How do I insert into a database with AJAX method

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Odin
- Joined 16 Mar, 2014
- 27 topics • 78 posts
1

11 May, 2015
Quote

I want to save one variable every 10 sec in the user row.

My php code:

<?php
	// Variables 
	$nombreUsuario = $_GET['nombreUsuario'];
	$passUsuario = $_GET['passUsuario'];
	$distanciaTotal = $_GET['distanciaTotal'];
	
	$con=mysqli_connect("mysql","database","userid","password");
	// Check connection
	if (mysqli_connect_errno())
	{
	 echo "Failed to connect to MySQL: " . mysqli_connect_error();
	 }
	$query =("INSERT INTO usuarios_aep (distanciaTotal) VALUES (".$distanciaTotal.") 
					WHERE nombreUsuario like ".$nombreUsuario." 
					AND passUsuario like ".$passUsuario." 
				");
	$mysqli->query($query);
	
	mysqli_close($con);
?>[/code:tb2wf9on]

event "every 10 sec"
[code:tb2wf9on]"http://fakefake.com/fakefake/php/insertDistanciaTotal.php?nombreUsuario='"&nombreUsuario&"'&passUsuario='"&passUsuario&"'&distanciaTotal='"&distanciaTotal&"'"[/code:tb2wf9on]

- lennaert
- - Joined 8 Oct, 2013
  - 94 topics • 2,110 posts
- 1
- 11 May, 2015
- Quote
After connecting, and after you have done the INSERT query, with your current approach with using a Ajax GET request, the following should work.

[quote:33zcixpd]$query =("UPDATE usuarios_aep SET distanciaTotal = '".$distanciaTotal."' WHERE nombreUsuario = '".$_GET['nombreUsuario']."' AND passUsuario = '".$_GET['passUsuario']."' ");
$mysqli->query($query);

I would suggest switching to post methods, otherwise it would become very easy to cheat

Odin
- Joined 16 Mar, 2014
- 27 topics • 78 posts
1

14 May, 2015
Quote

After connecting, and after you have done the INSERT query, with your current approach with using a Ajax GET request, the following should work.

[quote:2jh1h120]$query =("UPDATE usuarios_aep SET distanciaTotal = '".$distanciaTotal."' WHERE nombreUsuario = '".$_GET['nombreUsuario']."' AND passUsuario = '".$_GET['passUsuario']."' ");

$mysqli->query($query);

I would suggest switching to post methods, otherwise it would become very easy to cheat

I come back with more problems!!

Actually I have this php in my server, I tested it from chrome and it works, makes the UPDATE correctly.

<?php
	// Variables 
	$nombreUsuario = "XXuser";
	$passUsuario = "XXpass";
	$distanciaTotal = $_GET['distanciaTotal'];
	
	$con=mysqli_connect("blablabla.com","blebleble","bliblibli","blublublu");
	// Check connection
	if (mysqli_connect_errno())
	{
	 echo "Failed to connect to MySQL: " . mysqli_connect_error();
	 }
	$qz =("UPDATE usuarios_aep SET distanciaTotal = '".$distanciaTotal."' 
				WHERE nombreUsuario = '".$nombreUsuario."' 
				AND passUsuario = '".$passUsuario."' 
			");
			
	if (mysqli_query($con, $qz)) {
		echo "Record updated successfully";
	} else {
		echo "Error updating record: " . mysqli_error($con);
	}
	mysqli_close($con);
?>[/code:2jh1h120]

The problem appears when c2 has send (POST or GET method) the data to php, it seems that I am not sending data at all...  so I think that the problem is in the next images (the AJAX connection) but I don't know how to solve it.
[img="http://www.therealgeekgirl.com/server1.jpg"]
[img="http://www.therealgeekgirl.com/server2.jpg"]

- lennaert
- - Joined 8 Oct, 2013
  - 94 topics • 2,110 posts
- 1
- 14 May, 2015
- Quote
yeah remove the single qoutes in the data entry.

Resulting in : "distanceTotal="&distanciaTotal

To add more

"distanceTotal="&distanciaTotal&"&playerID="&SomeConstruct2Var

Result would look like

"distanceTotal=100&playerID=10"

Odin
- Joined 16 Mar, 2014
- 27 topics • 78 posts
1

14 May, 2015
Quote

yeah remove the single qoutes in the data entry.

Resulting in : "distanceTotal="&distanciaTotal

To add more

"distanceTotal="&distanciaTotal&"&playerID="&SomeConstruct2Var

Result would look like

"distanceTotal=100&playerID=10"

Hello again, first of all so many thanks you !!

It seems I can connect properly to send data to mysql db, my problem now is that I can't acces to data from the db!! My problem now is in the loging.

