Now my PHP can display Blob from MySQL on web browser with http://localhost/download.php?id=1
How to display this with webview on android?
<?php
if($_SERVER['REQUEST_METHOD']=='GET'){
$id = $_GET['id'];
$sql = "select * from loancontract where id = '$id'";
require_once('db_connect.php');
$r = mysqli_query($con,$sql);
$result = mysqli_fetch_array($r);
header('content-type: application/pdf');
echo base64_decode($result['data']);
mysqli_close($con);
} else {
echo "Error";
}
?>
Related
I have created API from my DB and the result appears in a PHP file, so the question is how I can read this result in a .json file?
<?php
$con = mysqli_connect("localhost","root","Password##","Mydb");
$response = array();
if($con){
$sql = "select * from limit_order";
$result = mysqli_query($con,$sql);
if($result){
header("Content-Type: JSON");
$i=0;
while($row = mysqli_fetch_assoc($result)){
$response[$i]['order_type'] = $row ['order_type'];
$response[$i]['from'] = $row ['from'];
$response[$i]['to'] = $row ['to'];
$response[$i]['quantity'] = $row ['quantity'];
$response[$i]['limit_price'] = $row ['limit_price'];
$response[$i]['created_at'] = $row ['created_at'];
$response[$i]['updated_at'] = $row ['updated_at'];
$response[$i]['status'] = $row ['status'];
$i++;
}
echo json_encode($response,JSON_PRETTY_PRINT);
}
}
else{
echo "Database connection failed";
}
?>
<?php
$con = mysqli_connect("localhost", "alsmwsk3", "skrktkzl12", "alsmwsk3");
$result = mysqli_query($con, "select * from NOTICE ORDER BY noticeDate DESC");
$response = array();
$num = 0;
while($row = mysqli_fetch_array($result))
{
array_push($response, array("noticeContent"=>$row[0],"noticeName"=>$row[1],"noticeDate"=>$row[2]));
echo $num;
}
//php를 읽었을떄 json타입이 아닐 readed php not but json?
//print (json_encode($response));
// echo json_encode(array("response"));
echo json_encode(array("response"=>$response));
mysqli_close($con);
?>
when I open this phpsource on browser there are no data on browser I think am i wrong some kind of miss?
and it is mysql setting
in mysql Datbase there is images stored using php Script (image got from a form.html/POST method) let's cal them (phpImages). and there is others stored using android application ( by converting Bitmap to String and using StringBuilder ). let's call them (androidImages).
with this php script i can load and display phpImages, but i cannot display androidImages.
<?php
$con = mysqli_connect("localhost","root","","othmane") or die(mysqli_error($con));
if($_SERVER['REQUEST_METHOD']=='GET'){
$id = $_GET['id'];
$sql = "SELECT image FROM images WHERE id = '$id'";
$r = mysqli_query($con,$sql) or die(mysqli_error($con));;
$result=mysqli_fetch_array($r);
header('Content-Type:image/jpeg');
echo ( $result['image']);
mysqli_close($con);
}
?>
with this php script i can load androidImages, but i cannot load phpImages :
<?php
$con = mysqli_connect("localhost","root","","othmane") or die(mysqli_error($con));
if($_SERVER['REQUEST_METHOD']=='GET'){
$id = $_GET['id'];
$sql = "SELECT image FROM images WHERE id = '$id'";
$r = mysqli_query($con,$sql) or die(mysqli_error($con));;
$result=mysqli_fetch_array($r);
header('Content-Type:image/jpeg');
echo base64_decode( $result['image'] );
mysqli_close($con);
}
?>
i wan't a php script that could display the both. because i want to load all images in a ListView of an android Apps.
**This is php script relied to android Application : **
<?php
if($_SERVER['REQUEST_METHOD']=='POST'){
$image = $_POST['image'];
$con=mysqli_connect("localhost","root","","othmane")or die(mysqli_error($con));
$sql = "INSERT INTO images (image,image_type) VALUES (?,'android')";
$stmt = mysqli_prepare($con,$sql);
mysqli_stmt_bind_param($stmt,"s",$image);
mysqli_stmt_execute($stmt);
$check = mysqli_stmt_affected_rows($stmt);
if($check == 1){
echo "Image Uploaded Successfully";
}else{
echo "Error Uploading Image";
}
mysqli_close($con);
}else{
echo "Error";
}
?>
this is php Script relied with Form.html post method :
<?php
echo ini_get( 'file_uploads' );
if(!isset($_POST['submit'])){
echo '<p>Please Select Image to Upload</p>';
}
else
{
try {
upload();
}
catch(Exception $e)
{
echo '<h4>'.$e->getMessage().'</h4>';
}
}
function upload(){
$imgfp = fopen($_FILES['photo']['tmp_name'], 'rb');
print_r($_FILES);
$dbh = new PDO("mysql:host=localhost;dbname=othmane", 'root', '');
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$stmt = $dbh->prepare("INSERT INTO images (image,image_type) VALUES (?,'php')");
$stmt->bindParam(1, $imgfp, PDO::PARAM_LOB);
$stmt->execute();
}
?>
Add a column called image_type in your table and pass one of the following values to determine what the source of the image is upon uploading: phpImage or androidImage
So you can do:
<?php
$con = mysqli_connect("localhost","root","","othmane") or die(mysqli_error($con));
if ($_SERVER['REQUEST_METHOD'] == 'GET'){
$id = $_GET['id'];
$sql = "SELECT image, image_type FROM images WHERE id = '$id'";
$r = mysqli_query($con,$sql) or die(mysqli_error($con));
$result = mysqli_fetch_array($r);
header('Content-Type: image/jpeg');
if ($result['image_type'] == 'phpImage') {
echo ( $result['image']);
} else if ($result['image_type'] == 'androidImage') {
echo base64_decode( $result['image'] );
}
mysqli_close($con);
}
?>
I know how to fetch user data from database with the code below. Now I want to navigate to another page (using onclick) and to display this user data by id. This would be like StackOverflow or Facebook, when you click on a photo or ID, and the site takes you to the user's profile page.
