PHP
<?php
$deleteImage = null;
if(isset($_POST["deleteImage"])){$deleteImage = $_POST["deleteImage"];}
unlink($deleteImage);
?>
I'm sending $deleteImage to the php page via ajax.and I got following
Jquery response OUTPUT
$deleteImage = '../pard_media/Upload/upload/23.jpg';
There is a image called 23.jpg in my upload directory. but images does to delete .Do i want to add absolute path to delete the image ?
My AJAX
$.ajax({
url: "../pard_media/Upload/delete.php",
type: "POST",
data: "deleteImage=" + data,
success: function (response) {
alert(response);
}
});
It's better, also safety wise to use an absolute path. But you can get this path dynamically.
E.g. using:
getcwd();
Depending on where your PHP script is, your variable could look like this:
$deleteImage = getcwd() . '/upload/23.jpg';
Related
What I did so far:
Okay, so i create a blob file (wav content) and it is stored under blob:http://localhost/cf6fefdc-352e-4cec-aef8-03af6d0d0ef6.
When i put this URL into my browser, it plays the file.
What i need to do
I want to convert the blob into an actual file, that i want to store on my webserver.
I'm handing the blob object over to a php file via AJAX
$.ajax ({
type: "POST",
url:"path/to/my/file.php",
data: {thefile : blob} ,
success: function(text) {
console.log(text);
},
error:function(){
console.log("Error")
}
});
That works. If I print_r the variable it returns [object Blob]
My PHP
<?php
$file = $_POST['thefile'];
print_r($file);
Can anyone tell me how I can convert and save the file to my server from there on?
In your file.php add this code
$filePath = 'uploads/' . $_POST['thefile'];
$tempName = $_FILES['thefile']['tmp_name'];
if (!move_uploaded_file($tempName, $filePath)) {
echo 'Problem saving file: '.$tempName;
die();
}
// success report
echo 'success';
I am trying to make an image upload where the JavaScript posts DataURI of an image via AJAX and the PHP receives it to decode it into an image.
The problem is, everything is working fine except that the end product is not an image file.
Please have a look at the following example code.
JavaScript:
dataString='encodedimg='+e.target.result.match(/,(.*)$/)[1]+'&type='+type;
$.ajax({
url: 'uploadhandler_ajax.php',
type: 'POST',
data: dataString,
success: function(data){
//print success message
});
PHP:
$encodedimg = $_POST['encodedimg'];
file_put_contents('asdf.png', base64_decode($encodedimg));
There is no problem with $_POST['encodedimg'] as it produces the right image using online base64 converter. So I am assuming that there is a misuse with file_put_contents() or base64_decode().
Appreciate the help!
To read image on PHP i used a function like this
function rcd($data) {
$p = strpos($data, ',');
$d = base64_decode(substr($data, $p+1));
$rfn = md5(mt_rand(1,123123123));
file_put_contents($rfn, $d, LOCK_EX);
return $rfn;
}
Usage example:
$img_file_name = rcd($_POST['image_data']);
On JS part it is tricky (different browsers, etc). First of all You need to have the image data. Now You do not precise how this is sourced and the code example does not give a hint. We can assume some options
Simple You get dataString properly populated by whatever means neccesary, then Your example should basically work
imgdata = .... // any means of getting the data
$.ajax({
url: 'uploadhandler_ajax.php',
type: 'POST',
image_data: imgdata,
success: function(data){
//print success message
});
Not so simple You have a Canvas object on the screen which was populated by any means and You want to send that data. Whatever above is true, however the way to get image data would be
var canv = document.getElementById('id_of_canvas');
imgdata = canv. toDataURL('image/jpeg', 0.88); // last arg is quality
However, as some browsers (mobiles) might not be so lucky to support this, you might want to find JPEGEncoder for JS and add it, along with the code below, to Your project.
var tdu = HTMLCanvasElement.prototype.toDataURL;
HTMLCanvasElement.prototype.toDataURL = function(type,param1)
{
var res = tdu.apply(this,arguments);
if(res.substr(0,11) != "data:image/")
{
var encoder = new JPEGEncoder();
return encoder.encode(this.getContext("2d").getImageData(0,0,this.width,this.height), (param1 ? param1*100 : 88));
}
else return res;
}
Hope this helps!
