mysql_query 不向数据库发送表单信息

Question

I can't seem to figure out why my form data isn't being sent to the database. I've tried multiple variations of coding and this is the only one where I could get a "result"

here's my code:

<?php
  #if the submit button has be selected...
  if(isset($_POST['submit_registration'])) {
  # assign variables to each form control to capture the values
    $first = $_POST['first_name'];
    $last = $_POST['last_name'];
    $email = $_POST['email'];
    $address1 = $_POST['address1'];
    $address2 = $_POST['address2'];
    $city = $_POST['city'];
    $state = $_POST['state'];
    $zipcode = $_POST['zipcode'];
    $phone = $_POST['phone'];
    $distance = $_POST['distance'];
  # assign null to values for use with isset function to identitfy required         fields with no value
    $nofirst = null;
    $nolast = null;
    $noemail = null;
    $noaddress1 = null;
    $noaddress2 = '';
    $nocityErr = null;
    $nostate = null;
    $nozipcode = null;
    $nophone = null;
    $nodistance = null;
  # if value of variable for required field is nothing, assign something other than null to $no variable
    if($first == "") {$nofirst = '';}
    if($last == "") {$nolast = '';           if($email == "") {$noemail = '';}
    if($address1 == "") {$noaddress1 = '';}
    if($address2 == "") {$noaddress2 ='';}
    if($city == "") {$nocity = '';}
    if($state == "") {$nostate = '';}
    if($zipcode == "") {$nozipcode = '';}
    if($phone == "") {$nophone = '';}
    else {
      $insertsql = "INSERT INTO `runner`(`fname`, `lname`, `email`, `address1`, `address2`, `city`, `state`, `postalcode`, `phone`, `distance`)
                VALUES ('$first','$last','$email','$address1','$address2','$city','$state','$zipcode','$phone','$distance')";
          echo $insertsql;
      mysql_query($lrconnect, $insertsql) or die("Insert failed ". mysql_error($lrconnect));
        echo "connected";
      $inserted = '';
    }
}
?>

Here's my the error I get:

INSERT INTO `runner`(`fname`, `lname`, `email`, `address1`, `address2`, `city`, `state`, `postalcode`, `phone`, `distance`) VALUES ('Crystal','Yang','cykher@gmail.com','55555 Avenue','','Chicago','FL','39485','5555555555','5K')

Insert failed

Answer 1

Using mysqli_query functions, you would put the connection first as you have. However, you are using mysql_query instead which puts the query first and the connection second.

http://php.net/manual/en/function.mysql-query.php

As mentioned in the comments, though, you really should not be just inserting data that users submit through $_POST directly into the database. Also mentioned in the comments, you should be using mysqli_* instead. This would have made your syntax correct.

There's not a lot of work to do to convert it to mysqli . Basically you just add an i on to the end of each place where you'd normally call mysql . You'd also switch the connection and query to have the connection come first, as you have in your code. Finally, your connection takes a 4th parameter now instead of 3, which is nice because you don't need a separate call to specify which database you want to use.

Here's an example:

<?php

$link = mysqli_connect('HOST', 'USER', 'PASS', 'DATABASE');

$q_test = "SELECT id FROM table";

$r_test = mysqli_query($link, $q_test) or trigger_error("Cannot Get ID: (".mysqli_error().")", E_USER_ERROR);

while ($row_test = mysqli_fetch_array($r_test)) {

    print "ID: ".$row_test['id'];

}

Answer 2

No need to use connection with mysql_query function.
If you are using mysqli_query function than you must have to use connection parameter in that function. so solution of your problem is.
mysql_query($insertsql) or die("Insert failed ". mysql_error($lrconnect));