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PHP - Show Part Of Image Without Distorting Image
Hi,
I have some images that I want to resize to a square dimention (40px 40px). However, not all image are square, so when they are resized, they lose their ration and look squashed. Is there a way I can display just a part of the original image, say 40px x 40px on the top middle of each image? I hope that makes sense Similar TutorialsI'll explain with an example: I have this image: http://www.fitnessvital.com/images/galeria/caballo-pastando-0.jpg I create a thumbnail of it: http://www.fitnessvital.com/images/galeria/thumbs/caballo-pastando-0.jpg You can see that it's a big image where the important thing is the horse. The grass is not important. So, the best thing to do it's to crop the extra grass and resizing the rest. So the question is: There's a way of recognizing the objects of an image and crop the rest ? Or identify the non important of an imagen and cropt it ? (the same rewrited) Thank you HI I'm thinking to do something like this , I've seen this in a website ,I'm a php designer but i don't know who something like this can be done . it's a website that sells sofas . we've the ability to change the color and fabric and see the result right away . for example ,this is the default sofa : then I can select a part of it and I can choose the fabric and color , then it make the result : thats it , How can I do that ? What should I do ? I'm really looking to hear from you King Regards Hi guys, I have page where it echos out the image url from mysql in a MAMP Server, however when i echo the image url out it seems to be fine but as soon as i put in a img scr it wont show the image but it shows the container. I have the code here Code: [Select] <div id="maincontentholderbottom"> <div id="maincontentholderbottom-index-left"> <div id="news-container"><ul> <?php $select=mysql_query("SELECT * FROM news"); while($get_news=mysql_fetch_array($select)){ $newsid=$get_news['id']; $title=$get_news['title']; $text=$get_news['text']; $newsdate=$get_news['date']; $newstime=$get_news['time']; $imagelink=$get_news['newsimagelink']; $newstext=substr($text, 0, 420); echo"<li><div id='title'>$title</div><div id='newsblock'>$newstext... <a href='http://localhost/mycomputer/create/news/index.php?id=$newsid'>Read More</a></div> <div id='newsimage'><img src='$imagelink' width='150' height='70'/></div> </li>"; } ?> </ul> </div> </div> </div> Do u know why is it like this? I appreciate your help in advance. Thanks! Hey guys i have a query echo (empty($row['deal'])? "empty": "not empty"); //result not empty but i need it to show the "not empty" if its "deal". by default it is a "1" when i select the row to be a deal it updates the table and inserts "Deal". Hey everyone, I could use some help with this one.. I need to create some php code to switch images when a certain page loads. For example: These are the pages, test1, test2,test3, so when you click on the test1 link in the navigation a certain image would show up : something like this: If test1 exists then show image1.jpg If test2 exists then show image2.jpg If test3 exists then show image2.jpg. I am doing this in wordpress and I can't figure out the if then statement.. Does anyone have any ideas?? thanks Hi again probably a very simple code but not working. I am trying to show a default image "fabric.jpg" if the recordset is empty. If not empty it shows the recordset image at a size of 100X100. This is the code I am using and have probably left something out, any ideas? <img src="<?php if ($row_Recordset3['fabricpicture']==null); echo "<img src='graphics/fabric.jpg'>"; ?