if condition in javascript taking data from xml not working

Question

I'm having a problem and i don't know what i'm doing wrong.Well,
I'm parsing an xml file with php.The if conditiondoesn't work for some reason. Talking about this line: if (( '<?php echo $item['temp'] ?>' >= "37.0 °C") ) {
I have no idea why this line is not working.

    <?php foreach ( $item_array as $item ) : ?> 

                            var latlng = new google.maps.LatLng(parseFloat(<?php echo $item['glat']; ?>),
                                                                parseFloat(<?php echo $item['glon']; ?>));

                            if (( '<?php echo $item['temp'] ?>' >= "37.0 °C") ) {
                                alert('<?php echo $item['temp']; ?>');
                                var contentString = '<?php echo $item['title']; ?>' + ' , ' + '<?php echo $item['temp']; ?>' ;
                                var marker1 = createMarker1( contentString,latlng,hot );
                            }
        var hot   = 'weather_icons/hot.png';
        function createMarker1( contentString,latlng,hot ) {
                            var marker1 = new google.maps.Marker({
                            position: latlng,
                            map: map,
                            icon: hot
                            });
                        google.maps.event.addListener( marker1, "click", function() {
                        if (infowindow) infowindow.close();
                            infowindow = new google.maps.InfoWindow({
                            content: contentString
                            });
                            infowindow.open(map, marker1);
                        });
                        return marker1;
        }
    <?php endforeach; ?>

A sample of the xml file:

<rss version="2.0">
<channel>    
<item>
    <title>Salonika</title>
    <temp>6.7 °C</temp>
    <glat>40.422726139672626</glat>
    <glon>22.93392777442932</glon>
</item>
</channel>
</rss>

Checking strings with the >= operator seems alien to me. Perhaps try parseInt('<?php echo $item['temp'] ?>', 10) >= '37.0 °C' — Tim Vermaelen
– Tim Vermaelen, Commented Apr 16, 2014 at 22:07
thanks a lot Tim . I'm gonna take a look on how to compare strings — marYOs
– marYOs, Commented Apr 16, 2014 at 22:11
@TimVermaelen: parseInt would chop off any fractional portion (parseInt("6.7", 10) is 6). Technically as the OP is comparing against a whole number using >= that doesn't matter, but parseFloat still seems like the more appropriate choice. — T.J. Crowder
– T.J. Crowder, Commented Apr 16, 2014 at 22:13
You are right and although my little comment may work if you just compare 37 >= 36 but then again not 37.0 >= 37.5 so parseFloat seems more on topic. — Tim Vermaelen
– Tim Vermaelen, Commented Apr 16, 2014 at 22:20

T.J. Crowder · Accepted Answer · 2014-04-16 22:13:51Z

2

You're comparing strings, not numbers. The string "6.7 °C" is greater than the string "37.0 °C". When comparing strings, JavaScript compares them "character" by "character" (technically, they're not characters but code units in UTF-16, but that's a technical detail). The character "6" is greater than the character "3", so the comparison doesn't have to look any further; it knows that the string starting with "6" is "greater than" the string starting with "3".

I would recommend modifying the PHP to output a number rather than a string. But if you want to leave the PHP as it, you can deal with this in the JavaScript:

if ( parseFloat('<?php echo $item['temp'] ?>') >= 37 ) {

parseFloat will parse the beginning of the string into a number (it assumes base 10), stopping at the first non-number character (in this case, the space). Note that I've changed what it's comparing to, as well, and gotten rid of the extra () (they were harmless, but pointless).

edited Apr 16, 2014 at 22:13

answered Apr 16, 2014 at 22:07

T.J. Crowder

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2 Comments

Yogi Over a year ago

+1 and a reference for parseFloat(): developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

marYOs Over a year ago

I have to leave the php.I use the line you wrote and it works great!Thanks!

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