Problems with JPGraph unable to display - php

I am trying to retrieve the data from a mysql data and show it in a PHP page running a session. The code runs in a separate page. And the data is retrieved as Image in another page.
The main problem is no matter what kind of coding that I put above the default graph coding, it will become a broken image. FYI I have already enabled php_gd2.dll and the extension in my php.ini file is set to the extension folder. Are there any other fixes that has to be made in any of the files?
<?php
session_start();
if($_SESSION['username'])
{
require "dbc.php";
echo "<h2>Patient Glucose Data</h2>";
$username = $_SESSION['username'];
$query_view = mysql_query("SELECT * FROM user_glucose WHERE username='$username'");
$numrows = mysql_num_rows($query_view);
if ($numrows != 0)
{
while ($rows = mysql_fetch_assoc($query_view))
{
$print_day1 = $rows['Day1'];
$print_day2 = $rows['Day2'];
$print_day3 = $rows['Day3'];
$print_day4 = $rows['Day4'];
$print_day5 = $rows['Day5'];
$print_day6 = $rows['Day6'];
$print_day7 = $rows['Day7'];
$print_day8 = $rows['Day8'];
$print_day9 = $rows['Day9'];
$print_day10 = $rows['Day10'];
$print_day11 = $rows['Day11'];
$print_day12 = $rows['Day12'];
$print_day13 = $rows['Day13'];
$print_day14 = $rows['Day14'];
}
}
require('/jpgraph-3.5.0b1/src/jpgraph.php');
require('/jpgraph-3.5.0b1/src/jpgraph_line.php');
// data
$ydata =
array($print_day1,$print_day2,$print_day3,$print_day4,$print_day5,$print_day6,$print_day7,$print_day8,$print_day9);
// Create the graph. These two calls are always required
$graph = new Graph(500,500);
$graph->SetScale('textlin');
$graph->xaxis->title->Set("Number of Days");
$graph->yaxis->title->Set("Glucose Concentration");
// Create the linear plot
$lineplot=new LinePlot($ydata);
$lineplot->SetColor('blue');
// Add the plot to the graph
$graph->Add($lineplot);
// Display the graph
$graph->Stroke();
}
?>

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craftcms 3 plugin which imports channel entries

I try to finish a plugin which imports data into my craftcms project. I already created a console based method/service, which I trigger/run in my shell. Inside my method(s) I receive data (XML or JSON) I parse my data and try to create and fill an entry of a specific channel I already created.
I tried "saveElement()" which doesn't work.
I found some tuts and informations for craftcms v2 (for example: https://docs.craftcms.com/api/v2/craft-entriesservice.html#public-methods)
Now i am stuck and i can not find any informations on how to solve this with craftcms v3.
Here is my last version of code after hours of different trys :(
$section = Craft::$app->sections->getSectionByHandle('testentry');
$entryTypes = $section->getEntryTypes();
$entryType = $entryTypes[0];
// Save Entry
//$entry = new EntryModel();
$entry = new \craft\elements\Entry();
$entry->sectionId = $section->id;
$entry->typeId = $entryType->id;
//$entry->locale = Craft::$app->i18n->getPrimarySiteLocaleId();
//$entry->authorId = 1; // TODO: Set author
$entry->enabled = true;
$entry->postDate = $post['post_date'];
$entry->slug = $post['post_name'];
// $entry->getContent()->title = $post['post_title'];
// $entry->setContentFromPost(array(
// 'body' => $postContent,
// 'categoryCareer' => NULL,
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if (Craft::$app->elements->saveElement($entry)) {
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echo 'Could not save the Job entry.'."\n";
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PHP - getting image from database

I want to save user's image to a database.
p9_member (member_no ,member_name , .... , member_photo)
Now, I am using a BLOB to save the image for "member_photo" column
What I want to do now is that:
1. I have an application that need to connect the database and get the image from the database.
2. I have also a website doing the same things.
I am implementing the application now. As I need to send the image with some information of the member like name or id. So I create an object to save all the things. This is my PHP code:
<?PHP
Class Friends{
var $no;
var $name;
var $photo;
var $status;
var $admin;
public function __construct($no,$name,$photo,$admin,$status){
$this->name = $name;
$this->photo = $photo;
$this->no = $no;
$this->status=$status;
$this->admin=$admin;
}
};
include ("../../server_config.php");
$friends = array();
$SQL = "SELECT p9_member.member_no,p9_member.photo AS PHOTO,p9_member.name,p9_member.photo FROM p9_member, p9_friends WHERE p9_member.member_no = p9_friends.member_no2 AND p9_friends.member_no1 =".$_POST['memberno'];
$result = mysqli_query($con,$SQL);
if (mysqli_num_rows($result)>0) {
while($row = mysqli_fetch_array($result)){
$friends[] = new Friends($row['member_no'],$row['name'],$row['photo'],null,null);
}
}else
$friends[] = 0;
echo json_encode($friends);
?>
The SQL CODE:
SELECT p9_member.member_no,p9_member.photo AS PHOTO
FROM p9_member, p9_friends
WHERE p9_member.member_no = p9_friends.member_no2
AND p9_friends.member_no1 =".$_POST['memberno']
I cannot get the data back from the application after json decode.
What is the best solution for me?
TRY the below code it will work..
$image_qry=mysql_query("SELECT * FROM tabel_name WHERE your condition");
$image=mysql_fetch_assoc($image_qry);
$get_image=$image['image_column']; //BLOB DATA TYPE
header("Content-type: image/jpeg"); // need to add the header content type in your code
echo $get_image;

