if !empty expression in echo

Question

I have code

 <?
 $query = mysql_query("SELECT * FROM poi_example");
 while ($row = mysql_fetch_array($query)){
 $name=$row['name'];
 $lat=$row['lat'];
 $lon=$row['lon'];
 $desc=$row['desc'];
 $pic1=$row['pic1'];
 echo ("addMarker($lat, $lon,'<b>$name</b><br/>$desc<br/><img src=\"$pic1\" style=\"width:250px;\">  ');\n");
 }
 ?>

And I want display 'img src...' only if $pic1 is not empty. But how I can add if !empty in echo?

Answer 1

You can use:

if(!empty($pic1))
   echo ("addMarker($lat, $lon,'<b>$name</b><br/>$desc<br/><img src=\"$pic1\" style=\"width:250px;\">  ');\n");
else
    echo ("addMarker($lat, $lon,'<b>$name</b><br/>$desc<br/>');\n");

Answer 2

Try this out
Insert this at line 9 below the $pic1=$row['pic1'];

IF($pic1 != '') $pic1 = '<img src="'.$pic1.'" style="width:250px;"> ';
Then change this line
echo ("addMarker($lat, $lon,'<b>$name</b><br/>$desc<br/><img src=\"$pic1\" style=\"width:250px;\"> ');\n");
Into:
echo "addMarker($lat, $lon,'<b>$name</b><br/>$desc<br/>$pic1');\n";

Answer 3

Here:

echo !empty($pic1) ? $string : '';

http://php.net/manual/en/language.operators.comparison.php#language.operators.comparison.ternary

Answer 4

Solution:

This will definitely solve your problem.

<?
 $query = mysql_query("SELECT * FROM poi_example");
 while ($row = mysql_fetch_array($query)){
 $name=$row['name'];
 $lat=$row['lat'];
 $lon=$row['lon'];
 $desc=$row['desc'];
 $pic1=$row['pic1'];
 echo ("addMarker($lat, $lon,'<b>"."$name"."</b><br/>"."$desc"."<br/>".if(!empty($pic1)){."<img src=\"$pic1\" style=\"width:250px;\">".}."  ');\n");
 }
 ?>

Answer 5

You can try this:-

if($pic1 != '')
       echo ("addMarker($lat, $lon,'<b>$name</b><br/>$desc<br/><img src=\"$pic1\" style=\"width:250px;\">  ');\n");
    else
        echo ("addMarker($lat, $lon,'<b>$name</b><br/>$desc<br/>');\n");