Make select default option from database

Question

i have issue with making default text in my select. (option list). So, my option is storage in database. Like varchar Test Option

Now if i make like this in my code

 <select name="category"> <option selected="selected"><?=$ad['category']?></option> <?php echo $options; ?> </select>

it working, but in my list be two same values then. my full code.

 <?php while($category = mysqli_fetch_array($q)):; ?> <option> value="<?php echo $category[0];?>" <?php echo $category[1];?> </option>" <?php endwhile; ?> <select name="category"> <option selected="selected"><?=$ad['category']?></option> <?php echo $options; ?> </select> <?php

My code for function q and pick value from database (all value from option is from database)

 $q2 = mysqli_query($con,$q22222); $options = ""; while($row2 = mysqli_fetch_array($q2)) { $options = $options."<option>$row2[1]</option>"; }

picture what explain situation more.

any idea how to solve it ? all help will be appreciated

Does the content from the pick list come from two different sources? Are $q and $q2 just revisions of the same code? Where does the selected value come from, is it user input? — Tim Morton
– Tim Morton, Commented Dec 6, 2020 at 22:50
Your code appears to make rather little sense to begin with - you output a couple of <option>…</option> in a loop first, and then directly after that comes a <select>? — C3roe
– C3roe, Commented Dec 7, 2020 at 9:32

BoredGeek · Accepted Answer · 2020-12-06 22:27:37Z

Well, the real situation is that you are fetching an array but you want that the option won't be printed. Then, you first have the selected option, that is:

<?php $ad['category']; ?>

Then the only thing that you have to do is that when you fetch the array, if the option is equal to the selected option won't be printed so:

 $q2 = mysqli_query($con,$q22222); $options = ""; while($row2 = mysqli_fetch_array($q2)) { if($row2[1]!=$selected){ $options = $options."<option>$row2[1]</option>"; } }

ısmaıl sahan · Accepted Answer · 2020-12-06 23:19:38Z

If you don't want to fetch same options again from the database you could use an SELECT DISTINCT statement in your select query.

And also handle the situation where the option is not same with the default one you chose.

while($row2 = mysqli_fetch_array($q2)) { if($row2[1] != $selected) $options = $options."<option>$row2[1]</option>"; }

Makwana Prahlad · Accepted Answer · 2020-12-07 04:49:28Z

<select > <option value="1" <?php if ($Defaultselection == 1) echo "selected"; ?>>A</option> <option value="2" <?php if ($Defaultselection == 2) echo "selected"; ?>>B</option> <option value="3" <?php if ($Defaultselection == 3) echo "selected"; ?>>C</option> <option value="4" <?php if ($Defaultselection == 4) echo "selected"; ?>>D</option> </select>

Try it

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