Create new thread via script vbulletin

I have a script to create new thread with via scritp in vbulletin
// Create a new datamanager for posting
$threaddm =& datamanager_init('Thread_FirstPost', $vbulletin, ERRTYPE_ARRAY, 'threadpost');
// Set some variable and information
$forumid = 87; // The id of the forum we are posting to
$userid = 2;
$_POST["title"] = $vinanghinguyen_title;
$_POST["content"] = $final_content; // The user id of the person posting
$title = $_POST["title"]; // The title of the thread
$pagetext = $_POST["content"]; // The content of the thread
$allowsmilie = '1'; // Are we allowing smilies in our post
$visible = '1'; // If the post visible (ie, moderated or not)
// Parse, retrieve and process the information we need to post
$foruminfo = fetch_foruminfo($forumid);
$threadinfo = array();
$user = htmlspecialchars_uni( fetch_userinfo($userid) );
$threaddm->set_info('forum', $foruminfo);
$threaddm->set_info('thread', $threadinfo);
$threaddm->setr('forumid', $forumid);
$threaddm->setr('userid', $userid);
$threaddm->setr('pagetext', $pagetext);
$threaddm->setr('title', $title);
$threaddm->set('allowsmilie', $allowsmilie);
$threaddm->set('visible', $visible);
// Lets see what happens if we save the page
$threaddm->pre_save();
if(count($threaddm->errors) < 1) {
// Basically if the page will save without errors then let do it for real this time
$threadid = $threaddm->save();
unset($threaddm);
} else {
// There was errors in the practice run, so lets display them
var_dump ($threaddm->errors);
}
/*======================================================================*\
It can create new thread with title, forumid, userid.....but it not insert tag. I want insert with this script. thank for help

get json data in function php

I am new in this json chapter.I have a file named mysql_conn.php .This file have a php function to call data from mysql database.So can anyone help me to create one json file to get data from mysql_conn.php.Below is my code
mysql_conn.php
function getWrkNoTest($wrkno){
$conf = new BBAgentConf();
$log = new KLogger($conf->get_BBLogPath().$conf->get_BBDateLogFormat(), $conf->get_BBLogPriority() );
$connection = MySQLConnection();
$getWrkNoTest ="";
$lArrayIndex = 0;
$query = mysql_query("
SELECT
a.jobinfoid,
a.WRKNo,
a.cate,
a.det,
a.compclosed,
a.feedback,
a.infoID,
b.callerid,
b.customername
FROM bb_jmsjobinfo a
LEFT JOIN bb_customer b
ON a.customerid = b.customerid
WHERE a.WRKNo = '$wrkno';"
);
$result = mysql_query($query);
$log->LogDebug("Query[".$query."]");
while ($row = mysql_fetch_array($result)){
$getWrkNoTest = array("jobinfoid"=>$row['jobinfoid'],
"WRKNo"=>$row['WRKNo'],
"cate"=>$row['cate'],
"det"=>$row['det'],
"compclosed"=>$row['compclosed'],
"feedback"=>$row['feedback'],
"infoID"=>$row['customerid'],
"customerid"=>$row['infoID'],
"callerid"=>$row['callerid'],
"customername"=>$row['customername']);
$iList[$lArrayIndex] = $getWrkNoTest;
$lArrayIndex = $lArrayIndex + 1;
}
$QueryResult = print_r($getWrkNoTest,true);
$log->LogDebug("QueryResult[".$QueryResult."]");
closeDB($connection);
return $iList;
}
json.php
if ($_GET['action']=="getJsonjms"){
$wrkno = $_GET["wrkno"];
if($wrkno != ""){
$jms = getWrkNoTest($wrkno);
if(!empty($jms)){
echo json_encode($jms);
}else{
echo "No data.";
}
}else{
echo "Please insert wrkno";
}
}
I dont know how to solve this.Maybe use foreach or something else.Sorry for my bad english or bad explanation.I'm really new in this json things. Any help will appreciate.Thanks
If I understand your question right, you want to convert the results you receive from your MySQL query into JSON and then store that data into a file?
If this is correct, you can build off of what you currently have in json.php. In this block here, you use json_encode():
if(!empty($jms)){
echo json_encode($jms);
}
We can take this data and pass it to file_put_contents() to put it into a file:
if (!empty($jms)) {
$json = json_encode($jms);
// write the file
file_put_contents('results.json', $json);
}
If this is a script/page that's visited frequently, you'll want to make the filename (above as results.json) into something more dynamic, maybe based on the $wrkno or some other schema.

Viewing Image from database

I have a file in which a user can view its stored files. I want that only the logged in user and it can only be viewed by that member who has stored that file. It works fine when I view other files like .html, .txt etc. But when I view any image, it doesn't work.
This is my script :
require ('config.php');
$id = intval($_GET['id']);
$dn2 = mysql_query('select authorid from files where uploadid="'.$id.'"');
while($dnn2 = mysql_fetch_array($dn2))
{
if(isset($_SESSION['username']) and ($_SESSION['userid']==$dnn2['authorid'] or $_SESSION['username']==$admin)){
$query = "SELECT data, filetype FROM files where uploadid=$id"; //Find the file, pull the filecontents and the filetype
$result = MYSQL_QUERY($query); // run the query
if($row=mysql_fetch_row($result)) // pull the first row of the result into an array(there will only be one)
{
$data = $row[0]; // First bit is the data
$type = $row[1]; // second is the filename
Header( "Content-type: $type"); // Send the header of the approptiate file type, if it's' a image you want it to show as one :)
print $data; // Send the data.
}
else // the id was invalid
{
echo "invalid id";
}
}
}
How can I view the image?

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