My English Languege is not good. sorry.
I Want Get My Variable Folder Name In mydiv Content and Embed to my php Code (in $flds)?
I do not want to use form.
please help me. thanks
<div id="mydiv">Variable Folder Name</div>
<?php
$flds = $mydivContent; //?????
$all_files = glob("uploads/$flds/*.*");
for ($i=0; $i<count($all_files); $i++) {
$image_name = $all_files[$i];
$supported_format = array('gif','jpg','jpeg','png');
$ext = strtolower(pathinfo($image_name, PATHINFO_EXTENSION));
if (in_array($ext, $supported_format)) {
echo '<img src="'.$image_name .'" alt="'.$image_name.'" />';
}
else {
continue;
}
}
?>
PHP is server side programming language, PHP executes first when you send request to server there is no way to call specific html element of same page by PHP.
Either you should already put value to variable $mydivContent in PHP
or you could use html <form>(s) to get response from client
Related
I'm using the below code to pull through user uploads into a portfolio. It's working great however one image (out of a list of many) is showing blank - yet displays on the next page, using the same path, but within a conventional <img /> tag.
<?php
$files= glob('./uploads/users/'. $item->user_id .'/'.$item->id.'/public_large/*');
$count = 0;
$user_avatar = '/themes/users/assets/img/noartwork.jpg';
foreach ($files as $file) {
$count++;
if ($count > 0 && is_file($file)) {
$user_avatar = $file;
}
}
?>
<div class="user-img" style="background-image:url('<?php echo site_url($user_avatar); ?>');"></div>
When I inspect element on the image that's not displaying, if I change the single quotes background-image:url('<?php echo site_url($user_avatar); ?>'); to double background-image:url("<?php echo site_url($user_avatar); ?>"); it pulls through, but this is not the case if I make this change in the actual code.
Any help would be great.
Thanks
have you tried removing single quotes? like this :
background-image:url(<?php echo site_url($user_avatar); ?>);
I found the fix for this. It was to do with the users file name containing a special character that I wasn't checking for e.g. my-user's-image.jpg breaks the code as the character needs to be escaped my-user\'s-image.jpg.
I amended the foreach to run a check as below and this is now working.
foreach ($files as $file) {
$count++;
if ($count > 0 && is_file($file)) {
$user_avatar = str_replace("'", "\'", $file);
}
}
Thanks
I'm just a persistent beginner and I've met another obstacle in my way, so I hope that you'll help me one more time... :) I've got that HTML:
<div class='hold'><?php include 'image.php'; ?></div>
<div class='refresh'>Get new image</div>
And that PHP code:
<?php
$dir = 'images/';
$img = array();
if(is_dir($dir)) {
if($od = opendir($dir)) {
while(($file = readdir($od)) !== false) {
if(strtolower(strstr($file, '.'))==='.jpg') {
array_push($img, $file);
}
}
closedir($od);
}
}
$id = uniqid();
$smth = array_rand($img, 1);
echo '<img src=' . $dir.$img[$smth] . '?' . $id . ' width="200px" height="50px" />';
echo '<input type="hidden" value=' . $id . ' />';
?>
So now when I'm looking at my page I see in the <div class='hold'></div> my img from the folder images, and it's allright. BUT when I click on the <div class='refresh'></div> I obviously want to get another img from my folder, but I dunno how to accomplish that trick correctly.
I know that first answer will be USE AJAX, and I know that I can do something like
function loadGraphic() {
$('.hold').load('image.php');
};
loadGraphic();
and then $('.refresh').click(loadGraphic); but when I'm trying to do that in response from server I get TWO THINGS: image.php and, of course, something like car.jpg?573c4e010c7f6... But I very-very wanna get just ONE looking like
car.jpg?573c4e010c7f6
or
image.php?573c4e010c7f6 - I don't care...
So... I hope you've got my concept... maybe I'm asking for miracle - I dunno! Any, absolutely any help will be appreciated :)
Try it the following way:
JS function:
function loadGraphic() {
$.get("image.php", function(data) {
$(".hold").html(data);
});
};
Html:
<div class='refresh' onclick='loadGraphic()'>Get new image</div>
so far I've successfully moved an uploaded image to its designated directory and stored the file path of the moved image into a database I have.
