CODEWITHSUNDEEP
phphtmlmysqlsqlselect
Seep Sooo
Seep Sooo

Reputation: 145

displaying values from sql in html select option not displaying values

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

Upvotes: 2

Views: 1778

Answers (3)

Arti Singh
Arti Singh

Reputation: 946

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>

Upvotes: 1

TsaiKoga
TsaiKoga

Reputation: 13404

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>

Upvotes: 2

Amit Sharma
Amit Sharma

Reputation: 1795

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

Upvotes: 1

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