I can't seem to get my image to display properly. Previously, I have used the following code snippet and it worked perfectly.
catalog.php (worked perfectly):
<p class="image">
<a href="synopsis.php?id=<?php echo $row['id']; ?>">
<img src="getImage.php?id=<?php echo $row['id']; ?>" alt="" width="175" height="200" />
</a>
</p>
synopsis.php (not displaying image at all):
<?php
$id = $_GET['id'];
...?>
<p class="image">
<img border="0" class="floatleft" src="getImage.php?id=<?php echo $row['id']; ?>" width="250" height="400" />
<?php echo $row['synopsis']; ?>
</p>
where getimage.php:
<?php
$id = $_GET['id'];
$link = mysql_connect("localhost", "root", "");
mysql_select_db("dvddb", $link);
$sql = "SELECT dvdimage_path FROM dvd WHERE id=$id";
$result = mysql_query($sql, $link);
$row = mysql_fetch_assoc($result);
mysql_close($link);
header("Content-type: image/jpeg");
echo file_get_contents($row['dvdimage_path']);
?>
Any idea why can't I display this image?
EDIT 1:
So after debugging, I got an error message:
Undefined index: id in C:\xampp\htdocs\synopsis.php on line 106
so i went to add the following code into the php code just before echo $row['id']:
<p>getImage.php?id=<?php error_reporting(0); echo $row['id']; ?></p>
However,
the paragraph i got was just getImage.php?id=.
Then, i went into synopsis.php -> <img border="0" class="floatleft" src="getImage.php?id=<?php echo $row['id']; ?>
and changed that into:
<img border="0" class="floatleft" src="getImage.php?id=2">
Again, same problem happens, where i can't get the specific image out.
I suspect something is wrong with my getimage.php file. However, this getimage.php file has been working fine for other pages when i use the snippet.
My requirements are very simple:
In catalog.php, i populate images and text from dvd database using a while loop. Then, each of these images has got their specific primary ID. when i click the the images, they will go to the link: synopsis.php?id="primaryid" Then, using this "primaryid" i should be able use getimage.php?"primaryid" to generate an image on synopsis.php page.
EDIT 2:
actually, i made a syntax error somewhere. So this line:
<img border="0" class="floatleft" src="getImage.php?id=2">
is working perfectly, this means the fault lies in somewhere that i cant echo 'id' out correctly.
EDIT 3:
I have included the links to the relevant source code:
catalog.php
synopsis.php
getimage.php
sortmenu.css
style.css
database in xml format
Questions to ask yourself:
Is it really required that you use a php script to mimic the image? If not, just use the image path.
Is there any output before the header(); function in the getimage.php file? Even just a space before the
Is the image actually a JPEG?
Are there any errors coming up when you go to getimage.php?id=ID in your browser?
In synopsis.php you are getting the id from the querystring but then trying to use a database value. I cant see all of your code so im posting solutions to cover two scenarios
src="getImage.php?id=<?php echo $id; ?>"
or
src="getImage.php?id=<?php echo $row[$id]; ?>"
Related
I have a database which contain blob images. I want to display them in a webpage. I'm using the following php code to get the image. But it didn't work properly.
<?php
while($row=mysql_fetch_array($results))
{?>
<tr>
<td><img src="<?php echo $row["image"]?>" height="200" width="250"></td>
</tr>
<?php
}?>
With this code I'm getting a webpage like this.
Where I have to do the correction to my code.
try to convert it to base64:
<img src="data:image/jpeg;base64,<?php echo base64_encode($row['image']); ?>" height="200" width="250">
Note that you should also store the image type in your DB, so you can dynamically change "data:image/jpeg".
I'm trying to display images by taking their file path from an sql table, but i'm having a lot of troubles.
Here is whats going on:
$image is a variable containing the text "itemimg/hyuna.png" which is path to an image.
$image = 'itemimg/hyuna.png';
I assumed I would be able to display the image outside of the php block like so:
<img src= "<? $image ?>" alt="test"/>
This doesn't work though for some reason.
So I thought maybe it's not able to read the variable outside the php block(i'm a beginner), so for testing i did:
<h1> "<? $image ?>" </h1>
It displays itemimg/hyuna.png as a h1 banner.
Meaning it's accessing the varible fine.
So I thought maybe the path is wrong. So I tried:
<img src= "itemimg/hyuna.png" alt="test"/>
This displays the image perfectly.
So now I'm stuck scratching my head why the first bit of code displays nothing but the text "test" from "alt="
Extra question:
How do I go about assigning a value from an sql cell to a variable?
I attempted the following with no luck:
$q = "select * from item where id=$id";
$results = mysql_query($q);
$row = mysql_fetch_array($results, MYSQL_ASSOC);
$image = ".$row['image'].";
item is a table with a collumn: image which contains file paths to images
First of all, you should not use PHP Shorttags.
When you use the PHP Shorttags you have to say:
<img src="<?=$image ?>" alt="test" />
But i would encourage to escape the Content off the variable like this:
<img src="<?php echo htmlspecialchars($image); ?>" alt="test" />
Your extra question:
This should lead to an syntax error because the string could not be parsed, just use $image = $row['image'];
Try this
<img src= "<?php echo $image ?>" alt="test"/>
try
<img src= "<?= $image ?>" alt="test"/>
or
<img src= "<? echo $image; ?>" alt="test"/>
try this
<img src= "<?php echo $image ?>" alt="test"/>
I am trying to display 4 images using PHP and MySQL database. I have to display the 4 images as rows.