Problem is, however, is that the img src I have echoed fails to read the variable containing the file path of the image. I've been spending time verifying the validity of my variables, the code syntax in echoing the img src, and the successful execution of the move/storing code, but I still get <img src='' when I refer to the view source of the page that is supposed to display the file path contained in the variable.
I believe the file path is stored within the variable because the variable was able to be recognized by the functions that both moved the image to a directory and the query to database.
My coding and troubleshooting experience is still very adolescent, thus pardon me if the nature of my question is bothersomely trivial.
Before asking this question, I've searched for questions within SOF but none of the answers directly addressed my issue (of the questions I've searched at least).
Main PHP Block
//assigning post values to simple variables
$location = $_POST['avatar'];
.
.
.
//re-new session variables to show most recent entries
$_SESSION["avatar"] = $location;
.
.
.
if (is_uploaded_file($_FILES["avatar"]["tmp_name"])) {
//define variables relevant to image uploading
$type = explode('.', $_FILES["avatar"]["name"]);
$type = $type[count($type)-1];
$chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
$rdn = substr(str_shuffle($chars), 0, 15);
//check image size
if($_FILES["avatar"]["size"] > 6500000) {
echo"Image must be below 6.5 MB.";
unlink($_FILES["avatar"]["tmp_name"]);
exit();
}
//if image passes size check continue
else {
$location = "user_data/user_avatars/$rdn/".uniqid(rand()).'.'.$type;
mkdir("user_data/user_avatars/$rdn/");
move_uploaded_file( $_FILES["avatar"]["tmp_name"], $location);
}
}
else {
$location = "img/default_pic.jpg";
}
HTML Block
<div class="profileImage">
<?php
echo "<img src='".$location."' class='profilePic' id='profilePic'/>";
?><br />
<input type="file" name="avatar" id="avatar" accept=".jpg,.png,.jpeg"/>
.
.
.
View Source
<div class="profileImage">
<img src='' class='profilePic' id='profilePic'/><br />
<input type="file" name="avatar" id="avatar" accept=".jpg,.png,.jpeg"/>
.
.
.
Alright, I've finally found the error and was able to successfully solve it!
Simply declare a avatar session variable to the $location variable after updating the table, update the html insert by replacing all $location variables with $_SESSION["avatar_column"] and you are set!
PHP:
$updateCD = "UPDATE users SET languages=?, interests=?, hobbies=?, bio=?, personal_link=?, country=?, avatar=? WHERE email=?";
$updateST = $con->prepare($updateCD);
$updateST->bind_param('ssssssss', $lg, $it, $hb, $bio, $pl, $ct, $location, $_SESSION["email_login"]);
$updateST->execute();
$_SESSION["avatar"] = $location; //Important!
if ($updateST->errno) {
echo "FAILURE!!! " . $updateST->error;
}
HTML:
<div class="profileImage">
<?php
$_SESSION["avatar"] = (empty($_SESSION["avatar"])) ? "img/default_pic.jpg" : $_SESSION["avatar"] ;
echo "<img src='".$_SESSION["avatar"]."' class= 'profilePic' id='profilePic'> ";
?>
.
.
.
Thank you!
try this code
<?php
error_reporting(E_ALL);
ini_set('display_errors','on');
$location = ""; //path
if($_POST && $_FILES)
{
if(is_uploaded_file())
{
// your code
if(<your condition >)
{
}
else
{
$location = "./user_data/user_avatars/".$rdn."/".uniqid(rand()).'.'.$type;
if(!is_dir("./user_data/user_avatars/".$rdn."/"))
{
mkdir("./user_data/user_avatars/".$rdn."/",0777,true);
}
move_uploaded_file( $_FILES["avatar"]["tmp_name"], $location);
}
}
else
{
$location = "img/default_pic.jpg";
}
}
?>
Html Code :-
<div>
<?php
$location = (empty($location)) ? "img/default_pic.jpg" : $location ;
echo "<img src='".location."' alt='".."'> ";
?>
</div>
If it helpful don't forget to marked as answer so another can get correct answer easily.
Good Luck..
I created a php page that print the barcode. Just to view it before i print it on an A4. Still in testing phase. The codes are as below.
