Syntax error with Escaping html in php code

Question

I am having syntax error with my following code

<?php 
If (!empty($_SESSION['LogedinStudentId'])) {
echo '<h3>Your Scholarship Applications:</h3>
<table width="100%" class="table table-bordered">
<tr>
<th scope="col">Sr.No.</th>
<th scope="col">Date of Application</th>
<th scope="col">Course Type</th>
<th scope="col">Course Description</th>
<th scope="col">Subject</th>
<th scope="col">Applied for Semester No.</th>
<th scope="col">Scholarship Status</th>
<th scope="col">View / Print</th>
</tr>
<tr>
<td>' . ++$serialno . '</td>
<td>' . if(empty($row_studentdashboard['DateofApplication'])) { 
echo '&nbsp;';
} else {
echo date("d-m-Y", strtotime($row_studentdashboard['DateofApplication']));
};
. '</td>
<td>' . $row_studentdashboard['CourseType'] .'</td>
<td>' . $row_studentdashboard['CourseDescriptionLong'] .'</td>
<td>' . $row_studentdashboard['Subject'] .'</td>
<td>' . $row_studentdashboard['ApplyForSemYear'] .'</td>
<td>' . $row_studentdashboard['ScholarshipStatus'] .'</td>
<td><a href="#">View</a> / <a href="#">Print</a></td>
</tr>
</table>';
} else {
echo '<h3>You do not have any application pending</h4>';
}
?>

I am getting syntax error on line no. 17 and 22. The second (nested) if statement is throwing syntax error. I can not judge what is wrong. If I run this second if statement outside of the html it is working fine. Can anyone point out what's wrong?

@Laser *"What syntax error?"* - Probably "unexpected T_IF - T_ECHO" or something along that line ;-) — Funk Forty Niner, Sep 07 '15 at 14:13
The other thing is for you to try and put constructs like increment in parenthesis when concatenating within strings ie. (++$var) — Peter Chaula, Sep 07 '15 at 14:16
@Fred-ii- I knew it. I just want people to be much clearer when asking questions. Its easier to look for errors that way — Peter Chaula, Sep 07 '15 at 14:17
possible duplicate of http://stackoverflow.com/questions/18050071/php-parse-syntax-errors-and-how-to-solve-them — Funk Forty Niner, Sep 07 '15 at 14:21

score 2 · Answer 1 · answered Sep 07 '15 at 14:01

2

You 're not supposed to concatenate an if statement to a string. That is what you did on line 17/18

answered Sep 07 '15 at 14:01

Peter Chaula

3,456
2
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Can you please elaborate on this? – Kamlesh Panchal Sep 07 '15 at 14:07
1

It's a sin to say: ' . if(empty($row_studentdashboard['DateofApplication'])) { – Peter Chaula Sep 07 '15 at 14:10

score 2 · Answer 2 · answered Sep 07 '15 at 14:15

Try :

<?php 
    if (isset($_SESSION['LogedinStudentId']) && !empty($_SESSION['LogedinStudentId'])) {
        $out = '<h3>Your Scholarship Applications:</h3>';
        $out .= '<table width="100%" class="table table-bordered">';
        $out .= '<tr>';
        $out .= '<th scope="col">Sr.No.</th>';
        $out .= '<th scope="col">Date of Application</th>';
        $out .= '<th scope="col">Course Type</th>';
        $out .= '<th scope="col">Course Description</th>';
        $out .= '<th scope="col">Subject</th>';
        $out .= '<th scope="col">Applied for Semester No.</th>';
        $out .= '<th scope="col">Scholarship Status</th>';
        $out .= '<th scope="col">View / Print</th>';
        $out .= '</tr>';
        $out .= '<tr>';
        $out .= '<td>' . ++$serialno . '</td>';
        $out .= '<td>';
        if(!isset($row_studentdashboard['DateofApplication']) || empty($row_studentdashboard['DateofApplication'])) { 
            $out .= '&nbsp;';
        } else {
            $out .= date("d-m-Y", strtotime($row_studentdashboard['DateofApplication']));
        };
        $out .= '</td>';
        $out .= '<td>' . $row_studentdashboard['CourseType'] .'</td>';
        $out .= '<td>' . $row_studentdashboard['CourseDescriptionLong'] .'</td>';
        $out .= '<td>' . $row_studentdashboard['Subject'] .'</td>';
        $out .= '<td>' . $row_studentdashboard['ApplyForSemYear'] .'</td>';
        $out .= '<td>' . $row_studentdashboard['ScholarshipStatus'] .'</td>';
        $out .= '<td><a href="#">View</a> / <a href="#">Print</a></td>';
        $out .= '</tr>';
        $out .= '</table>';
    } else {
        $out  = '<h3>You do not have any application pending</h4>';
    }
    echo $out;

You went through a lot of trouble writing this up. But your opening line of *"Try"*, doesn't say much as to where the OP made their mistakes. *Enlighten us* ;-) — Funk Forty Niner, Sep 07 '15 at 14:16
Lol i wanted to show an another way to write his code, more easy to read ,(i think ^^) — Fky, Sep 07 '15 at 14:20

score 2 · Accepted Answer · answered Sep 07 '15 at 14:16

What you are doing is an if-statement inside of echo-statement. It is wrong. Run second if-statement outside of html and create a variable that you later print in your html. A kind of this:

if(empty($row_studentdashboard['DateofApplication'])) { 
$text = '&nbsp;';
} else {
$text = date("d-m-Y", strtotime($row_studentdashboard['DateofApplication']));
}

.....

<td>' . ++$serialno . '</td>
<td>' . $text . '</td>

Syntax error with Escaping html in php code

3 Answers3