>" alt="" name="fabric" width="100" height="100" border="0" align="bottom" id="fabric" title="Selected Bottom Up Blind Fabric" /> I have tried switching the img scr both "" and ' ' but still no joy? I need to be able to create an image from user input. I have found a couple of samples that purport to do that but when I copy the code I can't get it to work. I would appreciate advice as to what I am doing wrong please. One example I tried is from a tutorial on this site: http://www.phpfreaks.com/tutorial/php-add-text-to-image Now perhaps there is something I am doing wrong when I call on the image that is supposed to display but given that I have tried the same code with other images and it works I can't see what I am doing wrong, unless there is something in the nature of scripts that requires some extra steps. Here is my code to see the image (note that 'myscript.php' in my code is in the same folder as the code I am calling: <html> <head> <title>Using Images Created by Scripts</title> </head> <body> <h1>Generated Image Below ...</h1> <img src="myscript.php"/> </body> </html> Can anyone help? Thank you I wish to get only the last part of each array 'g' And then show non duplicated 'g' but in the same order that it was in the original array. Code: [Select] <?php $formarray = array( 'name' => array( 'i1'=> array('a'=>'a1', 'b'=>'b1', 'c'=>'c1', 'd'=>'d1', 'e'=>'e1', 'f'=>'f1', 'g'=>'g1'), 'i2'=> array('a'=>'a2', 'b'=>'b2', 'c'=>'c2', 'd'=>'d2', 'e'=>'e2', 'f'=>'f2', 'g'=>'g2'), 'i3'=> array('a'=>'a3', 'b'=>'b3', 'c'=>'c3', 'd'=>'d3', 'e'=>'e3', 'f'=>'f3', 'g'=>'g3'), 'i4'=> array('a'=>'a4', 'b'=>'b4', 'c'=>'c4', 'd'=>'d4', 'e'=>'e4', 'f'=>'f4', 'g'=>'g4'), ) ); foreach ($formarray as $newarray => $a) { ?><strong><?=$newarray;?></strong><br><? foreach ($a as $key => $k) { ?>"<?=$key;?>", <? foreach ($k as $b) { //if ($k['DBfield'] != "") { ?>"<?=$b;?>", <? //} } ?><br><? } //end of second foreach ?><br><br><br><? } //end of first foreach ?> The script for creating a new file name for the image:
$validextensions = array("jpeg", "jpg", "png"); //Extensions which are allowed
$ext = explode('.', basename($_FILES['file']['name'][$i]));//explode file name from dot(.)
$file_extension = end($ext); //store extensions in the variable
$new_image_name = md5(uniqid()) . "." . $ext[count($ext) - 1];
$target_path = $target_path . $new_image_name;//set the target path with a new name of image
The script creates a new file like:
f6c9b8d9db05366c3504210cded9ddb2.jpgand moves the file to the "uploads" folder. And then the script also creates a thumbnail with the same file name and moves the file to the "thumbs" folder. The issue I am having is that the same ID code could happen again for a different image in the database, thus I would be calling a different original sized image than the thumbnail image. My question is: How to avoid this issue of the same ID code has happened again for a different file. What is the proper way to reference the anchor tag of the thumbnail image to its actual original sized image? With the script I have the thumbnail image would be coming from the "thumbs" folder and the anchor tag would get referenced to the "uploads" folder to get the original sized image. Edited by glassfish, 12 October 2014 - 05:51 AM. Hi, I've read a lot of places that it's not recommended to store binary files in my db. So instead I'm supposed to upload the image to a directory, and store the link to that directory in database. First, how would I make a form that uploads the picture to the directory (And what kinda directories are we talking?). Secondly, how would I retrieve that link? And I guess I should rename the picture.. I'd appreciate any help, or a good tutorial (Haven't found any myself). After image is drop into container , I want to move copy of the original image to be move when original image dragend within container. I tried but it display copy image each time when original image dragend. can anyone help me?