Here is my code so far:
<?php
$connect = mysql_connect("localhost","root","") or die(mysql_error());
$select = mysql_select_db("profile") or die(mysql_error());
$result = mysql_query("SELECT * FROM users order by id DESC");
$id = $_SESSION['id'];
while($row = mysql_fetch_array($result)){
if($row['id'] !== $id){
echo "<table id='suggest'><tr><td id='frienddata'><a href='http://localhost/profile/userprofile.php'>".$row['first'].' '. $row['last']."<a/></td><br></tr></table>";
}
}
?>
$id = $_GET['id'];
if(!isset($id))
{
$connect = mysql_connect("localhost","root","") or die(mysql_error());
$select = mysql_select_db("profile") or die(mysql_error());
$result = mysql_query("SELECT * FROM users WHERE id = '"$id"'");
if(!$result)
{
die('user_not_found');
}
mysqli_fetch_row( $result );
echo "<table id='sugest'><tr><td id='frienddata'><a href='http://localhost/profile/userprofile.php'>".$row['first'].' '. $row['last']."<a/></td><br></tr></table>";
suppose you are in a page before clicking on a user profile,the link should be some thing like this 'site.com/userprofile.php?id=5'.
now in userprofile.php:
$id = $_GET['id'];
if(!isset($id))
die('user not found');
$connect = mysql_connect("localhost","root","") or die(mysql_error());
$select = mysql_select_db("profile") or die(mysql_error());
$result = mysql_query("SELECT * FROM users where id='".$id."'");
if (!$result) {
die('user not found');
}
$row = mysql_fetch_row($result);
echo "<table id='sugest'><tr><td id='frienddata'><a href='http://localhost/profile/userprofile.php'>".$row['first'].' '. $row['last']."<a/></td><br></tr></table>";
I'm trying to fetch couple of single data in my server database but this is throwing some errors. The incoming data is correct. The search function just don't get completed.
Here's the code:
<?php
if($_SERVER['REQUEST_METHOD']=='POST'){
define('HOST','xxxxxxxxxxx');
define('USER','xxxxxxxxxxxx');
define('PASS','xxxxxxxxx');
define('DB','xxxxxxxxxx');
$con = mysqli_connect(HOST,USER,PASS,DB);
$post_id = $_POST['id'];
$buyer_mobile = $_POST['mobile'];
$buyer_name = $_POST['name'];
$sql = "select mobile from flatowner where id='$post_id'";
$res = mysqli_query($con,$sql);
$owner_mobile = $row['mobile'];
$sql = "select name from user where mobile='$owner_mobile'";
$r = mysqli_query($con,$sql);
$owner_name = $row['name'];
$sql = "INSERT INTO flat_booking (post_id,owner_mobile,owner_name,buyer_mobile,buyer_name) VALUES ('$post_id','$owner_mobile','$owner_name','$buyer_mobile','$buyer_name')";
if(mysqli_query($con,$sql)){
echo "Success";
}
else{
echo "error";
}
mysqli_close($con);
}else{
echo 'error1';
}
What am I doing wrong here? Maybe this:
$owner_mobile = $row['mobile'];
Thanks in advance!
create table flatower and add mobile column
$post_id = 1;
$sql = "select mobile from flatowner where id='$post_id'";
$res = mysql_query($con,$sql);
$row = mysql_fetch_array($res);
$owner_mobile = $row[0]['mobile'];
Your problem is this line:
$owner_mobile = $row['mobile'];
You have not created the $row variable. For this you would need to do something such as:
Do this first:
<?php
$row = array();
while ($result = mysqli_fetch_assoc($res))
{
$row[] = $result;
}
?>
This allows you to do this:
<?php
foreach ($row as $r)
{
var_dump($r); print "<br />"; // One row from the DB per var dump
}
?>