displaying values from sql in html select option not displaying values

Question

i have created a html form using php and sql. The form contains a field called spaces which is a select dropdown. The values are coming from an array which i declared in php. If the values in array are already present in database it should not display. So i have done the following code:

<select class="form-control" id="space" name="space">
  <option value="--Select--">--Select--</option>
  <?php
$select=mysqli_query($con,"select * from clients where Spaces IS NOT NULL");
while($menu1=mysqli_fetch_array($select))
      {
$filled =$menu1['name'];
$valuez = array("C101","C102","C103","C104");
if ($filled != $valuez) {


      ?>
    <option value="<?php echo $valuez;?>">
      <?php echo $valuez;?>
    </option>
    <?php
      }}
      ?>
</select>

but this is not making any values display. Can anyone please tell me what is wrong in my code. thanks n advance

TsaiKoga · Accepted Answer · 2020-01-06 07:03:13Z

2

Your $filled is string, cannot equals to array.

And you can just use whereIn, don't need to take all clients out:

<select class="form-control" id="space" name="space">
  <option value="--Select--">--Select--</option>
  <?php
    $valuez = "'C101','C102','C103','C104'";
    $select = mysqli_query($con, "SELECT * FROM clients WHERE Space IN ($valuez)");
  ?>

<?php while($menu1 = mysqli_fetch_array($select) ) : ?>
    <option value="<?php echo $menu1['name'];?>">
      <?php echo $menu1['name'];?>
    </option>
<?php endwhile; ?> 
</select>

edited Jan 6, 2020 at 7:03

answered Jan 6, 2020 at 6:38

TsaiKoga

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3 Comments

Seep Sooo Over a year ago

i want to display $valuez , if values in $valuez are not present in $filled

TsaiKoga Over a year ago

@SeepSooo Yes, you don't need to use array, that will slow down your query performance

Seep Sooo Over a year ago

still nothing bro

Amit Sharma · Accepted Answer · 2020-01-06 06:52:09Z

1

you are comparing a string with the array. you should use in_array like this

 <select class="form-control" id="space" name="space">
   <option value="--Select--">--Select--</option>
 <?php
  $select=mysqli_query($con,"select * from clients where Space IS NOT NULL");
 while($menu1=mysqli_fetch_array($select))
  {
  $filled =$menu1['Space'];
 $valuez = array("C101","C102","C103","C104");
 foreach($valuez as $value){
    if($value != $filled){ 
    ?>
        <option value="<?php echo $value;?>">
          <?php echo $value;?>
        </option>
    <?php 
    }
 }
}
?>

update the code

edited Jan 6, 2020 at 6:52

answered Jan 6, 2020 at 6:25

Amit Sharma

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6 Comments

Seep Sooo Over a year ago

i have used your code, still not displaying the values

Amit Sharma Over a year ago

you are echoing array here use a single value instead . SO I guess you need to use $filled here <option value="<?php echo $filled;?>"><?php echo $filled;?> updated the code as well

Seep Sooo Over a year ago

i want to display $valuez , if values in $valuez are not present in $filled

Amit Sharma Over a year ago

okay then you need to use foreach and iterate through values array

Seep Sooo Over a year ago

can you please update the code and show, it would be great

|

Arti Singh · Accepted Answer · 2020-01-06 07:41:51Z

1

Try this and please make sure the value you want to display as the options are from the column name or spaces. I used the name($menu1['name']) column as per your code.

<select class="form-control" id="space" name="space">
  <option value="--Select--">--Select--</option>
   <?php
     $select = mysqli_query($con,"select * from clients where Spaces NOT IN ('C101','C102','C103','C104')");
     while($menu1=mysqli_fetch_array($select)) {
       $filled = $menu1['name'];
       if (!empty($filled)) {
   ?>
   <option value="<?php echo $filled;?>">
     <?php echo $filled;?>
   </option>
  <?php
    }}
  ?>
</select>

answered Jan 6, 2020 at 7:41

Arti Singh

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