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>Prototype</title>
<script type="text/javascript" src="http://code.jquery.com/jquery.min.js"></script>
<script src="http://d3lp1msu2r81bx.cloudfront.net/kjs/js/lib/kinetic-v4.7.2.min.js"></script>
<script src="http://code.jquery.com/ui/1.9.2/jquery-ui.min.js"></script>
<style>
body{padding:20px;}
#container{
border:solid 1px #ccc;
margin-top: 10px;
width:350px;
height:350px;
}
#toolbar{
width:350px;
height:35px;
border:solid 1px blue;
}
</style>
<script>
$(function(){
var $house=$("#house");
$house.hide();
var $stageContainer=$("#container");
var stageOffset=$stageContainer.offset();
var offsetX=stageOffset.left;
var offsetY=stageOffset.top;
var stage = new Kinetic.Stage({
container: 'container',
width: 350,
height: 350
});
var layer = new Kinetic.Layer();
stage.add(layer);
var image1=new Image();
image1.onload=function(){
$house.show();
}
image1.src="http://vignette1.wikia.nocookie.net/angrybirds/images/b/b6/Small.png/revision/latest?cb=20120501022157";
$house.draggable({
helper:'clone',
});
$house.data("url","house.png"); // key-value pair
$house.data("width","32"); // key-value pair
$house.data("height","33"); // key-value pair
$house.data("image",image1); // key-value pair
$stageContainer.droppable({
drop:dragDrop,
});
function dragDrop(e,ui){
var x=parseInt(ui.offset.left-offsetX);
var y=parseInt(ui.offset.top-offsetY);
var element=ui.draggable;
var data=element.data("url");
var theImage=element.data("image");
var image = new Kinetic.Image({
name:data,
x:x,
y:y,
image:theImage,
draggable: true,
dragBoundFunc: function(pos) {
return {
x: pos.x,
y: this.getAbsolutePosition().y
}
}
});
image.on("dragend", function(e) {
var points = image.getPosition();
var image1 = new Kinetic.Image({
name: data,
id: "imageantry",
x: points.x+65,
y: points.y,
image: theImage,
draggable: false
});
layer.add(image1);
layer.draw();
});
image.on('dblclick', function() {
image.remove();
layer.draw();
});
layer.add(image);
layer.draw();
}
}); // end $(function(){});
</script>
</head>
<body>
<div id="toolbar">
<img id="house" width=32 height=32 src="http://vignette1.wikia.nocookie.net/angrybirds/images/b/b6/Small.png/revision/latest?cb=20120501022157"><br>
</div>
<div id="container"></div>
</body>
</html>
Edited by Biruntha, 08 January 2015 - 10:14 AM. How can i edit just one image at on time with a multiple image upload form? I have the images being stored in a folder and the path being stored in MySQL. I also have the files being uploaded with a unique id. My issue is that I want to be able to pass the values of what is already in $name2 $name3 $name4 if I only want to edit $name1. I don't want to have to manually update the 4 images. Here is the PHP: Code: [Select] <?php require_once('storescripts/connect.php'); mysql_select_db($database_phpimage,$phpimage); $uploadDir = 'upload/'; if(isset($_POST['upload'])) { foreach ($_FILES as $file) { $fileName = $file['name']; $tmpName = $file['tmp_name']; $fileSize = $file['size']; $fileType = $file['type']; if ($fileName != ""){ $filePath = $uploadDir; $fileName = str_replace(" ", "_", $fileName); //Split the name into the base name and extension $pathInfo = pathinfo($fileName); $fileName_base = $pathInfo['fileName']; $fileName_ext = $pathInfo['extension']; //now we re-assemble the file name, sticking the output of uniqid into it //and keep doing this in a loop until we generate a name that //does not already exist (most likely we will get that first try) do { $fileName = $fileName_base . uniqid() . '.' . $fileName_ext; } while (file_exists($filePath.$fileName)); $file_names [] = $fileName; $result = move_uploaded_file($tmpName, $filePath.