I am trying to show an image from the database to a new page using php. I am clicking on the image which needs to open in a new page. But everytime a same image is showing on the new page. I want an image in the new page which I clicked in last page. please help me.
<?php
$con = mysql_connect("localhost", "root" , "");
$sdb = mysql_select_db("display_images",$con);
$sql = "SELECT image FROM `tab1` WHERE 1";
$mq = mysql_query($sql) or die ("not working query");
$row = mysql_fetch_array($mq) or die("line 44 not working");
$s = $row['image'];
echo $row['image'];
echo '<img src="'.$s.'" alt="HTML5 Icon" style="width:800px;height:500px " >';
?>
Kindly use the below code:
<?php
$con=mysql_connect('localhost','root','');
$coo=mysql_select_db('your_db',$con);
if(!$coo)
{
echo 'error';
}
$get='select * from `your_table` where 1';
$get1=mysql_query($get);
$r=mysql_num_rows($get1);
for($j=1;$j<=$r;$j++){
$ans=mysql_fetch_array($get1);
?>
<img src="<?php echo $ans['your_field_name']?>" width="200" height ="200"
attr="<? php echo $ans['your_field_name']?>" class="image"/>
<?php }
?>
<script src="jquery.min.js" ></script>
<script>
$( document ).ready(function(){
$('.image').click(function(){
var new_image= $(this).attr('attr');
window.open( new_image );
});
});
</script>
The above code will do that just list out all the images from your table and then when you click a particular image that will displayed in the new window!!Is it your need??
mysql_query() is not preferable...Just for you!!
Related
am trying to display images page A but calling them from page B. images are contained in the images folder and database table images
my page a code is
<img src="addimage.php?image_id=<?php echo $row["code"]; ?>" />
page B code is
<?php
require_once "db.php";
if(isset($_GET['image_id'])) {
$sql = "SELECT image FROM images WHERE code=" . $_GET['image_id'];
$result = mysqli_query($mysqli, $sql) " . mysqli_error($mysqli));
$row = mysqli_fetch_array($result);
echo $row["image"];
}
mysqli_close($mysqli);
?>
i have store an image into my database.
but i am unable to display the image what i have done till now is under.... any help will be appreciated. thanks in advance!
<?
$query="SELECT * from testimonial";
$ret = mysqli_query($mysql,$query);
if (isset($ret) && $ret->num_rows>0)
{
while($row=mysqli_fetch_array($ret))
{
$body=$row['body'];
$name=$row['name'];
$image=$row['img'];
?>
<li>
<div class="frame-icon"><? echo "<img src=test_img.php?id=".$row['id']." width=150 height=150/>";?></div>
<p class="quote"><?php echo $body; ?><span><?php echo $name; ?></span></p>
</li>
<?php }
echo "</table>";
}
?>
and my test_img code is
<?
<?php if (isset($_GET['id'])){
$id=mysql_real_escape_string($_GET['id']);
$query=mysql_query("SELECT *FROM testimonial WHERE id='$id' ");
while($row = mysql_fetch_assoc($query))
{
$image=$row["img"];
}
header("content-type: image/png");?>
hi Abhik i have try this one but i get something like this
and in my case i have png image so just change the jpeg to png the code you have given
but i got this <img src="data:image/png;base64,iVBORw0KGgpcMFwwXDANSUhEUlwwXDBcMGRcMFwwXDBkCAZcMFwwXDBw4pVUXDBcMFwwCXBIWXNcMFwwCxNcMFwwCxMBXDCanBhcMFwwCk9pQ0NQUGhvdG9zaG9wIElDQyBwcm9maWxlXDBcMHjanVNnVFPpFj333vRCS4iAlEtvUhUIIFJCi4AUkSYqIQkQSoghodkVUcERRUUEG8igiAOOjoCMFVEsDIoK2AfkIaKOg6OIisr74Xuja9a89+bN/rXXPues852zzwfACAyWSDNRNYAMqUIeEeCDx8TG4eQuQIEKJHBcMBAIs2Qhc/0jAVww+H48PCtcIsAHvlwwAXjTCwhcMMBNm8AwHIf/D+pCmVxcAYCEAcB0kThLCIAUXDBAeo5CplwwQEYBgJ2YJlNcMKAEXDBgy2Ni41wwUC1cMGBcJ3/m01wwgJ34mXsBXDBblCEVAaCRXDAgE2WIRFwwaDtcMKzPVopFXDBY...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">
How about if you store the location of the image in DB then store the image itself in the file system? It's a better choice I think.