FOr #Marcin Gałczyński:
$.ajax({
url: 'uploadhandler_ajax.php',
type: 'POST',
image_data: imgdata,
success: function(data){
//print success message
}
})
I think jQuery.ajax didnt have image_data jQuery.ajax
So I'm making a website for a client, and the client has tons of photos from tons of different bands they photographed in the 80s and 90s that they would like to try and sell.
Instead of making a page for each band (theres over 100) like the previous site did, I am trying to make one page that uses Javascript/PHP to change the image directory to that band when the text for that band is clicked.
So far, I am able to use a PHP function to find photos in the slideshow folder, but I have been unable to update this function to search through a sub directory in the slideshow folder. (For example, when 'Metallica' is clicked, I empty #imageGal, and then I would like to append all the new metallica images from the metallica folder to the gallery).
My PHP code is:
<?php
$imagesDir = '';
$images = glob($imagesDir . '*.{jpg,jpeg,png,gif}', GLOB_BRACE);
echo json_encode($images);
?>
This PHP code seems to work great.
I get the images using this JQuery code:
$('#imageGal').empty();
$.ajax({
url: "slideshow/getimages.php",
dataType: 'json',
success: function(json){
for(var i=0;i<json.length;i++){
$('#imageGal').append('<img src="slideshow/' + json[i] + '">');
}
}, failure: function(json){
alert('Something went wrong. Please try again.');
}
});
When a user clicks on a band (ie Metallica), this code is executed.
$('.options').mousedown(function() {
var name = $(this).attr('id');
//$('#output').html(name);
$.ajax({
type: "POST",
url: "slideshow/getimages.php",
data: {
imageDir: name
}, success: function(msg){
alert( "Data Saved: " + msg );
}
});
$('#imageGal').empty();
$.ajax({
url: "slideshow/getimages.php",
dataType: 'json',
success: function(json){
for(var i=0;i<json.length;i++){
$('#imageGal').append('<img src="slideshow/' + json[i] + '">');
}
}, failure: function(json){
alert('Something went wrong. Please try again.');
}
});
});
I am unable to get the $imagesDir variable to change, but if I were to manually enter "Metallica" in $imagesDir = "Metallica" variable, it loads those images perfectly.
Can anyone offer any help/advice? I've been at this for a many hours now. Thanks for anything!
Unless you have register_globals on then you need to reference the variable through the global $_POST array. $_POST['imagesDir'] instead of $imagesDir.
However I would state in it's current form it would be a very bad idea to simply replace it as someone could attempt to exploit your code to list any directory on the server.
You should append the parent directory to prevent an exploit. Something like this:
EDIT you have to chdir() to the path before glob will work. I've updated my code below.
<?php
$imagesDir = $_SERVER['DOCUMENT_ROOT']; // this is the root of your web directory
$images = array();
// and this line ensures that the variable is set and no one can backtrack to some other
// directory
if( isset($_POST['imagesDir']) && strpos($_POST['imagesDir'], "..") === false) {
$imagesDir .= "/" . $_POST['imagesDir'];
chdir($imagesDir);
$images = glob('*.{jpg,jpeg,png,gif}', GLOB_BRACE);
}
echo json_encode($images);
?>
I'm not an ajax expert but you seem to be posting imageDir.
So your PHP code should be looking for $_POST['imageDir'].
<?php
$imagesDir = $_POST['imageDir'];
$images = glob($imagesDir . '*.{jpg,jpeg,png,gif}', GLOB_BRACE);
echo json_encode($images);
?>
Does this solve it?