$fileName); } if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); $filePath = addslashes($filePath); } $fileinsert[] = $filePath; } } $mid = mysql_real_escape_string(trim($_POST['mid'])); $cat = mysql_real_escape_string(trim($_POST['cat'])); $item = mysql_real_escape_string(trim($_POST['item'])); $price = mysql_real_escape_string(trim($_POST['price'])); $about = mysql_real_escape_string(trim($_POST['about'])); $fields = array(); $values = array(); $updateVals = array(); for($i = 0; $i < 4; $i++) { $values[$i] = isset($file_names[$i]) ? mysql_real_escape_string($file_names[$i]) : ''; if($values[$i] != '') { $updateVals[] = 'name' . ($i + 1) . " = '{$values[$i]}'"; } } $updateNames = ''; if(count($updateVals)) { $updateNames = ", " . implode(', ', $updateVals); } $update = "INSERT INTO image (mid, cid, item, price, about, name1, name2, name3, name4) VALUES ('$mid', '$cat', '$item', '$price', '$about', '$values[0]', '$values[1]', '$values[2]', '$values[3]') ON DUPLICATE KEY UPDATE cid = '$cat', item = '$item', price = '$price', about = '$about' $updateNames"; $result = mysql_query($update) or die (mysql_error()); Hello I am having problems uploading an image through a HTML form. I want the image to be uploaded to the server and the image name to be written to the mysql database. Below is the code I am using: Code: [Select] <?php if (isset($_POST['add'])){ echo "<br /> add value is true"; $name = $_POST['name']; $description = $_POST['description']; $price = $_POST['price']; $category_id = $_POST['category_name']; $image = $_FILES['image']['name']; //file path of the image upload $filepath = "../images/"; //mew name for the image upload $newimagename = $name; //new width for the image $newwidth = 100; //new height for the image $newheight = 100; include('../includes/image-upload.php'); mysql_query("INSERT INTO item (item_name, item_description, item_price, item_image) VALUES ('$name','$description','$price','$image')"); ?> Here is the image-upload.php file code: Code: [Select] <?php //assigns the file to the image $image =$_FILES["image"]["name"]; $uploadedfile =$_FILES["image"]["tmp_name"]; if ($image) { //retrieves the extension type from image upload $extension = getextension($image); //converts extension to lowercase $extension = strtolower($extension); //create image from uploaded file type if($extension=="jpg" || $extension=="jpeg") { $uploadedfile = $_FILES['image']['tmp_name']; $src = imagecreatefromjpeg($uploadedfile); }else if($extension=="png") { $uploadedfile = $_FILES['image']['tmp_name']; $src = imagecreatefrompng($uploadedfile); }else{ $src = imagecreatefromgif($uploadedfile); } //creates a list of the width and height of the image list($width,$height)=getimagesize($uploadedfile); //adds color to the image $tmp = imagecreatetruecolor($newwidth,$newheight); //create image imagecopyresampled($tmp,$src,0,0,0,0,$newwidth,$newheight,$width,$height); //set file name $filename = $filepath.$newimagename.".".$extension; $imagename = $newimagename.".".$extension; //uploads new file with name to the chosen directory imagejpeg($tmp,$filename,100); //empty variables imagedestroy($src); imagedestroy($tmp); } ?> Any help would be appreciated, fairly new to all this! Thanks!!! This topic has been moved to mod_rewrite. http://www.phpfreaks.com/forums/index.php?topic=351137.0 Can I get some help or a point in the right direction.
I am trying to create a form that allows me to add, edit and delete records from a database.
I can add, edit and delete if I dont include the image upload code.
If I include the upload code I cant edit records without having to upload the the same image to make the record save to the database.
So I can tell I have got the code processing in the wrong way, thing is I cant seem to see or grasp the flow of this, to make the corrections I need it work.
Any help would be great!
