使用SQL在下拉菜单>（名称，数字）内的一行中使用SQL带有两个不同查询的下拉PHP

Question

I want the dropdown to show the "client_code", "name" in one line. It almost works but not 100%. I am a beginner with php and SQL, can someone help me please?

Code that doesn't work

<form id="thirdForm" name="form1" action="" method="post">
<select id="klantWidth">
<?php
$queryKlant = "SELECT naam FROM klant";
$queryKlantCode = "SELECT klant_code FROM klant";
$resultKlant=mysqli_query($mysqli,$queryKlant);
$resultKlantCode=mysqli_query($mysqli,$queryKlantCode);

while($row=mysqli_fetch_array($resultKlant) && 
$row2=mysqli_fetch_array($resultKlantCode)  )
{
?>

<option><?php echo $row[0]. ", ". $row2[0];?></option>
<?php
}
?>
</select>
</form>

Code that only works with retrieving name in dropdown from database

<form id="thirdForm" name="form1" action="" method="post">
<select id="klantWidth">
<?php
$queryKlant = "SELECT naam FROM klant";
$res=mysqli_query($mysqli,$queryKlant);

while($row=mysqli_fetch_array($res))
{
?>
<option><?php echo $row[0]; ?></option>
<?php
}
?>
</select>
</form>

Answer 1

You can select more than one column from a table in the same select, and as both these columns live in the same table it makes producing this result much simpler.

<form id="thirdForm" name="form1" action="" method="post">
    <select id="klantWidth">

<?php
    $sql = "SELECT naam, klant_code FROM klant";

    $result = mysqli_query($mysqli,$sql);

    while($row=mysqli_fetch_array($result)){
?>

        <option><?php echo $row[0]. ", ". $row[1];?></option>
<?php
    }
?>  
    </select>
</form>

You probably want to do this with your <option> tag as well rather than put the name and code in the visible portion

    <option value="<?php echo $row[1];?>"><?php echo $row[0];?> </option>

And if you use mysqli_fetch_assoc() you can use the columns names so you know what you are putting where

while($row=mysqli_fetch_assoc($result){

    <option value="<?php echo $row['klant_code'];?>"><?php echo $row['naam'];?> </option>