CODEWITHSUNDEEP
phphtmlmysql
O San
O San

Reputation: 83

How to get the dropdown list value and insert into MySQL server?

Statement

I would like to get the value from the dropdown in main.php file into MySQL server using PHP file.

Condition

If the user didn't select any choice or a select default choice <option value="0">(Select Animal)</option> where option value = 0.

Then if the option value = 0, the value from dropdown should not be inserted into the MySQL server.

So, what should I need to add into add.php?

main.php

<html>
<?php
include 'connection.php';
require_once 'add.php';
?>
<body>
<form method="post" action="add.php">
     <select id="animal" name="animal">                      
            <option value="0">(Select Animal)</option> <!--Non select option-->
            <option value="Cat">Cat</option>
            <option value="Dog">Dog</option>
            <option value="Cow">Cow</option>
     </select>
     <button type="submit" name="Save">Save</button > <!--Send the selected value to add.php-->
</form>
</body>
</html>

Another PHP file (add.php)

<?php

include 'connection.php';
if(isset($_POST['Save']))
      {
              $animal = mysqli_real_escape_string($conn,$_POST['animal']; );            
              $conn->query("INSERT INTO animal (name) VALUES(?)");              
              header("location: main.php");
      }
?>

Upvotes: 0

Views: 351

Answers (3)

O San
O San

Reputation: 83

From @sony's answer, I change if($_POST['animal']!= 0) to if($_POST['animal']!= '0') make it works.

Change

  • 2 Single quotes are added.
<?php

include 'connection.php';
if(isset($_POST['Save']))
{
 if($_POST['animal']!= '0') //2 Single quotes are added.
{
              $animal = mysqli_real_escape_string($conn,$_POST['animal']);            
              $sql="INSERT INTO animal (name) VALUES(?) ";              
              header("location: main.php");
    }else{ 
      header("location: main.php");
    }
      }
?>

Upvotes: 0

Adam
Adam

Reputation: 1304

Make sure you know the array key exists before checking if it is set or you may get an error/warning.

Then ensure that the animal value is greater than 0... and include it in your query.

<?php
    include 'connection.php';
    if(array_key_exists('Save', $_POST)) {
      if(array_key_exists('animal', $_POST) && $_POST['animal'] > 0) {
        $animal = mysqli_real_escape_string($conn, $_POST['animal']);

        //Prepare statement
        $conn->prepare("INSERT INTO animal (name) VALUES (?)");

        //Include animal value in string query
        $conn->bind_param("s", $animal);

        //Run query
        $conn->query($sql);    

        header("location: main.php");
      } else {
        //Fail gracefully
        header("location: error.php"); //Or whereever you want to redirect to.
      }  
    }
    ?>

Bare in mind, you are not actually running the query here.

You're just creating it as a string, nor are you adding the value of animal to it.

Also, remove that semicolon after $_POST['animal'] when doing mysqli_real_escape_string().

EDIT: I've updated mine with your update.

EDIT 2: I have also added what you need to correctly prepare the statement.

Upvotes: 1

user9925311
user9925311

Reputation:

you can use Javascript validation for the dropdown or Keep a condition for $_POST of animal like below:

<?php

include 'connection.php';
if(isset($_POST['Save']))
      {
 if($_POST['animal']!=0){
              $animal = mysqli_real_escape_string($conn,$_POST['animal']);            
              $sql="INSERT INTO animal (name) VALUES(?) ";              
              header("location: main.php");
    }else{ 
      header("location: main.php");
    }
      }
?>

Upvotes: 0

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