PHP External Links

PHP
1

Hi,

I have a peice of code which publishes an image with a link from my database. However I cant get it to use external links. My code is:

echo "<a href=\\"" .$link . "\\"><img src=\\""   .$image ."\\" />  </a><BR />";

I have tried all the options I can think of but I cant get it work. Can anyone advise please?

2

Hi,

I dont understand why would there be an error?

Does it use target=“_blank”

I thought I would a php command to open the image link in a new page.

3

try this:


echo "<a href=\\"" .$link . "\\"><img src=\\""   .$image ."\\" />".$image. "</a><BR />";

or


echo "<a href=\\"" .$link . "\\"><img src=\\""   .$image ."\\" />click here</a><BR />";
4

Hi,

I am trying to open a link from my database into a black tab.

Can I try this: “target=\”_blank\“>”

Will this use the " without messing up the PHP script?

5

I think you may use it. Please let us know the html output of your php code.
Regards,
zico

6

Hi,

It wouldn’t work. I am publishing images from a database with links but I am struggling to get the links to open new windows.

Is there a standard method for doing this?

7

Please let us know the html output of your php code.

8

Hi,

What do you mean the HTML output?

Isn’t it the parts below. I’m new to mixing PHP and HTML.

echo “<a href=\”" .$link . “\”><img src=\“” .$image .“\” /> </a><BR />";

9

I think he is looking to see a sample result of your page. Show what the echo puts on your page at the end.

10

Hi,

it displays an image fitted with a link.

11

By html output I mean the source of the html page displayed in your browser as a result of running php code on server.
For Firefox: View–>Page source (or Ctrl-U)
For IE: View–>Source

Reply with the relevant html code to see where your problem is.

If you consider usefull here is a piece of code from one of my older projects. It works for me. Please note that $file is the file name and $thumb_name the thumb name already generated. If you are not using automatically generated thumbs you may replace with what you want.


<a href="/images/<?php echo ($file); ?>"><img src="/images/<?php echo ($thumb_name); ?>"  border='0' target='_blank'">Click here to view full size picture.</a>

Regards,
zico

12

Assuming by ‘external link’ you mean new tab/window like your 2nd post describes, the following would be simple enough…

echo("<a href='$link' target='_blank'><img src='$image' /></a>");

Because you’re using " it will parse the variables ($link, $image) without going in and out of the string (“…”).

Also by using ’ instead of " inside the echo, we don’t need to escape it each time.