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How to determine if a link is clicked or not staying on the same page in PHP?

Posted on 2011-09-30
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Last Modified: 2012-05-12
Hi,

I have a link on the page, say X, saying "Connect" clicking on which user is redirected to another page, say Y. If everything goes fine on page Y, the user is redirected to page X but instead of link, there is a text saying "Account Connected". But if something goes wrong on page Y, then I need to catch the error and display it on page X and let the link still remain active. How do I do this is in PHP? Since I am not an expert in PHP, I dont know how to do this and I need quick answers as I m stuck with my project.

Thanks in advance
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Question by:ishani_v
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3 Comments
 
LVL 13

Accepted Solution

by:
Andrew Derse earned 2000 total points
ID: 36891476
You would send variables back and forth through the URL using PHP.

On page X you will need code at the top to catch if there is anything from Page Y.

That code would look like this:
 
<?php

if (isset($_GET['test']))
{
$x = $_GET['test'];
} else {
$x = no;
}

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Then you need to pass the test variable from Y to X like this:
 
<a href="page-x.php?test='$results';>Submit</a>

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The: test = $result is set to have $result being the message you want to send back.  It can be whatever you want, but it needs to be something logical.

Back to the X page we finish the logical check to display the X or Account Connected bit.
 
if ($test = 'OK') {
$buttonText = 'Account Connected";
} else {
$buttonText = "X";
}

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That's the gist of it.  Please let me know if you need more assistance with this.
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LVL 13

Expert Comment

by:Andrew Derse
ID: 36891482
Wow, sorry, I didn't close the strings properly...

Third code, line 2 needs to be:
 
$buttonText = "Account Connected";

Open in new window


0
 
LVL 13

Expert Comment

by:Andrew Derse
ID: 36891490
And one more thing...make the Third code: line 1

if ($test == 'OK') {

It needs to have 2 equal signs, not one...

(Still too early in the morning for me...sorry about that)
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