I'm setting up a small signup form that should be inserting the data to an existing table, return the unique ID and insert that value to a secondary table.
Currently, I presume that the the first query is not running because the unique ID returned is 0.
I've gone over this small portion of code, that works on all my other developments and can't seem to figure out where it's blocking.
mysqli_query ($db_conx, "INSERT INTO Users
(first_name, last_name, dob, email, user_type, company, start_date, end_date, create_timestamp, last_modify_timestamp, create_matricule, pwd_hash)
VALUES
('$first_name','$last_name','$dob','$email','$user_type','$company','$start_date','$end_date','$create_timestamp','$create_timestamp','$create_matricule','$password_hash')");
//Now select this user to retreive the matricule
$matricule = mysqli_insert_id($db_conx);
$sql2 ="INSERT INTO Permissions (matricule) VALUES ('$matricule')";
$query2 = mysqli_query($db_conx, $sql2);
if(!$query){
echo json_encode(['message'=>"Error : ". mysqli_error($db_conx), 'code'=>500]);
}
echo json_encode(['message'=>"Matricule ".$matricule." created", 'code'=>200]);
}
The console returns the following :
Object { message: "Matricule 0 created", code: 200 }
javascript.js:125:13
success javascript.js:126:13
ReferenceError: matricule is not defined[Learn More]
So therefore the creation to the second table works just fine but not the first.
My database is configured as follows:
CREATE TABLE `Users` (
`matricule` int(6) NOT NULL,
`first_name` varchar(35) NOT NULL,
`last_name` varchar(35) NOT NULL,
`dob` date NOT NULL,
`email` varchar(50) NOT NULL,
`user_type` varchar(3) NOT NULL,
`company` varchar(30) NOT NULL,
`start_date` date NOT NULL,
`end_date` date NOT NULL,
`pwd_hash` varchar(500) NOT NULL,
`pwd_reset_hash` varchar(100) NOT NULL,
`otp_token` varchar(6) NOT NULL,
`create_timestamp` datetime NOT NULL,
`create_matricule` varchar(7) NOT NULL,
`manager_matricule` varchar(7) NOT NULL,
`sq_1` varchar(2) NOT NULL,
`sa_1` varchar(50) NOT NULL,
`sq_2` varchar(2) NOT NULL,
`sa_2` varchar(50) NOT NULL,
`last_modify_timestamp` datetime NOT NULL,
`last_modify_matricule` varchar(7) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
Thanks for any input!!
I think you mispelled some variable names.
You are fetching $query2 :
$query2 = mysqli_query($db_conx, $sql2);
then checks $query :
if(!$query){
echo json_encode(['message'=>"Error : ". mysqli_error($db_conx), 'code'=>500]);
}
There is most probably an issued in the structure of you database or the JavaScript you haven't shared.
First, you must make the quotation marks as follows.
$sql="INSERT INTO Users (first_name) VALUES ('".$first_name."')";
Or you can use prepared statements
Related
I have this code and for some reason it won't get inserted into my database. It's basically taking an array, turning it into a string and then submit the values.
