above code doesn't do much, but this is how far I can get without problems. I refresh the joomla page and I see inside the article's text:
...
This text is placed through PHP!
Connected successfully
...
Seems ok, the connection obviously established (I verified this by stopping the sybase service and getting the "Could not connect" message). Then I added one more line, just below the $sql assignment.
$rs = odbc_exec($conn,$sql);
I refresh and ...I see nothing coming from the script (not even the "This text is placed through PHP!").
Obviously, I see nothing if I include code to echo the contents of $rs. I also tried this but in vain.
if (!$rs)
{exit("Error in SQL");}
Once I add the odbc_exec command, the entire script ceases working.
In php.ini I read:
; Windows Extensions
; Note that ODBC support is built in, so no dll is needed for it.
Do you have any idea what is going wrong?
UPDATE
Connecting to a MySQL database using code like this, works like a charm.
To answer your question
Is this the correct way to get data
from another database in to an article
or should I use some plug-in to do
that?
The correct way is to use JDatabase object which you get by JFactory::getDBO() or JDatabase::getInstance(), see Joomla JDatabase documentation.
$db = JDatabase::getInstance( $databasConfigArray );
$db->setQuery('your query');
$data = $db->loadObjectList();
Here is a good tutorial showing how to connect to multiple databases in Joomla, there is even source code for helper class.
Also look at this thread Connecting to 3rd party databse in Joomla!?