This is my html form where i fetch a dropdown value from a table. i want to insert this dropdown value to a another table.
JavaScript
x
from_rate:<input type="text" name="from_rate" placeholder="$">
to_rate:<input type="text" name="to_rate" placeholder="$">
<label for="receive">From Gateway:</label>
<?php
$option="";
$query="SELECT * FROM settings WHERE TRIM(usd_gateway)> '' ";
$result=mysqli_query($con,$query);
?>
<select name='from_gateway' class='form-control col-8'>
<?php
while($row = mysqli_fetch_assoc($result)) {
$currency=$row['usd_gateway'];
echo "<option value='$currency'>$currency</option>";
}
?>
<input type="submit" value="Submit" name="submit">
my php code is following
JavaScript
if (isset($_POST['submit'])){
if (empty($_POST['from_rate']) or empty ($_POST['to_rate'])){
$err= "All field required";
}
else {
$from_rate=$_POST['from_rate'];
$to_rate=$_POST['to_rate'];
$from_gateway=$_POST['from_gateway'];
$to_gateway=$_POST['to_gateway'];
$status='1';
}
if (empty($err)){
$sql="INSERT INTO gateway (from_rate, to_rate, from_gateway, status) VALUES ('$from_rate','$to_rate','$from_gateway','$to_gateway','$status')";
$result=mysqli_query($con,$sql);
echo "Gateway direction setup completed";
}
}
everything works perfectly when i removed the dropdown from the form.
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Answer
I have just solved the problem by using foreach function. this may be helpful for someone who visiting this page.