Reputation: 1085
I have a link that passes variable to another page but I want to display that link only on some condition so how to place it inside php
code.
<a href="page2.php?Id=<?php echo $Id; ?>">Product</a>
<?php
echo '<a href="page2.php?Id=<?php echo $Id; ?>">Product</a>';
?>
This doesn't work.
Upvotes: 1
Views: 2658
Reputation: 460
You dont have to print echo inside3 echo so u hv to do this way!
<?php
echo '<a href="page2.php?Id="'.$Id.'" ">Product</a>';
?>
Upvotes: 1
Reputation: 1835
Your echo field should be.
<a href="page2.php?Id=<?php echo $Id; ?>">Product</a>
<?php
echo '<a href="page2.php?Id='.$Id.'">Product</a>';
?>
Upvotes: 1
Reputation: 23978
Simply, put your anchor link in if
condition:
<?php
if (YOUR_CONDITION_HERE) {
echo '<a href="page2.php?Id='.$Id.'">Product</a>';
}
?>
OR
<?php
if (YOUR_CONDITION_HERE) {
?>
<a href="page2.php?Id=<?php echo $Id;?>">Product</a>
<?php
}
?>
Above condition will display your link only when your condition is TRUE
.
Otherwise, it will hide.
Upvotes: 1
Reputation: 1345
Your second line is incorrect why are you opening your php tags again while you're in a echo ?
Please try the following :
<?php
echo '<a href="page2.php?Id='.$Id.'">Product</a>';
?>
Upvotes: 1