<?php
include('include/conn.php');
include('include/Barcode39.php');
$sql="select * from barcode where b_status = 'NOT-PRINTED'";
$result=mysqli_query($conn,$sql);
echo mysqli_num_rows($result);
$i=0;
while($row=mysqli_fetch_assoc($result)){
$acc_no = $row["b_acc_no_code"];
$bc = new Barcode39($row["b_acc_no_code"]);
echo $bc->draw();
$bc->draw($acc_no.$i.".jpg");
echo '<br /><br />';
$i++;
}
?>
Without the while loop, it can be printed, but only one barcode. How to make it generate, for example in the database have 5 values, it will print 5 barcode in the same page. Thanks in advance
Try to use another bar code source. Because It is generate only one bar code per page. Can't able to create multiple bar code per page.
I know this is an older post but comes up in searches so is probably worth replying to.
I have successfully used the Barcode39 to display multiple barcodes. The trick is to get base64 data from the class and then display the barcodes in separate HTML tags.
The quickest way to do this is to add a $base64 parameter to the draw() method:
public function draw($filename = null, $base64 = false) {
Then, near the end of the draw() method, modify to buffer the imagegif() call and return the output in base64:
// check if writing image
if ($filename) {
imagegif($img, $filename);
}
// NEW: Return base 64 for the barcode image
else if ($base64) {
ob_start();
imagegif($img);
$image_data = ob_get_clean();
imagedestroy($img);
return base64_encode($image_data);
}
// display image
else {
header("Content-type: image/gif");
imagegif($img);
}
Finally, to display multiples from the calling procedure, construct the image HTML in the loop and display:
// assuming everything else has been set up, end with this...
$base64 = $barcode->draw('', true); // Note the second param is set for base64
$html = '';
for ($i = 0; $i < $numBarcodes; $i++) {
$html .= '<img src="data:image/gif;base64,'.$base64.'">';
}
die('<html><body>' . $html . '</body></html>');
I hope this helps anyone else facing this challenge.
Hello there i have a php file with the included:
The image shows properly when i access the PHP file, however when I try to show it in the HTML template, it shows as the little img with a crack in it, so basically saying "image not found"
<img src="http://konvictgaming.com/status.php?channel=blindsniper47">
is what i'm using to display it in the HTML template, however it just doesn't seem to want to show, I've tried searching with next to no results for my specific issue, although I'm certain I've probably searched the wrong title
adding code from the OP below
$clientId = ''; // Register your application and get a client ID at http://www.twitch.tv/settings?section=applications
$online = 'online.png'; // Set online image here
$offline = 'offline.png'; // Set offline image here
$json_array = json_decode(file_get_contents('https://api.twitch.tv/kraken/streams/'.strtolower($channelName).'?client_id='.$clientId), true);
if ($json_array['stream'] != NULL) {
$channelTitle = $json_array['stream']['channel']['display_name'];
$streamTitle = $json_array['stream']['channel']['status'];
$currentGame = $json_array['stream']['channel']['game'];
echo "<img src='$online' />";
} else {
echo "<img src='$offline' />";
}
The url is not an image, it is a webpage with the following content
<img src='offline.png' alt='Offline' />
Webpages cannot be displayed as images. You will need to edit the page to only transmit the actual image, with the correct http-headers.
You can probably find some help on this by googling for "php dynamic image".
Specify in the HTTP header that it's a PNG (or whatever) image!
(By default they are interpreted as text/html)
in your status.php file, where you output the markup of <img src=... change it to read as follows
$image = file_get_contents("offline.png");
header("Content-Type: image/png");
echo $image;
Which will send an actual image for the request instead of sending markup. markup is not valid src for an img tag.
UPDATE your code modified below.
$clientId = ''; // Register your application and get a client ID at http://www.twitch.tv/settings?section=applications
$online = 'online.png'; // Set online image here
$offline = 'offline.png'; // Set offline image here
$json_array = json_decode(file_get_contents('https://api.twitch.tv/kraken/streams/'.strtolower($channelName).'?client_id='.$clientId), true);
header("Content-Type: image/png");
$image = null;
if ($json_array['stream'] != NULL) {
$channelTitle = $json_array['stream']['channel']['display_name'];
$streamTitle = $json_array['stream']['channel']['status'];
$currentGame = $json_array['stream']['channel']['game'];
$image = file_get_contents($online);
} else {
$image = file_get_contents($offline);
}
echo $image;
I suppose you change the picture dynmaclly on this page.
Easiest way with least changes will just be using an iframe:
<iframe src="http://konvictgaming.com/status.php?channel=blindsniper47"> </iframe>