(If you need me to edit to show my whole code, I will do so)
Code I am having issues with down below
$array = array($RaceNumber,$Track,$Num,$HorseName,$Odds,$Color,$Jockey,$Trainer,$PostTime,$Course,$RaceDistance,$Win,$Place,$Show);
for ($a=0; $a<$Num; $a++) {
$dataArray=array($RaceNumber[$a],$Track[$a],$Num[$a],$HorseName[$a],$Odds[$a],$Color[$a],$Jockey[$a],$Trainer[$a],$PostTime[$a],$Course[$a],$RaceDistance[$a],$Win[$a],$Place[$a],$Show[$a]);
$dataArray--;
for ($j=0; $j<$Num; $j++) {
$RaceNumber=$dataArray[0];
$Track=$dataArray[1];
$Num=$dataArray[2];
$HorseName=$dataArray[3];
$Odds=$dataArray[4];
$Color=$dataArray[5];
$Jockey=$dataArray[6];
$Trainer=$dataArray[7];
$PostTime=$dataArray[8];
$Course=$dataArray[9];
$RaceDistance=$dataArray[10];
$Win=$dataArray[11];
$Place=$dataArray[12];
$Show=$dataArray[13];
$sql="INSERT INTO `$Date` (RaceNumber,Track,HorseNum,HorseName,Odds,Color,JockeyName,TrainerName,PostTime,Course,RaceDistance,Win,Place,Show) VALUES ('$RaceNumber','$Track','$Num','$HorseName','$Odds','$Color','$Jockey','$Trainer','$PostTime','$Course','$RaceDistance','$Win','$Place','$Show')";
echo $sql;
mysqli_query($query2,$sql);
}
}
when I echo my $sql I get
INSERT INTO 2018-09-20 (RaceNumber,Track,HorseNum,HorseName,Odds,Color,JockeyName,TrainerName,PostTime,Course,RaceDistance,Win,Place,Show) VALUES ('1','FingerLakes','1','','','Red','','','','Dirt','','none','none','none')
But when I do my query, it isn't inserting into database.
Part of my code where I create the datatable
<?php
if(isset($_POST['submit'])) {
$Date = $_POST['date'];
$sql = "CREATE TABLE IF NOT EXISTS `$Date` (
`Id` int NOT NULL AUTO_INCREMENT PRIMARY KEY,
`RaceNumber` varchar(255) NOT NULL,
`Track` varchar(255) NOT NULL,
`HorseNum` varchar(255) NOT NULL,
`HorseName` varchar(255) NOT NULL,
`Odds` varchar(255) NOT NULL,
`Color` varchar(255) NOT NULL,
`JockeyName` varchar(255) NOT NULL,
`TrainerName` varchar(255) NOT NULL,
`PostTime` varchar(255) NOT NULL,
`Course` varchar(255) NOT NULL,
`RaceDistance` varchar(255) NOT NULL,
`Win` varchar(255) NOT NULL,
`Place` varchar(255) NOT NULL,
`Show` varchar(255) NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=utf8"
;
$query2 = mysqli_connect('localhost','root','','Races');
$z= mysqli_query($query2, $sql) or die("Table already exist.. please try again");
echo "Your Table ".$Date." is successfully created <br/>";
$RaceNum = $_POST['RaceNum'];
$i=1;
I am receiving in my error log of
2018-09-20 16:00:59 9444 [ERROR] Incorrect definition of table mysql.column_stats: expected column 'max_value' at position 4 to have type varbinary(255), found type varchar(255).
You are using a column named Show that's a reserved keyword in mysql, add backquotes to it and the insert query should work.
It's worth noting that you shouldn't name your table with only digits and hyphens.
For reference here is the complete list of the reserved keywords:
https://dev.mysql.com/doc/refman/8.0/en/keywords.html
I have a database table with two columns that have been set as boolean. Whenever I store something in the table everything works correctly except the boolean columns are always false. The columns are featured post and published.
I am using php 7.1.1 and MariaDB 10.1.21 on my xampp local installation.
