again i am seeking for you help! I am new to NuSOAP, so please bear with me. I am trying to send multiple records, with multiple columns to my service, split this into records and writting them into my sql table.
For example, i have a batch that updates my variable like so (records are separated with ";", columns inside records are separated with "¤")
Name1¤Surname1¤Date1¤number1;Name2¤Surname2¤Date2¤number2;Name3¤Surname3¤Date3¤number3 ...
I have a simple function which accepts this variable. (1st of all i dont know if sending a string is optimal ... I read that i should be sending an xml document ...)
So if i declare a new variable inside my script and past the exact value that my program sets up, execute the script, everything works! Records are written in a table without any problem. I wrote up to 500 records. The problem is when i call my webservice ... In that case i get an error:
"SOAP Fault: error in msg parsing: XML error parsing SOAP payload on line 1: Invalid character:"
I think i am sending a to many chars in my variable ... Again i am new to NuSOAP and i am trying to figure this out based on an example i found online ...
When i was sending just text with only 1 delimeter, i was able to sent and write 500 records. The variable was setup like so:
TEST001;TEST002;TEST003; ...;TEST500
And the web service recieved the variable and wrote all 500 records to the table. Can someone please help me out or tell me the correct way of doing this?
Regards,
HEki
<?php
require 'lib/nusoap.php';
$server = new nusoap_server();
$server->configureWSDL("test"."urn:test");
$server->register(
"service",
array("variable_text"=>'xsd:string'),
array("return"=>"xsd:string")
);
function service($variable_text)
{
$mysql_hostname = "localhost";
$mysql_user = "root";
$mysql_password = "root";
$mysql_database = "service";
$today = date("Y-m-d");
$response='START';
// Connect to database server
$con = mysqli_connect($mysql_hostname, $mysql_user, $mysql_password, $mysql_database);
if (mysqli_connect_errno()) {
echo "ERROR: " . mysqli_connect_error();
}
$token = strtok($variable_text, ";");
while ($token !== false)
{
$data = explode('¤', $token);
$data0 = $data[0];
$data1 = $data[1];
$data2 = $data[2];
$data3 = (int)$data[3];
$strSQL = "INSERT INTO test (column1,column2,column3,column4) VALUES ('".$data0."','".$data1."','".$data2."', '".$data3."')";
mysqli_query($con,$strSQL);
$response='NEW';
$token = strtok(";");
}
// Close the database connection
mysqli_close($con);
return $response;
}
$HTTP_RAW_POST_DATA = isset($HTTP_RAW_POST_DATA) ? $HTTP_RAW_POST_DATA : '';
$server->service($HTTP_RAW_POST_DATA);
?>
Related
I'm currently working on a project. It's almost done, there's only one big problem. I tested my code all the time with a xamp server on my computer, which worked perfectly fine. the goal is to run it (apache server, mysql database) on my raspberry pi. Now my project is finished, I came figured out the problem why my code doesn't work on my raspberry (at least not as I expected).
I turned on error reporting in PHP and came to this error message:
Notice: Trying to get property of non-object in /var/www/html/test.php on line 41
I use this function for all my SQL queries. Can someone provide a solution so I don't have to rewrite the whole code? Thanks in advance!
PS: this is just a piece of the code (the function where I pull the data out of the database + example of one of my queries)
<?php
// Enable debugging
error_reporting(E_ALL);
ini_set('display_errors', true);
$servername = "localhost";
$username = "root";
$password = "*****"; // I just dont want to give my sql database password its nothing wrong ;)
$dbname = "test";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} else {
print_r("ok connection");
function sqlquery ($sql, $conn, $naamtabel) {
global $myArray;
global $stateLoop;
$stateLoop = "0";
$result = $conn->query($sql);
if ($result->num_rows > 0) { //line 41 in my code ==> do a while loop to fetch all data to an array
// output data of each row
while($row = $result->fetch_assoc()) {
$myArray[] = $row["$naamtabel"]; //alle data van kolom "tijd" in een array
}
$stateLoop = "1";
}
else { // if there are no results
}
}
$sql1 = "SELECT stopTijd FROM gespeeldeTijd WHERE naam = 'thomas' ORDER BY ID DESC LIMIT 1"; // get data with SQL query
sqlquery($sql1,$conn,"stopTijd");
if ( $stateLoop == "1") {
print_r("ok loop");
$date1 = $myArray["0"];
print_r($date1);
$myArray = [];
$stateLoop == "0";
}
}
?>
It pretty much looks like you have some sql error in your query; check if your field names in your database match those on the raspberry.