Proper way to display the blob images stored in DB is
echo '<img src="data:image/jpeg;base64,' . base64_encode($row['img']) . '">';
I have a comics website which loops through all images in a db and displays them as thumbnails.
The user can click on one of those images to see it in normal size on a viewComic.php template.
I'd like to allow users to press left and right arrows to navigate images.
So, my idea is:
pagination.php handles image display on correct pages (by offsetting) by looping through database result array. The user can click on a result (below) to go to that specific image on the viewcomic.php template.
'<br />IMG: <a href="./templates/viewcomic.php?id=' . $row['imgid'] . '&image=' . $imgpath.$row['imgname'] . '">
Now on viewcomic.php, I get the id and image, and display the image
$imgid = $_GET['id'];
$imgpath = $_GET['image'];
<center><img src=".<?php echo $imgpath ?>" /></center>
The user can press left and right arrows to navigate through images...
My goal was to somehow increment the image id to move to the next image, but that doesn't seem to be working...
<script type="text/javascript">
$(document).ready(function() {
$(document).keydown(function (e) {
if (e.which == 39) { //get next image
<?php
$count = 0;
$count++;
echo "<img src=" . $imageArray[$count] . "/>";
?>
}
});
});
</script>
Any ideas?
EDIT: I'm going to go through an image array passed in from pagination.php.
So, in my viewcomic.php file, I've updated my jquery script (see above).. but the jquery doesn't seem to like the embedded php, even though it's all in a php file.
Here's a picture of page source vs code:
Here is what i would do:
assuming that an imagepath is surrounded by quotes:
echo $imageArray[0]; // 'imagepath/image'
Script:
<script type="text/javascript">
var imgArray = [<?php echo implode(',',$imageArray) ?>];
// now the image array have the list of all your images.
$(document).ready(function() {
var img = document.getElementById("theImage");
imgIndex = 0;
$(document).keydown(function (e) {
if (e.which == 39) { //get next image
img.src = imgArray[imgIndex++]
}
... /* Logic to check if at the end of imageArray */ ...
});
});
</script>
The Html:
<center><img src="" id="theImage"/></center>
How about:
$(document).ready(function() {
$(document).keydown(function (e) {
if (e.which == 39) {
var nextId = $_GET['id'] + 1;
window.location = "./templates/viewcomic.php?id=" + nextId;
}
});
});
In this case your page is submit on every request, You can also handle this at client site.
