CODEWITHSUNDEEP
phphtml
Gokul Shinde
Gokul Shinde

Reputation: 965

Drop Down of float values not showing option as selected

I am displaying a drop down of height and it content float values. When I displaying this drop down on edit form, not showing old value of height as selected in it.

I want to show 5.6 ft height value as selected in my drop down having values from 4, 4.1, 4.2....6.10, 6.11, 7 etc.

Following is the code that I used

<select name="height">
    <?php for($height=(4.0); $height <= 7; $height=($height+0.1) ): ?>
        <option value='<?php echo $height;?>' <?php if((5.6) == ($height)) echo "selected=selected"; ?> ><?php echo $height;?> ft</option>
    <?php endfor;?>                     
</select>

Is any one know solution to this issue ? Please help.

Upvotes: 4

Views: 1063

Answers (2)

tomahaug
tomahaug

Reputation: 1476

Comparing floating points in PHP can be quite a pain. A solution to the issue could be to do the following comparison instead of 5.6 == $height:

abs(5.6-$height) < 0.1

This will result in true for 5.6 and false for the other values in question.

Full solution:

<select name="height">
    <?php for($height=(4.0); $height <= 7; $height=($height+0.1) ): ?>
        <option value='<?php echo $height;?>' <?php if(abs(5.6-$height) < 0.1) echo "selected=selected"; ?> ><?php echo $height;?> ft</option>
    <?php endfor;?>                     
</select>

Upvotes: 1

Jamie Taylor
Jamie Taylor

Reputation: 4775

As Mark said in his comment, this is a floating point precision issue. You can solve this by using round() on your $height like this:

<select name="height">
    <?php for($height=(4.0); $height <= 7; $height=($height+0.1) ): ?>
        <option value='<?php echo $height;?>' <?php if(5.6==round($height,2)) echo "selected=selected"; ?> ><?php echo $height;?> ft</option>
    <?php endfor;?>                     
</select>

More information can be found here: A: PHP Math Precision - NullUserException

Upvotes: 1

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