Here is the form add.php code
<?php require_once ("dbconnection.php");
$id="";
$venue_name="";
$address="";
$city="";
$post_code="";
$country_code="";
$url="";
$email="";
$description="";
$img_url="";
$tags="";
if(isset($_GET['id'])){
$id = $_GET['id'];
$sqlLoader="Select from venue where id=?";
$resLoader=$db->prepare($sqlLoader);
$resLoader->execute(array($id));
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>Add Venue Page</title>
<link href='http://fonts.googleapis.com/css?family=Droid+Sans' rel='stylesheet' type='text/css'>
<link rel="stylesheet" href="//maxcdn.bootstrapcdn.com/bootstrap/3.2.0/css/bootstrap.min.css">
<script src="//maxcdn.bootstrapcdn.com/bootstrap/3.2.0/js/bootstrap.min.js"></script>
</head>
<body>
<div class="container">
<?php
$sqladd="Select * from venue where id=?";
$resadd=$db->prepare($sqladd);
$resadd->execute(array($id));
while($rowadd = $resadd->fetch(PDO::FETCH_ASSOC)){
$v_id=$rowadd['id'];
$venue_name=$rowadd['venue_name'];
$address=$rowadd['address'];
$city=$rowadd['city'];
$post_code=$rowadd['post_code'];
$country_code=$rowadd['country_code'];
$url=$rowadd['url'];
$email=$rowadd['email'];
$description=$rowadd['description'];
$img_url=$rowadd['img_url'];
$tags=$rowadd['tags'];
}
?>
<h1 class="edit-venue-title">Add Venue:</h1>
<form role="form" enctype="multipart/form-data" method="post" name="formVenue" action="save.php">
<input type="hidden" name="id" value="<?php echo $id; ?>"/>
<div class="form-group">
<input class="form-control" type="hidden" name="id" value="<?php echo $id; ?>"/>
<p><strong>ID:</strong> <?php echo $id; ?></p>
<strong>Venue Name: *</strong>
<input class="form-control" type="text" name="venue_name" value="<?php echo $venue_name; ?>"/><br/>
<br/>
<strong>Address: *</strong>
<input class="form-control" type="text" name="address" value="<?php echo $address; ?>"/><br/>
<br/>
<strong>City: *</strong>
<input class="form-control" type="text" name="city" value="<?php echo $city; ?>"/><br/>
<br/>
<strong>Post Code: *</strong>
<input class="form-control" type="text" name="post_code" value="<?php echo $post_code; ?>"/><br/>
<br/>
<strong>Country Code: *</strong>
<input class="form-control" type="text" name="country_code" value="<?php echo $country_code; ?>"/><br/>
<br/>
<strong>URL: *</strong>
<input class="form-control" type="text" name="url" value="<?php echo $url; ?>"/><br/>
<br/>
<strong>Email: *</strong>
<input class="form-control" type="email" name="email" value="<?php echo $email; ?>"/><br/>
<br/>
<strong>Description: *</strong>
<textarea class="form-control" type="text" name="description" rows ="7" value=""><?php echo $description; ?></textarea><br/>
<br/>
<strong>Image Upload: *</strong>
<input class="form-control" type="file" name="image" value="<?php echo $img_url; ?>"/>
<small>File sizes 300kb's and below 500px height and width.<br/><strong>Image is required or data will not save.</strong></small> <br/><br/>
<strong>Tags: *</strong>
<input class="form-control" type="text" name="tags" value="<?php echo $tags; ?>"/><small>comma seperated vales only, e.g. soul,hip-hop,reggae</small><br/>
<br/>
<p>* Required</p>
<br/>
<input class="btn btn-primary" type="submit" name="submit" value="Save">
</div>
</form>
</div>
</body>
</html>
Here is the save.php code
<?php
error_reporting(E_ALL);
ini_set("display_errors", 1);
include ("dbconnection.php");
$venue_name=$_POST['venue_name'];
$address=$_POST['address'];
$city=$_POST['city'];
$post_code=$_POST['post_code'];
$country_code=$_POST['country_code'];
$url=$_POST['url'];
$email=$_POST['email'];
$description=$_POST['description'];
$tags=$_POST['tags'];
$id=$_POST['id'];
if(is_uploaded_file($_FILES['image']['tmp_name'])){
$folder = "images/hs-venues/";
$file = basename( $_FILES['image']['name']);
$full_path = $folder.$file;
if(move_uploaded_file($_FILES['image']['tmp_name'], $full_path)) {
//echo "succesful upload, we have an image!";
var_dump($_POST);
if($id==null){
