$_GET command help

[phpfreak007]

I am having difficulty working out how to grab from a URL within a URL a link to an image and display this image on a webpage.

example:

 

http://domainA.com/page1.php?file_image=http://domainB.com/image.jpg

 

I know little of this the code below but this is not working:

 

<?php echo $_GET['file_image']; ?>

 

this just copies the url, whereas I want to display this image.jpg from domainB using image source by grabing it.

<img src=" ">

 

I hope you understand me.

[Maq]

Whatever value is after the equals sign is exactly what you're going to get.

file_image=http://domainB.com/image.jpg

 

Either pass in the correct value, or use some string functions to extract it.

 

[dcro2]

Do you mean this?

echo '<img src="'.$_GET['file_image'].'">';

[phpfreak007]

dcro2 thank you that's exactly what I wanted many thanks to all and a nice forum 

[phpfreak007]

I have a new query with reference to above slightly modified.

 

(reference from above: http://domainA.com/page1.php?file_image=http://domainB.com/image.jpg)

 

I now would like to know coding how to insert a hidden field within a simple href URL so that on clicking that URL it goes to the next page with the contents of 'file_image' copied to the end of the URL of the next page so e.g it becomes:

 

http://domainA.com/page2.php?file_image=http://domainB.com/image.jpg

 

I tried the code below but no luck:

 

<a href='page2.php' input type="hidden" name="file_image" value="<?php echo $_GET['file_image']; ?>">page 2</a>

 

Hope it makes sense.

[wigwambam]

<a href='page2.php?file_image=<?php echo $_GET['file_image']; ?>'>File image</a>

 

[phpfreak007]

Thank You

[smoseley]

Just wanted to add $_GET is not a command, it's a collection of parameters sent via the HTTP "get" request method.

 

$_GET can be used like any array in PHP.