Click link to see demo about rotate link using JavaScript. : Link Rotate using javascript
I am trying to capture the contents of my php page using output buffering:
<?php
function connect() {
$dbh = mysql_connect ("localhost", "user", "password") or die ('I cannot connect to the database because: ' . mysql_error());
mysql_select_db("PDS", $dbh);
return $dbh;
}
session_start();
if(isset($_SESSION['username'])){
if(isset($_POST['entryId'])){
//do something
$dbh = connect();
$ide = $_POST['entryId'];
$usertab = $_POST['usertable'];
$answertable = $usertab . "Answers";
$entrytable = $usertab . "Entries";
$query = mysql_query("SELECT e.date, q.questionNumber, q.question, q.sectionId, a.answer FROM $answertable a, Questions q, $entrytable e WHERE a.entryId = '$ide' AND a.questionId = q.questionId AND e.entryId = '$ide' ORDER BY q.questionNumber ASC;") or die("Error: " . mysql_error());
if($query){
//set variables
$sectionOne = array();
while($row=mysql_fetch_assoc($query)){
$date = $row['date'];
$sectionOne[] = $row;
}
}else{
//error - sql failed
}
}
?>
<?php
ob_start();
?>
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<script src = "jQuery.js"></script>
<script>
$(document).ready(function(){
$("#export").click(function(e){
//post to html2pdfconverter.php
$("#link").val("<?php echo(ob_get_contents()); ?>"); //THIS DOESN'T WORK
$("#nm").val("Entry Report.pdf");
$("form#sendanswers").submit();
});
});
</script>
<title>Personal Diary System - Entry Report - <?php echo($date); ?></title>
</head>
<body>
<h1>Entry Report - <?php echo($date); ?></h1>
<div id = "buttons">
<form id = "sendanswers" name = "sendanswers" action="html2pdfconverter.php" method="post">
<input type = "hidden" name = "link" id = "link" value = "">
<input type = "hidden" name = "nm" id = "nm" value = "">
<input type = "button" name = "export" id = "export" value = "Export As PDF"/>
</form>
</div>
<h3>Biological Information</h3>
<?php
echo('<p>');
$i = 0;
foreach($sectionOne as &$value){
if($i == 1 || $i == 3){
$image = "assets/urine".$i.".png";
echo("<br/>");
echo($value['question']." <br/> "."<img src = \"$image\"/>");
echo("<br/>");
}else{
echo($value['question'].' : '.$value['answer']);
}
echo("<br/>");
$i++;
}
echo('</p>');
?>
</body>
</html>
<?php
}
$contents = ob_get_contents(); //THIS WORKS
ob_end();
?>
I assign the contents of ob to $contents using ob_get_contents(); This works, and echoing $contents duplicates the html page.
However, in my jQuery, I am trying to assign this to a hidden text field ('link') using:
$("#link").val("<?php echo($contents); ?>");
This doesn't work however..And I have a feeling its because I am accessing $contents too eraly but not too sure...any ideas?
$("#link").val("<?php echo(ob_get_contents()); ?>"); //THIS DOESN'T WORK
at the point you do that ob_get_contents call, you've only output about 10 lines of javascript and html. PHP will NOT reach back in time and magically fill in the rest of the document where you do this ob_get_contents().
You're basically ripping the page out of the laser printer the moment the page starts emerging, while the printer is still printing the bottom half of the page.
I fail to see why you want to embed the contents of your page into an input field. If you want to somehow cache the page's content in an input field, you can just use JS to grab the .innerHTML of $('body').
Well, you have two problems.
The first is what you suspect. You can't access that stuff until later. The second problem which you may not realize is that you will have quoting issues in JavaScript even if you manage to find a way to reorder this and make it work. It's recursive, in a bad way.
What you should do instead is change your $('#export').click handler to do an Ajax call, render the HTML you need to appear in the link on the server in a separate PHP script (no output buffering necessary) and then have your code inject the result of that call into the page the way you're trying to do in your click handler now.
guys.
I tried to load image stored in mysql blob field with php, but the image does not show correctly. In firebug, I got these infos: get-image.php Dimensions0 × 0File size5.35KBMIME typeimage/jpeg
Here is my code
HTML
<html>
<head>
<title>Demo of Database Image in a page</title>
</head>
<body>
Here is your picture:<br>
img src=get-image.php?id=1 width=400 height=300><br>
</body>
</html>
PHP
<?php
include "db.php";
$conn = OpenDbConnection();
$key = $_GET["id"];
$tkey = "" . $key . "";
$strsql = "SELECT * FROM `images` WHERE `image_id` = " . $tkey;
$rs = mysql_query($strsql, $conn) or die(mysql_error());
if (!($row = mysql_fetch_array($rs))) {
die("File not exists.");
}
header("Content-type: image/jpeg");
echo $row["content"];
mysql_free_result($rs);
mysql_close($conn);
?>
Please someone tell me what is wrong with my code?
Please try this code.
Instead of
echo $row["content"];
Use this code
?>
<img scr="<?php echo $row["content"];?>" />
<?php
Thanks,
Kanji
Maybe it's because of blog type. Whenever you upload an image which exceed the limit of blob, then image not displayed correctly. Try to change type from blob to long blob.