javascript if else statement problem

Question

<?php $genderdb = "Male"?>
<script type="text/javascript">
        var genderdb = "<?php echo $genderdb; ?>";
        if ((genderdb == "Female") || ($genderdb == "Shemale")) {
            var himher = "her";
        } else {
            var himher = "him";
        }
        alert (himher);
</script>

If I change php $genderdb = "Female", then it can alert the value successful. But if $genderdb = "Male", the page won't have alert. Why? Where is the error?

@CD Sanchez: That's an answer. Post it, and I'll vote it up. Good eye. @zac1987: JavaScript expressions are short-circuited. When genderdb == "Female", the if condition is true and so the second half doesn't have to be evaluated at all. If genderdb != "Female", the second half of that condition has to be evaluated, resulting in an undefined symbol error. You should be seeing it in the JavaScript console of whatever browser you're using. — T.J. Crowder
– T.J. Crowder, Commented Jun 17, 2011 at 23:21
For reference, you can use <?=$genderdb?> instead of <?php echo $genderdb; ?> — Jamie Wong
– Jamie Wong, Commented Jun 17, 2011 at 23:22
@Jamie: (a) only if you have short_tags enabled; (b) please do not. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Jun 17, 2011 at 23:22
@Jamie: Only if the shorttags option is turned on, which it isn't necessarily. — T.J. Crowder
– T.J. Crowder, Commented Jun 17, 2011 at 23:23

Lightness Races in Orbit · Accepted Answer · 2011-06-17 23:20:59Z

4

if ((genderdb == "Female") || ($genderdb == "Shemale")) {

should be

if ((genderdb == "Female") || (genderdb == "Shemale")) {

It works when genderdb is "Female", because then the (broken) second comparison doesn't come into the equation.

In addition, using var inside the conditional blocks is suspect. Javascript actually doesn't have block scope so it works, but it's something to consider.

answered Jun 17, 2011 at 23:20

Lightness Races in Orbit

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3 Comments

T.J. Crowder Over a year ago

Re var: Yes, it's specifically allowed by the spec to repeat a var within an execution context (logic flow has no effect whatsoever on var, it's evaluated upon entry to a context), but it's not a good idea.

zac1987 Over a year ago

what is blocks? Do you mean I must not use var? If so, what should I use?

Lightness Races in Orbit Over a year ago

@zac1987: Since it's not a problematic issue here, and quite complex to explain in a comment, I think I'll actually leave the topic for another day.

Wayne · Accepted Answer · 2011-06-17 23:21:02Z

2

You have an error in this line:

if ((genderdb == "Female") || ($genderdb == "Shemale")) {

Namely, the second reference to genderdb uses the name of the PHP variable, instead of the JavaScript variable.

answered Jun 17, 2011 at 23:21

Wayne

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Comments

Town · Accepted Answer · 2011-06-17 23:21:58Z

0

Remove the $ from your 'shemale' comparison.

That's causing your script to fail as it's looking for a non-existent variable.

        var genderdb = "<?php echo $genderdb; ?>";
        if ((genderdb == "Female") || (genderdb == "Shemale")) {
            var himher = "her";
        } else {
            var himher = "him";
        }
        alert (himher);

answered Jun 17, 2011 at 23:21

Town

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