Populate the results of my query properly in mysqli

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PHP

<?php
        $con = mysqli_connect('server','username','password','DBName');
        // Check connection
        if (mysqli_connect_errno())
           {
           echo "<option>Failed to connect to the DB</option>" . mysqli_connect_error();
           }
         mysqli_query($con,"SELECT Category FROM Category");
         
         // HERE IS WHERE I NEED TO STORE THE VALUES & CREATE THE WHILE LOOP 
         // PRODUCING OPTION TAGS & VALUES BASED UPON THE RESULTS OF THE QUERY


         mysqli_close($con);
?>

Can someone help me complete this. My database.table only has one column "Category". I just need to print the results of this query with a while loop that will allow me to add options to a select box in HTML

Posted 3-Aug-14 15:06pm

Member 10964144

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Peter Leow · Accepted Answer · 2014-08-03T15:46:00

Try this, watch up for typo, this came straight from my head:

XML

<?php

// create database connection
$mysqli = mysqli_connect('server','username','password','DBName');
// Check connection
if ($mysqli->connect_errno) {
    die("Failed to connect to MySQL: ".$mysqli->connect_errno);
}

$sql = "SELECT Category FROM Category";

$stmt = $mysqli->prepare($sql);

$stmt->execute();

$resultset = $stmt->get_result()

while ($row = $resultset->fetch_assoc()) {
    $category[$row['category']] = $row['category'];
}

$stmt->close();

$resultset->close();

$mysqli->close();
?>

<select name="category" id="category">

<?php // populate a combo box with $category
        foreach( $category as $key => $value){
?>
<option value="<?php echo $key ?>"><?php echo $value ?></option>
<?php
        }
?>
</select>

Member 10964144 · Accepted Answer · 2014-08-04T05:35:00

PHP

$con = mysqli_connect(,,,,);
// Check connection
if (mysqli_connect_errno())
   {
   echo "<option>Failed to connect to MySQLi</option>";
   }
 $result = mysqli_query($con,"SELECT * FROM Category");
 while($row = mysqli_fetch_array($result)) {
  echo "<option value="".$row[" category="]."">".$row['Category']."</option>";
  }
 // Free result set
 mysqli_free_result($result);
 mysqli_close($con);
?>

Using what you provided I was able to come up with this, which worked flawlessly. Thanks.

Populate the results of my query properly in mysqli

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