i want to have an iframe, which displays the page given as parameter in the url.
I'm using ASP.NET MVC4
so I wanna do something like this:
updated it... but still not right i think
<?php
if(!isset($_GET['link']){
$link = $_GET['link'];}
?>
<iframe name="inlineframe" src="<?php $link ?>" frameborder="0" scrolling="auto" width="500" height="180" marginwidth="5" marginheight="5" ></iframe>
but i can't figure out the right code for this. can anyone help?
I also tried echoing php, but that doesnt seem to work for me.
The problem with your code snippet is this:
src="<?php $link ?>"
You're calling the variable, but doing nothing with it
To write the variable in the src attribute, use echo:
src="<?php echo $link ?>"
You should also remove the ! in your if statement
Try this:
<?php
$link = '';
if(isset($_GET['link']){
$link = $_GET['link'];
}
?>
Also echo your link variable.
src="<?php echo $link; ?>"
Related
I want to keep src attribute in a variable. I am working on php and tried the following method but it doesn't display the image on html page.
Code:
<?php $path="C:/horizontal.jpg"; ?>
<image src="<?php echo $path; ?>" style="width:304px;height:228px" />
I think you can use this technique:
<?php $path="http://" . $_SERVER["SERVER_NAME"];?>
<img src="<?php echo $path.'/projectName/image.jpg'; ?>" style="width:304px;height:228px" />
Where:
$_SERVER["SERVER_NAME"]: to get the server name, for example: www.example.com.
Also you can use $_SERVER["SERVER_ADDR"], in this case you can get the address IP of your server, for example: "127.0.0.1".
I hope this information helps you.
Good Luck.
<?php $path="localhost/projectName/";?>
<img src="<?php echo $path.'image.jpg'; ?>" style="width:304px;height:228px" />
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 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]; ?>"
Here is my problem.
If I do:
$imagePath = "images/spalt_images/body_images/525A.JPG";
?>
<img src="<?php $imagePath ?>" alt="front image" class="productImage"/>
<?php
Then my image does not show up.
However, if I do:
<img src="images/spalt_images/body_images/525A.JPG" alt="front image" class="productImage"/>
Then my image shows up just fine. Why would it not work with php?
Thanks
You need to echo it:
<?php echo $imagePath ?>
or use a short tag (not recommended, but that's another discussion):
<?=$imagePath ?>
You need to echo $imagepath otherwise it won't 'print' it out.
Change <?php $imagePath ?> in your code to <?=$imagePath?>, or add an echo if your server doesn't allow shorttags.
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"/>