My construct call is this:

"http://myweb.com/myproject/php/login.php?nombreUsuario=" & nombreUsuario & "&passUsuario=" & passUsuario[/code:3tqmklo3]

An this is my php wich is actually working properly (i tested it):

[code:3tqmklo3]<?php
	// Variables 
	$nombreUsuario = $_GET['nombreUsuario'];
	$passUsuario = $_GET['passUsuario'];
	
	$con=mysqli_connect("blebleble","bliblibli","blobloblo","blublublu");
	// Check connection
	if (mysqli_connect_errno())
	{
	 echo "Failed to connect to MySQL: " . mysqli_connect_error();
	 }
	$qz = "SELECT idUsuario FROM usuarios_aep where nombreUsuario='".$nombreUsuario."' and passUsuario='".$passUsuario."'" ;
	$qz = str_replace("\'","",$qz);
	$result = mysqli_query($con,$qz);
	
	if ($result->num_rows > 0) {
		while($row = $result->fetch_assoc()){
			echo $row["idUsuario"];
		}
	} else {
		echo "Error al acceder a los datos: " . mysqli_error($con);
	}
	mysqli_close($con);
?>[/code:3tqmklo3]

I try to get as data "idUsuario" and I save this id in a variable, but when I run my project in debug mode the content of that variable is NaN (not a Number). I'm not sure what is happening, any idea??

thanks.

Odin
- Joined 16 Mar, 2014
- 27 topics • 78 posts
1

14 May, 2015
Quote

yeah remove the single qoutes in the data entry.

Resulting in : "distanceTotal="&distanciaTotal

To add more

"distanceTotal="&distanciaTotal&"&playerID="&SomeConstruct2Var

Result would look like

"distanceTotal=100&playerID=10"

Hello again, first of all so many thanks you !!

It seems I can connect properly to send data to mysql db, my problem now is that I can't acces to data from the db!! My problem now is in the loging.

My construct call is this:

"http://myweb.com/myproject/php/login.php?nombreUsuario=" & nombreUsuario & "&passUsuario=" & passUsuario[/code:3dvx20px]

An this is my php wich is actually working properly (i tested it):

[code:3dvx20px]<?php
	// Variables 
	$nombreUsuario = $_GET['nombreUsuario'];
	$passUsuario = $_GET['passUsuario'];
	
	$con=mysqli_connect("blebleble","bliblibli","blobloblo","blublublu");
	// Check connection
	if (mysqli_connect_errno())
	{
	 echo "Failed to connect to MySQL: " . mysqli_connect_error();
	 }
	$qz = "SELECT idUsuario FROM usuarios_aep where nombreUsuario='".$nombreUsuario."' and passUsuario='".$passUsuario."'" ;
	$qz = str_replace("\'","",$qz);
	$result = mysqli_query($con,$qz);
	
	if ($result->num_rows > 0) {
		while($row = $result->fetch_assoc()){
			echo $row["idUsuario"];
		}
	} else {
		echo "Error al acceder a los datos: " . mysqli_error($con);
	}
	mysqli_close($con);
?>[/code:3dvx20px]

I try to get as data "idUsuario" and I save this id in a variable, but when I run my project in debug mode the content of that variable is NaN (not a Number). I'm not sure what is happening, any idea??

thanks.

- Fireche
- - Joined 23 Mar, 2015
  - 54 topics • 230 posts
- 1
- 15 May, 2015
- Quote
NaN means Not a Number. Save the output in a text variable and check what it says. Also I would suggest to use PDO rather than mysqli.
- Odin
- - Joined 16 Mar, 2014
  - 27 topics • 78 posts
- 1
- 15 May, 2015
- Quote
NaN means Not a Number. Save the output in a text variable and check what it says. Also I would suggest to use PDO rather than mysqli.

If I change the variable to text the return is "" too, I think the problem is in the request but I can't find the error.
- lennaert
- - Joined 8 Oct, 2013
  - 94 topics • 2,110 posts
- 1
- 15 May, 2015
- Quote
If you still use GET methods you should be able to use that link:

"http://myweb.com/myproject/php/login.php?nombreUsuario=" & nombreUsuario & "&passUsuario=" & passUsuario

as

http://myweb.com/myproject/php/login.ph ... sUsuario=1

in the browser and see if you get the results you are expecting. (just replace 10 and 1 with actual numbers)

You could perhaps enable error reporting in PHP, or add a : echo mysql_error(); below the query.

But, if you switched from GET methods to POST methods, you should update your PHP to reflect that.

ie

the $_GET['meh']

should become

$_POST['meh']
- gumshoe2029
- - Joined 4 Mar, 2014
  - 31 topics • 1,416 posts
- 1
- 15 May, 2015
- Quote
Make sure you use PreparedStatements, please!

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- Odin
- - Joined 16 Mar, 2014
  - 27 topics • 78 posts
- 1
- 16 May, 2015
- Quote
Finally i found the error. The php code need "header("Access-Control-Allow-Origin: *");" in their header

Rookie mistake , thanks for the help.