Seeing through your code it seems like you are pretty new to programming (which is no bad thing, I was once, too). So I made a few more modifications to your code showing you the prettiness of PHP
use "return" in function sqlquery instead of globals
check for errors after executing the code
use only one variable to check if data was loaded
I commented everything I changed
<?php
// Enable debugging
error_reporting(E_ALL);
ini_set('display_errors', true);
$servername = "localhost";
$username = "root";
$password = "*****";
$dbname = "test";
// Your function with some modifications
function sqlquery($sql, $conn, $naamtabel) {
$result = $conn->query($sql);
// Check for errors after execution
if(!$result)
die('mysqli error: '. htmlentities(mysqli_error($con)));
// If we have no data, we simply return an empty array
if($result->num_rows == 0)
return array();
// This is a variable we store the data we processed in
// We will return it at the end of our function
$myArray = null;
// Read all field data and store it $myArray
while($row = $result->fetch_assoc())
$myArray[] = $row[$naamtabel]; // if you use "$naamtabel" here, PHP first needs to interpret the string (= slower)
return $myArray;
}
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error)
die("Connection failed: " . $conn->connect_error);
// Because we use "die" above we don't need an "else"-clause
print_r("ok connection");
$sql = "SELECT `stopTijd` FROM `gespeeldeTijd` WHERE `naam` = 'thomas' ORDER BY `ID` DESC LIMIT 1";
$data = sqlquery($sql, $conn, "stopTijd");
// $data will contain $myArray (see "return $myArray" in function sqlquery)
// Instead checking for $stateLoop being "1" we check if $data contains any values
// If so, we fetched some data
if(sizeof($data) >= 1) {
print_r("ok loop");
$date1 = $data[0]; // No "0", because we are trying to get index 0
print_r($date1);
$data = array(); // Are you sure this is nessecary?
} else {
echo 'No data returned from query!';
}
?>
Note: code tipped on my smartphone -> untested!
If you don't want to adapt the code I wrote, the important part for this question is:
if(!$result)
die('mysqli error: '. htmlentities(mysqli_error($con)));
Your error Notice: Trying to get property of non-object means "you are trying to get num_rows from $result, but $result is not an object, so it can't contain this property".
So to figure out why $result is not an object, you need to get the error from $conn->query - my code above probably won't fix your error, but it will display you one you can work with (+ it's too long for a comment)
If you have a more detailed error message and you can't solve it on your own, feel free to comment; I will update my answer!
I've been having a bit of trouble with my PHP code.
I'm trying to insert a new row in table gebruikers.
I'm using a JSON API to post the data from my C# Android app to the server.
running the code returns an invalid request error.
PHP:
function registerUser($api_data)
{
// connection
$servername = "xxx";
$username = "xxx";
$password = "xxx";
$database = "test";
$mysqli = new mysqli($servername, $username, $password, $database);
//check connection
if(mysqli_connect_errno())
{
API_Response(true, 'connection error');
}
$register_data = json_decode($api_data);
$leerlingnummer = intval($register_data->leerlingnummer); //passed as string, int in database
$wachtwoord = $register_data->wachtwoord; //string
$email = $register_data->email; //string
$result = $mysqli->query("INSERT INTO `gebruikers` (`Leerlingnummer`, `Wachtwoord`, `Email`) VALUES ({$leerlingnummer}, {$wachtwoord}, {$email})");
if ($result == false)
{
API_Response(true, "{$mysqli->error}");
}
else
{
API_Response(false, 'SUCCESS');
}
$mysqli->close();
}
database is looking as follows:
database layout
I've never felt this stupid before, but the error came from the fact that I was still referencing to an older .php file. I was so focussed on the PHP script that I didn't notice this error in my app before.
the quotes advised by Sean and fuso were needed later on though so thanks for that.
Problem solved, sorry for wasting some of your time :/
You should quote your data in the insert query.
... VALUES ('{$email}','{$other}')
Here's the PHP code:
<?php
$servername = "***";
$username = "*****";
$password = "*****";
$database = "*****";
try {
$conn = new PDO('mysql:host='.$servername.';dbname='.$database, $username, $password);
console.log('yes!');
}
catch(PDOException $e) {
print "Error!:" . $e->getMessage(). "<br/>";
die();
}
if (isset($_POST['submit']))
{
//$name = $_POST['name'];
//$day = $_POST['day'];
//$acctName = $_POST['acctName'];
//$acctType = $_POST['acctType'];
//$location = $_POST['location'];
//$prospect = $_POST['prospect'];
//$notes = $_POST['notes'];
$name = 'sally sue';
$day = 'monday';
$acctName = 'Account Uno';
$acctType = 'Cold Call';
$location = 'Location';
$prospect = 'Prospect';
$notes = 'These are notes! Notey notey notes';
$order = "INSERT INTO `schedule`(`id`, `name`, `day`, `acctName`, `acctType`, `location`, `prospect`, `notes`) VALUES ('$name', '$day', '$acctName', '$acctType', '$location', '$prospect', '$notes')";
$stmt = $conn->prepare($order);
$stmt->execute();
}
?>
Here's the deal. I have an HTML form that I use jQuery to grab the variables, and AJAX to post the form to this PHP file. I feel confident that everything is fine up to the point where it gets to the PHP file.
I commented out the POST variables and hard-coded my own to make it a little simpler. I'm not getting a 500 Internal Server Error. I've ran my code through a PHP syntax validator (and fixed a billion errors haha). Obviously I'm still doing something wrong, but I cannot find it for the life of me. I'm hoping that someone here has some insight?
