douxie1692 2013-03-17 22:02
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INSERT INTO MySQLi语句第一次没有工作

I have a pretty standard MySQL insert into statement that doesn't seem to be working. I'm posting an HTML form to this page and it's translating the information and inserting into the table (pending_jobs). It worked the first time and hasn't worked since. I'm not looking to UPDATE the row but add a brand new one every time this code is submitted so I figured this would be pretty simple. Here's the code:

<?php
include('includes/connect.php');
$id=$mysqli->real_escape_string($_POST['id']);
$sql = "SELECT * FROM users WHERE id='$id'";
$result = $mysqli->query($sql);
$row = mysqli_fetch_array($result);

$user_id = $_SESSION['user_id'];

$company_name = $row['company_name'];
$date_registered = $row['date_registered'];
$job_title=$mysqli->real_escape_string($_POST['jobtitle']);
$job_description=$mysqli->real_escape_string($_POST['jobdescription']);
$salary=$mysqli->real_escape_string($_POST['salary']);
$industry_sector=$mysqli->real_escape_string($_POST['industry']);
$number_people=$mysqli->real_escape_string($_POST['num_people']);

if(isset($_POST['payment']) && $_POST['payment']=='yes'){
$payment_received = 'Yes';
}
else{
$payment_received = 'No';
}

$sql_add = "INSERT INTO pending_jobs (company_name, job_title) VALUES ('$company_name', '$job_title')";
$result_add = $mysqli->query($sql_add);

if(isset($result_add)){
  session_start();
  $id = $row['id'];
   $_SESSION['user_id'] = $user_id;
  header('location:profile.php?id='.$id);
} else {
echo "<script>alert('".$job_title."');</script>";
}
?>

It is redirecting to the profile.php page so I know that $result_add is actually going through but when I check the table in the database there is no new row. Any ideas on why it wouldn't be creating a new row?

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3条回答 默认 最新

  • douhandie6615 2013-03-17 22:20
    关注

    Change your query to

    $result_add = $mysqli->query($sql_add) or die($mysqli->error);
    

    and you will see the problem at once. Also

     if (isset($result_add)) {
    

    will always be true, because $result_add will be either true or false for your insert.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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