Parse error: syntax error, unexpected T_IF while parsing if statement in xml sitemap

Question

i am trying to create xml sitemap i have a column album_image in my database. in start day's i did't upload album image and the column store '0' as value after years ago i start uploading album image. now i want to create sitemap for it. if album_image == 0 display static image else display album iamge. how can i use if statment in $xml .=

$xml .= '<image:image>';
        $xml .= '<image:loc>'.SITE_URL. if($row->album_image == '0'){'images/video_icon.png'}else{'media/'.$row->cat_id.'/images/'.$row->album_image.'}</image:loc>';
        $xml .= '<image:caption>'.$row->album_name.'</image:caption>';
        $xml .= '</image:image>';

i am getting this error " Parse error: syntax error, unexpected T_IF in E:\wamp\www\site\sitemap-album.php on line 15 "

You can't put an if statement there. you need to separate it. — Farkie, Jun 20 '16 at 08:57
Probably the wrong website for you... but I've answered below. — Farkie, Jun 20 '16 at 09:01

Farkie · Accepted Answer · 2016-06-20T09:33:46.597

1

You need to separate the if condition:

$xml .= '<image:image>';
$xml .= '<image:loc>'.SITE_URL;

if($row->album_image == '0') {
   $xml .= 'images/video_icon.png';
 } else {
   $xml .= 'media/'.$row->cat_id.'/images/'.$row->album_image;
 }

 $xml .= '</image:loc>';
 $xml .= '<image:caption>'.$row->album_name.'</image:caption>';
 $xml .= '</image:image>';

edited Jun 20 '16 at 09:33

answered Jun 20 '16 at 08:58

Farkie

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score 0 · Answer 2 · answered Jun 20 '16 at 09:02

you can user this also $xml .= '';

 $xml .= '<image:loc>'.SITE_URL. ($row->album_image == '0')?'images/video_icon.png':'media/'.$row->cat_id.'/images/'.$row->album_image.'</image:loc>';
    $xml .= '<image:caption>'.$row->album_name.'</image:caption>';
    $xml .= '</image:image>';

score 0 · Answer 3 · answered Jun 20 '16 at 09:02

the full code is bellow . now i m geting "Parse error: syntax error, unexpected '=' in"

$sql = "SELECT * from albums  LIMIT 0, 1";
$query = mysql_query($sql);
if(mysql_num_rows($query) > 0){
    while($row = mysql_fetch_object($query)){
        $xml .= '<url>';
        $xml .= '<loc>'.SITE_URL.$row->album_url.'-album-1</loc>';
        $xml .= '</url>';
        $xml .= '<image:image>';
        $xml .= '<image:loc>'.SITE_URL;

        if($row->album_image == '0') {
           $xml . = 'images/video_icon.png';
         } else {
           $xml .= 'media/'.$row->cat_id.'/images/'.$row->album_image.';
         }

         $xml .= '</image:loc>';
         $xml .= '<image:caption>'.$row->album_name.'</image:caption>';
         $xml .= '</image:image>';
    }
}
$xml .= '</urlset>';
echo $xml;

I've fixed my code. Use that one instead. You shouldn't post an answer as an update. — Farkie, Jun 20 '16 at 09:34

Pupil · Answer 4 · 2016-06-20T09:51:04.393

Every PHP statement must end with semi colon (;).

It has an exception: Last statement in the file.

echo statement after some code needs that previous statement is closed.

Concatenation can work, but, for echo, closing previous statement is the only option.

Use ternary operator and store result in variable.

Use that variable to make code readable.

Complete code:

$videoImage = ($row->album_image == '0') ? 'images/video_icon.png' : 'media/'.$row->cat_id.'/images/'.$row->album_image;
$xml .= '<image:image>';
$xml .= '<image:loc>'.SITE_URL . $videoImage . '</image:loc>';
$xml .= '<image:caption>'.$row->album_name.'</image:caption>';
$xml .= '</image:image>';

Parse error: syntax error, unexpected T_IF while parsing if statement in xml sitemap

4 Answers4