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Insert record failing! php mysqli

开发者 https://www.devze.com 2022-12-11 05:42 出处:网络
i have a mysql database... CREATE TABLE `applications` ( `id` int(11) NOT NULL auto_increment, `jobref` int(11) NOT NULL,

i have a mysql database...

CREATE TABLE `applications` (
`id` int(11) NOT NULL auto_increment,
`jobref` int(11) NOT NULL,
`userid` int(11) NOT NULL,
`q1` text NOT NULL,
`q2` text NOT NULL,
`q3` text NOT NULL,
`sub_q1` text,
`sub_q2` text,
`sub_q3` text,
`timestamp` timestamp NOT NULL default CURRENT_TIMESTAMP on update CURRENT_TIMESTAMP,
`printed` int(11) NOT NULL default '0',
PRIMARY KEY  (`id`),
KEY `jobref` (`jobref`),
KEY `applications_ibfk_2` (`userid`),
CONSTRAINT `applications_ibfk_1` FOREIGN KEY (`jobref`) REFER开发者_C百科ENCES `jobs` (`id`) ON     DELETE NO ACTION ON UPDATE NO ACTION,
CONSTRAINT `applications_ibfk_2` FOREIGN KEY (`userid`) REFERENCES `users` (`id`) ON  DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB DEFAULT CHARSET=latin1 COMMENT='InnoDB free: 9216 kB; (`jobref`) REFER `iwcjobs/jobs`(`id`) '

and this php file, using mysqli for data operations.

<?php

require_once '../includes/constants.php';

if(isset($_POST['submit'])) {
$q1 = $_POST['question1'];
$q1a = $_POST['ifNoQuestion1'];
$q2 = $_POST['question2'];
$q2a = $_POST['ifNoQuestion2'];
$q3 = $_POST['question3'];
$q3a = $_POST['ifNoQuestion3'];
$JobRef = $_POST['jobref'];
$UserRef = $_POST['id'];

$mysql = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD, DB_NAME) or die('There was a problem connecting to the database');

if($stmt = $mysql->prepare('INSERT INTO `applications` VALUES (NULL,?,?,?,?,?,?,?,?,NULL,NULL)')) {
  $stmt->bind_param('iissssss',$JobRef,$UserRef,$q1,$q2,$q3,$q1a,$q2a,$q3a);
  $stmt->execute();
  $stmt->close();

  header('location: ../myApps.php?apr=y');
  //echo ("<h2>success</h2> - $q1 . $q2 . $q3 . $q1a . $q2a . $q3a . $JobRef . $UserRef");

} else {
  echo 'error: ' . $mysql->error;
}

} else {

echo 'errorStage2: ' . $mysql->error;

}

?>

The script is grabbing the correct values from the previous form, but not inserting them into the database. Any ideas people? Thx in advance, Aaron.


the way your insert query is currently written, you are trying to set id to NULL, which is a NOT NULL field.

it'd be better to structure the insert would be like this:

INSERT INTO 'applications' (jobref,userid,etc..) VALUES (?,?etc..)

this way, you not try to change id, and just let auto_increment do it's thing.

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