PHP - Create An Invisible Text In An Image

I know with a submit button I can have a new record created when clicking it, but is it possible to do with a text link?

Hello Friends...

I am creating a HTML webpage and connecting with MYSQL.

If the client will check the conditions, it will pass it to PHP and get values from MYSQL and print it in the webpage as a HTML table format.

Now i want to create a SAVE button and save the HTML TAble values as TXT tab demilited file.

I tried of creating a TXT file by using fopen() and write the values by using fwrite() function. But it is working only in the putty. While running the program from the browser, its not working.

Will somebody help me to solve this prob?

I also was searched for creating a TXT file from HTML table. But i wouldnt get any codes reg this....

Thanks
Rose

Hi all

I am trying to write a piece of code to take an image, resize it and centre it on a canvas 300 pixels tall by 400 pixels high. I know I need to use imagecopymerged but how do I add it to the below code:

Code: [Select]
$imagelarge = $_FILES['image']['tmp_name'];
$imagelargemain = $_FILES['image']['name'];
$src = imagecreatefromjpeg($imagelarge);

list($width,$height)=getimagesize($imagelarge);
$newwidth=300;
$newheight=($height/$width)*$newwidth;
$tmp=imagecreatetruecolor($newwidth,$newheight);
imagecopyresampled($tmp,$src,0,0,0,0,$newwidth,$newheight,$width,$height);

$filename = WEB_UPLOAD."images/right-images/". $_FILES['image']['name'];
imagejpeg($tmp,$filename,100);
imagedestroy($src);
imagedestroy($tmp);

Many thanks for your help!
Pete

Hi! my code uploads an image and saves it to an upload folder, but I want to create a thumbnail of it.

For example, if I upload a 1400x1000 JPG, I would like to resize the image and it has to respect the proportion given a maximum size.

if I say the maximum width/height is 250px, the image should be thumbnailed according to the max height/width, and in proportion with the width/height

Thanks in advance

Fernando

Hi, I have a script that uploads an image to a directory and then saves the fill path to a field in a table for later use. The only problem is people are uploading huge images and then when I produce a catalogue of my listings it takes forever to load because the images are so big. What I am after is an add in script to create an ADDITIONAL image 100 x 75px, I don't really want to change my upload script. Any ideas? Thanks in advace.

Here is what I have:

Code: [Select]

<?php
$idir = "../fleet/";   // Path To Images Directory


if (isset ($_FILES['fupload'])){

//upload the image to tmp directory
$url = $_FILES['fupload']['name'];   // Set $url To Equal The Filename For Later Use 
if ($_FILES['fupload']['type'] == "image/jpg" || $_FILES['fupload']['type'] == "image/jpeg" || $_FILES['fupload']['type'] == "image/pjpeg") { 
$file_ext = strrchr($_FILES['fupload']['name'], '$account.');   // Get The File Extention In The Format Of , For Instance, .jpg, .gif or .php 
$copy = copy($_FILES['fupload']['tmp_name'], "$idir" . $_FILES['fupload']['name']);   // Move Image From Temporary Location To Perm  
}
}
$fleetimage1 = mysql_real_escape_string("$idir" . $_FILES['fupload']['name']);

//then insert sql code below...
?>

Hi guys, I have a simple code below which allows to me to upload image  to a file directory and save the location of my file in database

the only thing i dont understand is how to 1-resize the images  2- create a thumbnail while uploading 3- allow only jpeg & JPG files to be uploaded with 200kb max

also, could you tell me what is the best size i can store the images in?

thanks in advance

 //image1
$nameone=$_FILES['myfileone']['name'];
$tmp_name= $_FILES['myfileone']['tmp_name'];
if ($nameone)

{
$locationone="images/$nameone";
move_uploaded_file($tmp_name, $locationone);
$image1 = mysql_query ("UPDATE user SET image1='$locationone'");

}

Not sure if this is the right place to post this.

I have  PHP form that I use to upload a document, PDF or Word Doc, I would also like the form to create a thumbnail of the document when it is uploaded, is this possible?

I need to upload an image using php and store the image as blob in mysql. I understand the standard way is to create a form for the user, let the user choose the image to upload, click submit.

But I need to create a php, for example, uploadImage.php?image="c:\myPhoto.jpg"
and when I type http:\\www.myhome.com\uploadImage.php?image="c:\myPhoto.jpg"
from a web browser in my pc, I can automatically upload myPhoto.jpg to the myhome.com server?

Thank you for your kind help.

I have a Php application that already has 2 buttons with input on the index page that perform some functions and output to different Php pages. Works good. How do I add a button with no input that would just perform a function and then just display a popup window showing the contents of a text file? I want to add several of these buttons for different functions and just display the contents of the resulting text files in popup windows. Any help would be greatly appreciated. Thanks in advance.

I have some code that displays 1 large image and then 5 small ones, you click on a small one and it becomes the large one "trades places"

I then fixed the sizes so they don't distort when changing from large to small and then back.

My problem is when the image is clicked the large image size does not change, the image does

I that my issue lies here
Code: [Select]
<img name='picture' src=uploads/harkly_1.jpg ";echo imageResize($myImg1[0], $myImg1[1], 300); echo " border=0></td> but no clue on how to fix it

this part Quote

$myImg1[0], $myImg1[1

needs to change based on the selected image, can someone help me figure out how to make this happen??