I use the table cars with fields (id_car, car_image1, car_image2, car_image3, car_image4), with all the images being blob datatype.
$id = $_GET['id'];
$link = mysql_connect("localhost", "root", "");
mysql_select_db("cars_database");
$sql = "SELECT car_image1, car_image2, car_image3, car_image4 FROM cars WHERE id_car='$id'";
$result = mysql_query("$sql");
mysql_close($link);
while($row=mysql_fetch_array($result))
{
header('Content-type: image/jpeg');
echo $row['car_image1'];
echo $row['car_image2'];
echo $row['car_image3'];
echo $row['car_image4'];
}
I can only display 1 image and not the other images. Since I am a newbie to this technology I need help.
That's how HTTP works (at least, current version): you cannot have a URL that points to more than one resource simultaneously. Your script needs to send one image.
Simply, call it four times:
<img src="/get-image.php?id=1">
<img src="/get-image.php?id=2">
<img src="/get-image.php?id=3">
<img src="/get-image.php?id=4">
You need to define path for images.
<?php
while($row=mysql_fetch_array($result))
{
header('Content-type: image/jpeg');
?>
<img src="path<?php echo $row['car_image1']; ?>" alt="" />
<img src="path<?php echo $row['car_image2']; ?>" alt="" />
<img src="path<?php echo $row['car_image3']; ?>" alt="" />
<img src="path<?php echo $row['car_image4']; ?>" alt="" />
<?php
}
?>
I am trying to display image from a blob field of a MySQL table. Looks like I have some sort of error in the following line. As soon as I put "header("Content-type: image/jpeg")" things get messed up and instead of displaying webpage, all source code of the page is displayed.
Please let me know how to correct.
<div class="image" align="left">
<a href="<?php header("Content-type: image/jpeg"); echo $rec['image']; ?>">
<img src="<?php echo $rec['image']; ?>" width="150" border="0"/>
</a>
</div><!-- image -->
You normally don't put the actual image contents in the src= attribute of the image tag. Instead, you point to the URL of an image file.
(There are ways to include the image source directly in the HTML, but it doesn't work consistantly with all browsers, and you still won't have your <a> link working properly.
Instead, the best way to do this is to create a separate PHP file to serve the image.
Your HTML:
<div class="image" align="left">
<img src="myimage.php?key=<?php echo($key) ?>" width="150" border="0"/>
</div><!-- image -->
myimage.php:
<?php
header("Content-type: image/jpeg");
$key = $_GET['key'];
// todo: load image for $key from database
echo $rec['image'];
You're trying to put the image data inline inside the content. The only feasible way to do this is via a Data URI data URI. Something like:
<img src="data:image/jpeg;base64,<?= base64_encode($rec['image']) ?>" width="150" border="0" />
However, what you probably want to do is put it into a separate script. So your HTML would be:
<img src="showimage.php?id=XXX" width="150" border="0" />
And your showimage.php script would be:
<?php
// Get $rec from database based on the $_GET['id']
header('Content-Type: image/jpeg');
echo $rec['image'];
?>
I've done something like that retrieving blob from my database in another way that you may find useful, here is the code example.. see if it suits your needs and if you needed anymore help let me know.
while ($row = mysql_fetch_array($hc_query2)) {
$title = $row['title'];
$text = $row['text'];
$image = $row ['image'];
$output ='<div class="HCInstance"><img src="data:image/jpeg;base64,' . base64_encode($image) . '" alt="High Council" width="100px" height="100px"/>
<div class="HCHeader"><h2>'.$title.'</h2></div><br/><div class="HCDetails"><p>'.$text.'</p></div></div>';
echo $output;
}
I have a PHP echo function inside of a HTML link, but it isn't working. I want to have an image location, defined in img src, be in part of the clickable link of the image. The page will have multiple images doing the same thing, so I am trying to use PHP to automate this.
<a href="http://statuspics.likeoverload.com/<?php echo $image; ?>">
<img src="<?php $image=troll/GrannyTroll.jpg?>" width="100" height="94" />
</a>
Turn
<?php $image=troll/GrannyTroll.jpg?>
into
<?php echo "troll/GrannyTroll.jpg"; ?>
?
Or provide more details on what you are trying to achieve.
Also, you might consider urlencode-ing some of those URL parameters.
Edit:
So you might try setting the variable beforehand:
<?php $image = "troll/GrannyTroll.jpg"; ?>
<img src="<?php echo $picture; ?>" width="100" height="94" />
So now i understand what you are trying to do.
One error is that you didn't enclose $image=troll/GrannyTroll.jpg with quotes like this:
$image = 'troll/GrannyTroll.jpg';
The second error is that you do it in the wrong order, you have to define $image first, before you use it.
That's what I believe you want to do:
<?php
$image = "troll/GrannyTroll.jpg";
?>
<img src="<?php echo $image; ?>" width="100" height="94"/>