Not able to insert stuff from an input field into a DB.

How do I insert the email_alert from the input into the DB? I tried

$email_alert = $_POST['email_alert'];

$job_alert = $job_name;
$area_alert = $city_name;

$conn = mysqli_connect($host, $user, $pass,$dbname);  
if(!$conn){  
  die('Could not connect: '.mysqli_connect_error());  
}  
echo 'Connected successfully<br/>';  
$alerts = "INSERT INTO jobalerts (email_alert, job_alert, area_alert) VALUES ('$email_alert', '$job_alert', '$area_alert')";  
if(mysqli_query($conn, $alerts)){  
 echo "Record inserted successfully";  
}else{  
echo "Could not insert record: ". mysqli_error($conn);  
}  
mysqli_close($conn);  

Code (markup):

<form method="post" name="job_alerts" action="/index.php">  
  Email Address <span class="error">* </span>: <br><input type="text" name="email_alert" value=""> <input type="submit" name="submit" value="Submit">  
</form>

Code (markup):

 

I got it working. That action="/index.php" should be the URL whether the submit form is on.

 

I've got another question. How can I avoid inserting duplicate entries?

I know in my DB I can set the email_alert as Primary, but that would prevent from entering the same email again. I want the same email to be entered as long as job_alert or area_alert are different from what has been entered.

$alerts = "INSERT INTO jobalerts (email_alert, job_alert, area_alert) VALUES ('$email_alert', '$job_alert', '$area_alert')";

Code (markup):

I tried INSERT IGNORE, but that didn't do anything.

 

Using SELECT? How do you count and mach records? I see a few solutions using SELECT and then inserting based on that but as many things as I tried, it didn't work for me. It was either still inserting the same records or stopped inserting anything.

 

What if statements do I use to make the insert do what it's supposed to do?

 

http://lmgtfy.com/?q=mysql+insert+if+not+exists+in+another+table (no, not a spam link ). Srsly, have you even tried to search for this?

 

Do a SELECT _first_ - and if you get a return of 1 (which means the entry exist) don't insert - if you get 0, then insert.
So, something like this:

SELECT COUNT(*) FROM jobalerts WHERE email_alert = '$email_alert' AND job_name = '$job_name' AND city_name = '$city_name';
//although you should use prepared statements, not inserting blind variables into a query

PHP:

Then use the result of that query to decide whether or not to insert a new entry

 

I got it working.

    $sql2 = "SELECT email_alert, job_alert, area_alert FROM jobalerts WHERE email_alert='$email_alert' AND job_alert='$job_name' AND area_alert='$city_name'";
    $result = $link->query($sql2);
    if ($result->num_rows > 0) {
}else{
$sql = "INSERT INTO jobalerts (email_alert, job_alert, area_alert) VALUES ('$email_alert', '$job_name', '$city_name')";

}

Code (markup):

 

Pop I am sanitizing them with $email_alert = mysqli_real_escape_string($link, $_REQUEST['email_alert']); etc.

 

Fair enough - it's still an outdated method, but it works, mostly I would still suggest using prepared statements, but your way works too.