$sql="INSERT INTO venue(venue_name,address,city,post_code,country_code,url,email,description,img_url,tags)values(:venue_name,:address,:city,:post_code,:country_code,:url,:email,:description,:img_url,:tags)";
$qry=$db->prepare($sql);
$qry->execute(array(':venue_name'=>$venue_name,':address'=>$address,':city'=>$city,':post_code'=>$post_code,':country_code'=>$country_code,':url'=>$url,':email'=>$email,':description'=>$description,':img_url'=>$full_path,':tags'=>$tags));
}else{
$sql="UPDATE venue SET venue_name=?, address=?, city=?, post_code=?, country_code=?, url=?, email=?, description=?, img_url=?, tags=? where id=?";
$qry=$db->prepare($sql);
$qry->execute(array($venue_name, $address, $city, $post_code, $country_code, $url, $email, $description, $full_path, $tags, $id));
}
if($success){
var_dump($_POST);
echo "<script language='javascript' type='text/javascript'>alert('Successfully Saved!')</script>";
echo "<script language='javascript' type='text/javascript'>window.open('index.php','_self')</script>";
}
else{
var_dump($_POST);
echo "<script language='javascript' type='text/javascript'>alert('Successfully Saved!')</script>";
echo "<script language='javascript' type='text/javascript'>window.open('index.php','_self')</script>";
}
} //if uploaded
else{
var_dump($_POST);
echo "<script language='javascript' type='text/javascript'>alert('Upload Recieved but Processed Failed!')</script>";
echo "<script language='javascript' type='text/javascript'>window.open('index.php','_self')</script>";
}
} //move uploaded
else{
var_dump($_POST);
echo "<script language='javascript' type='text/javascript'>alert('Successfully Updated.')</script>";
echo "<script language='javascript' type='text/javascript'>window.open('index.php','_self')</script>";
}
?>
Thanks in advance!
Edited by hankmoody, 12 August 2014 - 05:15 PM. Hello, I've had success creating a cool vanity QR code with rounded edges for my organization by generating a standard QR code, then applying a noise -> median effect in Photoshop to get the edges rounded among some other effects. Photoshop is great, but I want to automate this. I found a great PHP library to generate QR codes. The part I don't get is the rounding of the hard edges. I've seen other sites do it like this one. So far google searches are yielding rounded corner tutorials. Any thoughts on how to do this with GD or ImageMagic? Best regards, Chris Would like to be able to click on a radio button that represents an image. Once selected and submitted, have that image display on another page. I have an idea, but need some guidance. BTW, is using php only doable? Is there a simpler or more elegant way to do this? Thanks all! Hello, Five images will be displayed inside a division. There will be a previous and next button/link. If someone click the next button the next image will be added in that div and the first image will be gone from that div. The previous button/link will do the same thing. Is it possible with php? I am confused if it's a javascript or ajax question. Thanks. Code: [Select] <? $out = preg_replace('/^(.{701}[^.]*).*/i','$1.',$detrsltnewsrow[news_desc]); echo $out; ?> </td></tr><tr><td colspan="2" class="para" style="padding-left:10px;"> <?= substr(stripslashes(trim($detrsltnewsrow[news_desc])),701) ?> </td></tr> I have the above snippet.. The first php statement, basically grasp the first 701 characters with the closet next stop "." character and out puts it. then out puts the HTML tags I have a problem with the second statement. I want to output anything after what has been outputted by: Code: [Select] <? $out = preg_replace('/^(.{701}[^.]*).*/i','$1.',$detrsltnewsrow[news_desc]); echo $out; ?> So need the correct syntax for Code: [Select] <?= substr(stripslashes(trim($detrsltnewsrow[news_desc])),701) ?> Currently it breaks at exactly the 701 character, want it to continue from the sentence the first code ended in. |
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