Parse error Unexpected T_IF

Question

What i want to do is when semester is equals to "1" it will print "st Semester S.Y" else if semester is equals to "2" it will print "nd Semester S.Y". But the problem is i get an error of Parse error: syntax error, unexpected 'if' (T_IF). Can someone help me about this?

here's my code.

if(isset($_POST['loadsem'])){
      $stud = $_POST['stud'];
        $output = '';  
    $sql = "SELECT DISTINCT sch_year,semester FROM grades WHERE stud_no ='$stud'";
    $result = mysqli_query($con,$sql);
    $output .= '
                  <div class="table-responsive">
                    <table class="table table-bordered">
                      <tr>
                    <th>Semesters Attended</th>
                      </tr>';
      while($row = mysqli_fetch_array($result))
      {
        $sem = $row['semester'];
        $year = $row['sch_year'];
        $output .= '<tr>
                    <td><a href="">'.
                     if($sem == "1"){
                      $row['semester']. "st Semester S.Y" .$row['sch_year']
                    }
                    else{
                      $row['semester']. "nd Semester S.Y" .$row['sch_year']
                    }
                    .'</a></td>
                    </tr>
          ';
      }
          $output .= '</table>
            </div>';
   echo $output;


}

You cannot concatenate conditional operator into a string. instead break your variable — Thamilhan, Dec 20 '16 at 04:54
You can't do a `if`/`else` block in the middle of a variable concatenate - `$output .= '...'. if(){ } else{ } .'... ';` — Sean, Dec 20 '16 at 04:55

Praveen Kumar Purushothaman · Accepted Answer · 2016-12-20T05:17:32.157

3

You cannot concatenate an if statement.

$output .= '<tr>
    <td><a href="">';
if($sem == "1") {
  $output .= $row['semester']. "st Semester S.Y" .$row['sch_year'];
} else {
  $output .= $row['semester']. "nd Semester S.Y" .$row['sch_year'];
}
$output .='</a></td>
    </tr>';

If you still wanna concatenate, you need to use ternary operator: (condition) ? true : false.

edited Dec 20 '16 at 05:17

answered Dec 20 '16 at 04:54

Praveen Kumar Purushothaman

164,888
24
203
252

1

Unexpecter { sir. – nethken Dec 20 '16 at 05:04
1

You forgot the `;` maybe? hehe. Thankyou :) – nethken Dec 20 '16 at 05:05
@nethken Thanks. Approved your change boss. `:D` – Praveen Kumar Purushothaman Dec 20 '16 at 05:08
Thankyou sir. Im just a beginner. Can you recommend some site to read and know some rules to php and html? – nethken Dec 20 '16 at 05:10
2

@nethken Go for php.net. Or I personally chose PHP 5 for Dummies, but now it is outdated. – Praveen Kumar Purushothaman Dec 20 '16 at 05:12

score 2 · Answer 2 · answered Dec 20 '16 at 05:03

You cant do a function in if statesments so allways when u canna do php code end your print or varable. and add the new code

exemple 1:

echo 'hello'; 
if($world == true){
 echo 'foo';
}else{
 echo 'bar;'
}

exemple 2:

$i = 'hello'; 
if($world == true){
$i .='foo';
}else{
$i .='bar;'
}

in both exemple i end.

.= means add so you know

LF00 · Answer 3 · 2016-12-20T05:13:13.400

1

you cannot use if else like that, but you can use ?: in (). Here is a simple demo

change you if else

if($sem == "1"){
  $row['semester']. "st Semester S.Y" .$row['sch_year']
}
else{
  $row['semester']. "nd Semester S.Y" .$row['sch_year']
}

to: (?:)

($sem == "1" ? $row['semester']. "st Semester S.Y" .$row['sch_year'] : $row['semester']. "nd Semester S.Y" .$row['sch_year'])

edited Dec 20 '16 at 05:13

answered Dec 20 '16 at 04:58

LF00

27,015
29
156
295

Parse error Unexpected T_IF

3 Answers3