Is there someway to create an if statement for this?
Code: [Select]
If (something){
echo "$myImg1[0], $myImg1[1]";
}
else {
echo "$myImg2[0], $myImg2[1]";
} 

My full code


Code: [Select]
<?php 

// to change the image size within the web page
function imageResize($width, $height, $target) {
//takes the larger size of the width and height and applies the formula accordingly...
//this is so this script will work dynamically with any size image 

if ($width > $height) {
$percentage = ($target / $width); 
} 
else { 
$percentage = ($target / $height); 
} 

//gets the new value and applies the percentage, then rounds the value 

$width = round($width * $percentage); 
$height = round($height * $percentage); 

//returns the new sizes in html image tag format...
//this is so you can plug this function inside an image tag and just get the 

return "width='$width' height='$height'"; 
} 
?>

<HEAD>
<SCRIPT LANGUAGE="JavaScript">
<!-- Original:  Ronnie T. Moore, Editor -->
<!-- Web Site:  The JavaScript Source -->

<!-- Begin
var photo_1 = new Image();
var photo_2 = new Image();
var photo_3 = new Image();

photo_1.src = "uploads/harkly_1.jpg";
photo_2.src = "uploads/harkly_2.jpg";
photo_3.src = "uploads/harkly_3.jpg";

function doButtons(picimage) {
eval("document['picture'].src = " + picimage + ".src");

}
//  End -->
</script>
</HEAD>

<?php //get the image size of the picture and load it into an array 

$photo_1="uploads/harkly_1.jpg";
$photo_2="uploads/harkly_2.jpg";
$photo_3="uploads/harkly_3.jpg";

$myImg1 = getimagesize($photo_1);
$myImg2 = getimagesize($photo_2);
$myImg3 = getimagesize($photo_3);


echo "

<BODY>
<center>
<table border=1>
<tr><td>

<p>
<li><a href = '' onmouseover = \"doButtons('photo_1')\"><img name='photo_1' src='$photo_1' ";echo imageResize($myImg1[0], $myImg1[1], 55); echo " border=0><p>

<li><a href = '' onmouseover = \"doButtons('photo_2')\"><img name='photo_2' src='$photo_2' ";echo imageResize($myImg2[0], $myImg1[1], 55); echo " border=0><p>

<li><a href = '' onmouseover = \"doButtons('photo_3')\"><img name='photo_3' src='$photo_3' ";echo imageResize($myImg3[0], $myImg1[1], 55); echo " border=0><p>

<td width=440 height=300> 
<img name='picture' src=uploads/harkly_1.jpg ";echo imageResize($myImg1[0], $myImg1[1], 300); echo " border=0></td> 

</tr>
</table>
</center>
";
?>

Hi,

So, I have two sessions. One's for displaying an array.

Is there some kind of function to make a session invisible?

Because when I display an error for not filling in, the previous words of the array show up behind the error message.
Like this:

$error = 'Please enter a word.'

Output: Please enter a word.word word word

So is there a way to make my Session Invisible (not removing any words from array)?

Hello
ive Googled around for ages for a PHP script that Shows the yahoo user statuts .. but the only thing i ve found is this link from yahoo thats saying the Online/Offline statuts  http://opi.yahoo.com/online?u=
ive heared that theres some way of sending a bot to the id and see if its on invisible ..
dos anyone know a way to do it ?

Hi all . I have a button now when i click the button an mypages is valid can i make that button in the page invisible in postback with php??

I  know you can in C# by saying button1.Visible = false;

Thanks for input...

This code is creating broken image.please check it
png and gd is enabled.
i also remove the header but still getting broken image.
there is no space before/after php tags.thanks you


my code
Code: [Select]
<?php
// Set the content-type
header('Content-Type: image/png');
// Create the image
$im = imagecreatetruecolor(400, 30);
// Create some colors
$white = imagecolorallocate($im, 255, 255, 255);
$grey = imagecolorallocate($im, 128, 128, 128);
$black = imagecolorallocate($im, 0, 0, 0);
imagefilledrectangle($im, 0, 0, 399, 29, $white);

// The text to draw
$text = 'Testing...';
// Replace path by your own font path
$font = 'arial.ttf';

// Add some shadow to the text
imagettftext($im, 20, 0, 11, 21, $grey, $font, $text);

// Add the text
imagettftext($im, 20, 0, 10, 20, $black, $font, $text);

// Using imagepng() results in clearer text compared with imagejpeg()
imagepng($im);
imagedestroy($im);
?>

Hi

I've got this database I created with fields

ProductId   ProductName  Image

I've managed to get it to list the ID,productname, and Image urls
in a list.

My next step is to have the image field actually display
an image and make it clickable:

heres what I've done so far:

Code: [Select]
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("productfeed", $con);

$result = mysql_query("SELECT * FROM productfeeds");

echo 
"<table border='0'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Image</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['ProductID'] . "</td>";
  echo "<td>" . $row['ProductName'] . "</td>";
  echo "<td>" . $row['ImageURL'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?>
Heres what I want to do:

Code: [Select]
while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['ProductID'] . "</td>";
  echo "<td>" . $row['ProductName'] . "</td>";

 // my changes beneath
  echo "<td>" . <a href="<?php echo $row['ImageURL'];?>"> <img src="<?php echo $row['LinkURL']; ?>"> </a>. "</td>";
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?>
Can you guys point me in the right direction?

Many thanks