I want to update "profile of a user" in php. There is a repetition of one value for two times in dropdown list. for example i take language value='Punjabi' from database but there is also a value placed in dropdown with name of 'Punjabi'. The issue is simply that there is a repetition of value which i don't want.
<?php $result=mysqli_query($conn, "select * from profile where id=$firstPerson");
while($queryArray=mysqli_fetch_array($result)){ ?>
<select name="language" id="language" >
<option value='<?php echo $queryArray["language"];?> '> <?php echo $queryArray["language"]; ?></option>
//for example, the value from database is "Punjabi"
<option value="Hindi">Hindi</option>
<option value="Punjabi">Punjabi</option>
<option value="Urdu">Urdu</option>
</select>
<?php } ?>
when a value='Punjabi' from database is selected in dropdown list, the dropdown should not show the value='Punjabi' that is already placed in dropdown. Remember: i have more than 1000 values in my dropdown(html) list.
Instead of creating a new option according to the user data, Check if existing options are equal to user data:
<select name="language" id="language" >
<option value="Punjabi" <?php if ($queryArray["language"]=="Punjabi"){echo 'selected="selected"'} ?>>Punjabi</option>
<option value="Hindi" <?php if ($queryArray["language"]=="Hindi"){echo 'selected="selected"'} ?>>Hindi</option>
<option value="Urdu" <?php if ($queryArray["language"]=="Urdu"){echo 'selected="selected"'} ?>>Urdu</option>
</select>
If there are large number of options and you don't want to hard code these conditions, you can remove the second option using javascript on DOM ready:
$(document).ready(function(){
$('option[value="<?php echo $queryArray["language"] ?>"]').eq(1).remove();
})