I want to send an array from javascript to php via ajax function.
And I don't want show the result as callback, but immediately open the target php file and show the images.
I mean, I want open the php file directly on the server side.
I think this is quite simple, but I just have no idea.
My javascript looks like:
var stringArray = new Array("/images/1.jpg", "/images/2.jpg", "/images/3.jpg");
$.ajax({
url: 'test.php',
data: {stringArray:stringArray},
success: function() {
window.open('test.php'); // It opens test.php in a window but shows nothing!
},
});
the test.php file:
$stringArray = $_GET['stringArray'];
foreach($stringArray as $value) {
echo "<img src=" . $value . "></img>";
}
Thanks for any help!
Likely the window.open command is being called without the POST data needed.
Now, I don't know what you want to do here but, why send an Ajax request when you don't really want to make use of it?.
Edit: just to make it a bit clearer, you seem to be calling the php file with no data trough POST. There is no clean way of opening a window in JS with POST data, just try GET for no critical information. Let is know how it goes.
As I can see, You are sending data by post method and accessing in test.php
but when u open file via window.location it doesn't get POST data hence no data get populated
You can achieve it via $_SESSION.
in test.php
session_start();
if(!empty($_POST['stringArray'])) {
$_SESSION['stringArray'] = $_POST['stringArray'];
}
$stringArray = (isset($_SESSION['stringArray']) && $_SESSION['stringArray'] != '') ? $_SESSION['stringArray'] : $_POST['stringArray'];
foreach($stringArray as $value) {
echo "<h3>" . $value . "</h3>";
}
Hope this will work for you...
you should write
var stringArray = new Array("apple", "banana", "orange");
$.ajax({
type: 'post',
url: 'test.php',
data: {stringArray:stringArray},
success: function(message) {
window.open(message); // It will open a window with contents.
},
});
I doesn't show the output because when you open the page on the callback you are not sending anything through to the test page through the post variable. Why you would want to do this I don't know. Send the information through the url, get.
I'm looking for a php script server side that load an image in a div on client side.
first shot:
ajax.php
if((isset($_POST['id'])) && ($_POST['id']=="loadphoto") && (ctype_digit($_POST['idp']))) {
$query=mysql_query("SELECT articleid, photoid FROM articles_photos WHERE id='".$_POST['idp']."'", $mydb);
if(mysql_num_rows($query)!=0){
while($row=mysql_fetch_array($query, MYSQL_NUM)) {
$path="./articles/photos/".$row[0]."/".$row[1];
}
echo $path;
}
}
myjs.js
function loadPhoto(mexid) {
$.ajax({
type: 'POST',
cache: false,
url: './auth/ajax.php',
data: 'idp='+escape(mexid)+'&id=loadphoto',
success: function(msg) {
$('.articleviewphoto1').find("img").each(function(){
$(this).removeClass().addClass('photoarts');
});
$('#'+mexid).removeClass().addClass('photoartsb');
$('#visualizator').html('<img src="'+msg+'" class="photoartb" />');
}
});
return false;
}
Are you looking to host a list of possible images in the PHP file and load different images based on the id you send? If so, you can set:
$Img = $_POST['idp'];
And use a switch statement to pass back the image file path:
switch ($Img){
case (some_value):
echo 'images/some_file.jpg';
break;
case (another_value):
echo 'images/another_file.jpg';
break;
}
Then place the file path in an tag and load that into $('#visualizator').html();
Am I on track at least with what you want? If so, I can flesh the answer out a bit.
I'm assuming the image already exists somewhere and isn't being generated on the fly. In the PHP script, you can access the variables with $_POST['idp'] and $_POST['id']. Do whatever processing you need, and then simply echo the URL. The variable msg in the javascript will be everything echo'd by the PHP script. Then, you could just do this:
$('#visualizator').html('<img src="' + msg + '" />');