EDIT: Also, the username, password, database, and table name are ALL correct. I've double checked them several times. The only thing I'm not sure of is the server name, which is 'localhost' since the DB is on the same server as this web page.
EDIT 2: I've changed the MySql insert statement back to the original, which had the back ticks. I copied it straight from phpMyAdmin console on the server which it resides. It was that way originally, but I changed it due to desperation. It still is not updating my database. Any further ideas?
Thanks in advance!
I've been trying to get my python code to transfer SQL query data from client running python to web server running PHP, then input that data into a MySQL database.
The below is the python test code simulating a single row:
#!/usr/bin/python
import requests
import json
url = 'http://192.168.240.182/insert_from_json.php'
payload = {"device":"gabriel","data_type":"data","zone":1,"sample":5,"count":0,"time_stamp":"00:00"}
headers = {'content-type': 'application/json'}
response = requests.post(url, data=dict(payload=json.dumps(payload)), headers=headers)
print response
The below is the PHP script on the server side:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "practice";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connection made...";
$payload_dump = $_POST['payload']
//$payload = '{"device":"gabriel","data_type":"data","zone":1,"sample":4,"count":0,"time_stamp":"00:00"}';
$payload_array = json_decode($payload_dump,true);
//get the data_payload details
$device = $payload_array['device'];
$type = $payload_array['data_type'];
$zone = $payload_array['zone'];
$sample = $payload_array['sample'];
$count = $payload_array['count'];
$time = $payload_array['time_stamp'];
$sql = "INSERT INTO data(device, data_type, zone, sample, count, time_stamp) VALUES('$device', '$type', '$zone', '$sample', '$count', '$time')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
If I comment out $payload_dump and instead use $payload and run the script the row is added correctly. But when I call the python script on the client I get a "response 200" but the record was not added. Something either my JSON string in python isn't formatted correctly or I am not passing the value correctly.
1) What is the syntax/procedure on the python code that would allow me to see what has actually been received by the PHP code. (i.e. convert the "$payload_dump" variable contents to a string and send back to the python script?
2) How do I resolve the above code?
Your first problem is using json headers which you shouldn't use since you aren't sending a raw json:
your data look like
{payload = {'key1': 'value1',
'key2': 'value2' }
}
and a raw json data should be in the form:
{'key1': 'value1',
'key2': 'value2'}
So you need to remove the headers from the request
response = requests.post(url, data=dict(payload=json.dumps(payload)))
Second you need to fix that missing semicolon
$payload_dump = $_POST['payload']
to
$payload_dump = $_POST['payload'];
Better solution
You can send the json data directly using requests
response = requests.post(url, json=payload)
then grab it in php with
$payload_array = json_decode(file_get_contents('php://input'), true);
The BLOB field (pic) is turning out as 0 Bytes when trying to send ByteArray through as3 to PHP, so i assume the PHP script or the HTTP_RAW_POST_DATA isn't working.
I think the Flash part is working, I have set a trace() to see if the bitmapdata is coming through and it seems it is, so I'm assuming its my php side. I'll post both parts of the code in hope someone here can fix it for me. Thanks.
AS3
private function export():void
{
var bmd:BitmapData = new BitmapData(600, 290);
bmd.draw(board);
var ba:ByteArray = PNGEncoder.encode(bmd);
trace(ba);
var _request:URLRequest = new URLRequest ("http://site.com/readimage.php");
var loader: URLLoader = new URLLoader();
_request.contentType = "application/octet-stream";
_request.method = URLRequestMethod.POST;
_request.data = ba;
loader.load(_request);
}
PHP
<?php
$username = "images";
$password = "password";
$host = "localhost";
$database = "images";
$link = mysql_connect($host, $username, $password);
if (!$link) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db ($database);
$query ="INSERT INTO main (pic) VALUES ('".$GLOBALS["HTTP_RAW_POST_DATA"]."')" or die(mysql_error());
$results = mysql_query($query, $link);
?>
$blob = file_get_contents('php://input');
This should work for you. This accesses PHP's raw input stream. It's more likely to work in some cases, apparently:
php://input allows you to read raw data from the request body. In case of POST requests, it preferrable to $HTTP_RAW_POST_DATA as it does not depend on special php.ini directives. Moreover, for those cases where $HTTP_RAW_POST_DATA is not populated by default, it is a potentially less memory intensive alternative to activating always_populate_raw_post_data.
You'll also want to ensure that you properly escape this data when placing it in the database:
$query = "INSERT INTO main (pic) VALUES ('" . mysql_real_escape_string($blob) . "')";
(It is also possible that $HTTP_RAW_POST_DATA's magic only works when you reference it directly instead of through the $GLOBALS array.)
Try breaking apart your whole process - if it's not working, start stripping away things before you get as far the sql insert...
First off, open up firebug or chrome/safari console and log your data being passed to your php page - then perhaps just start seeing what's being passed:
foreach (getallheaders() as $name => $value) {
echo "$name: $value\n";
}
If you have the console open, it should log the echo's to that.