Parse Error in css/html when they are inside PHP code

Question

Trying to do this but it giving parse error... is there any way?

<?php
  if($profileIdType==1)
  {
    <li class="col-md-6">
        <label>First Name</label>
        <input type="text" class="form-control validate[required]"
               name="first_name" id="first_name" placeholder="First name" readonly="true"
               data-toggle="tooltip" title="First name">
    </li>
  }
  else
  {

  }
?>

?> <- close php before the html and then start it again <?php before the bracket. — Rizier123
– Rizier123, Commented Jul 12, 2015 at 20:24

jedrzej.kurylo · Accepted Answer · 2015-07-12 20:26:17Z

2

If you want to output HTML/CSS from your PHP script you need to close/reopen PHP tags:

<?php
if ($profileIdType == 1) {
?>
    <li class="col-md-6">
        <label>First Name</label>
        <input type="text" class="form-control validate[required]"
               name="first_name" id="first_name" placeholder="First name" readonly="true"
               data-toggle="tooltip" title="First name"
            >
    </li>
<?php
} else {

}

answered Jul 12, 2015 at 20:26

jedrzej.kurylo

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1 Comment

Milon Sarker Over a year ago

Thanks ... Stackoverflow is the best thing in Internet... and its users too... @jedrzej.kurylo

Peyman Mohamadpour · Accepted Answer · 2015-07-12 20:30:16Z

1

You need to separate scripting from markup.

<?php
if ($profileIdType == 1) {?>
<li class="col-md-6">
    <label>First Name</label>
    <input type="text" name="first_name">
</li><?php
} else {?>
<li class="col-md-6">
    <label>Last Name</label>
    <input type="text" name="last_name">
</li><?php
}?>

answered Jul 12, 2015 at 20:30

Peyman Mohamadpour

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