Confused with how to retrieve an value from table register in page1

PHP
1

I want to formulate an if statment for when a store is registered or not.

In the stores table I have build an register row.

register int(11) NOT NULL,

then it is going to be equal to 1 for registered and 0 for Not Registered. Now in page1 I am trying to send the value to page2 and formulate an if condition in page2


If (1== $_GET['register']){ echo"Registered" }else{Not Registered}

That how it would look in page2 Now How can I retrieved the 1 or 0 value in page one from register table in page1 to send it to page2?

I am kind of confused help!

thank you

2

query the database to retrieve the value, then use echo to produce the appropriate link

<a href=“page2.php?register=1”>foo</a>
<a href=“page2.php?register=0”>foo</a>

3

This is the final result for a Registered, Not registered or Register now Script!

page1.


query= "SELECT register FROM stores"
$result = mysql_query($query) or die ("Error in query: $query. ".mysql_error()); 
if (mysql_num_rows($result) > 0) { 
$row = mysql_fetch_array($result); {
echo'<a href="page2.php?register=". $content['register']. "';
}}

page2.

div id="registered">
Store is: <?php If (1==$_GET['register']){ echo'<span style="color: #0000FF">Registered</span>'; }else{echo'<span style="color:#FF0000">Not Registered!</span>';}?> 
</div>
<div id="registernow">
<?php If (1==$_GET['register']){ '';}else{ echo'<a href="registrationform.php">Register now!</a>'; }?></div>

Thank you Crmalibu.

4 
query= "SELECT register FROM stores"
$result = mysql_query($query) or die ("Error in query: $query. ".mysql_error()); 
if (mysql_num_rows($result) > 0) { 
$row = mysql_fetch_array($result); {
echo'<a href="page2.php?register=". $content['register']. "';
}}

is this a real code or just dummy code(for example purpose)

5

that will be dummy code for example purposes.

6

Better if you had mentioned such(for example dummy code, just for example, untested code etc.) while presenting your code.