两个人不断互相改变

Question

I'm trying to update a database from two id's. One arrives with an other page. And one gets send by an element on the page. The problem is that both are send by GET so they keep changing each other. Is there anyway to fix this?

<?php
$id1 = $_REQUEST["id"];
echo $id1;
if($_GET["Action"] == "Wijzig")  
{  
    include 'Includes/database_connection.php';
    $strSQL = "UPDATE pairings SET sim_id='".$_GET["id"]."'";  
    $strSQL .="WHERE unit_id = '".$id1."' ";  
    $objQuery = mysql_query($strSQL);  
}

\\Action on current page
    <li><a href="JavaScript:if(confirm('Heeft u de goede geselecteerd?')==true){window.location='<?=$_SERVER["PHP_SELF"];?>?Action=Wijzig&id=<?=$row["id"];?>';}"><?php echo $row['phonenr'];?></a></li>
\\Action on previous page
<a href="#" onclick="window.open('test.php?id=<?php echo $row['id']; ?>', 'newwindow', 'width=5000, height=5000'); return false;" ><img src="images/arrowUpDown.png"/></a>

Answer 1

Try this to avoid the GET and make it a POST:

• put a hidden form on the page named actionform containing hidden type form fields "Action" and "id".
• <li><a href="#" onclick="if(confirm('Heeft u de goede geselecteerd?')){document.forms.actionform.elements['id'].value='<?php echo $row['id'];?>';} return false;"><?php echo $row['phonenr'];?></a></li>

This should put the id in the form, submit the form, and return false to indicate that the default action of the a is not to be executed.

Regarding your duplication problem, explain it better, but maybe the return false helps.