Output of show table:
CREATE TABLE `posts` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`title` varchar(255) NOT NULL,
`content` text NOT NULL,
`seo_description` varchar(255) NOT NULL,
`featured_post` tinyint(1) NOT NULL DEFAULT '0',
`published` tinyint(1) NOT NULL DEFAULT '0',
`seo_title` varchar(255) NOT NULL,
`post_type` enum('blog','product') NOT NULL,
`featured_image_id` int(11) NOT NULL,
`category_id` int(11) NOT NULL,
PRIMARY KEY (`id`),
UNIQUE KEY `title` (`title`),
UNIQUE KEY `seo_title` (`seo_title`)
) ENGINE=InnoDB AUTO_INCREMENT=10 DEFAULT CHARSET=latin1 ROW_FORMAT=COMPACT
Php code defining onr of the variables, the other is exactly the same:
if(isset($_POST["published"])){
$published = sanitizeInput($_POST["published"]);
if($published){ //test to see if correct values where being sent and they were
echo json_encode("boolean true");
} else {
echo json_encode("boolean false");
}
} else {
$published = false;
}
$query = "insert into posts "
. "(title, content, seo_description, featured_post, published, seo_title, post_type, category_id) "
. "values "
. "(:title, :content, :seo_description, :featured_post, :published, :seo_title, :post_type, :category_id)";
$stmt = $pdo->prepare($query);
$stmt->bindValue("featured_post", $featuredPost, PDO::PARAM_BOOL);
$stmt->bindValue("published", $published, PDO::PARAM_BOOL);
$stmt->bindValue("title", $title);
$stmt->bindValue("content", $content);
$stmt->bindValue("seo_description", $seoDescription);
$stmt->bindValue("seo_title", $seoTitle);
$stmt->bindValue("post_type", $postType);
$stmt->bindValue("category_id", $categoryId);
$stmt->execute();
Any help would be much appreciated.
Thanks in advance
I restarted the MySQL service and I attempted to use my PHP programs delete function to delete an existing row but I'm finding although the delete queries were counted the row was not deleted. I tried applying on delete cascade to the foreign key of the child table but that did not seem to have an effect. I'm wondering why the delete would be doing nothing.
CREATE TABLE `customers` (
`idcustomers` int(11) NOT NULL AUTO_INCREMENT,
`firstname` varchar(45) DEFAULT NULL,
`lastname` varchar(45) DEFAULT NULL,
`address1` varchar(45) DEFAULT NULL,
`address2` varchar(45) DEFAULT NULL,
`city` varchar(45) DEFAULT NULL,
`state` varchar(45) DEFAULT NULL,
`zip` varchar(45) DEFAULT NULL,
`phone` varchar(45) DEFAULT NULL,
`email` varchar(45) DEFAULT NULL,
`cell` varchar(45) DEFAULT NULL,
PRIMARY KEY (`idcustomers`),
UNIQUE KEY `idcustomers_UNIQUE` (`idcustomers`)
) ENGINE=InnoDB AUTO_INCREMENT=54 DEFAULT CHARSET=latin1
CREATE TABLE `events` (
`idevents` int(11) NOT NULL AUTO_INCREMENT,
`title` varchar(250) DEFAULT NULL,
`start` datetime DEFAULT NULL,
`end` datetime DEFAULT NULL,
`allday` varchar(50) DEFAULT NULL,
`url` varchar(1000) DEFAULT NULL,
`customerid` int(11) NOT NULL,
`memo` longtext,
`dispatchstatus` varchar(45) DEFAULT NULL,
PRIMARY KEY (`idevents`),
KEY `FK_events` (`customerid`),
CONSTRAINT `FK_events` FOREIGN KEY (`customerid`) REFERENCES `customers` (`idcustomers`) ON DELETE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=18 DEFAULT CHARSET=latin1
Com_delete 2
The PHP looks like this:
<?php
session_start();
date_default_timezone_set("America/Los_Angeles");
if($_SESSION['loggedin'] != TRUE)
{
header("Location: index.php");
}
require_once('../php.securelogin/include.securelogin.php');
$mysqli = new mysqli($ad_host, $ad_user, $ad_password, "samedaycrm");
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
$customerid = $_SESSION['customer_id'];
$tSQL = "delete from events where customerid = \"$customerid\"";
$result = $mysqli->query($tSQL);
$tSQL = "delete from customers where idcustomers = \"$customerid\"";
$result = $mysqli->query($tSQL);
echo $mysqli->error;
?>
Assuming that the customerid and idcustomers columns are both numeric it should be fine. You should not need to quote the variables in those queries btw, then you wouldnt need to escape them. You may try:
$tSQL = "delete from events where customerid = $customerid";
but it should not be any different than what you used already. Of course if you are not sure of the type of the column you can use:
$tSQL = "delete from events where customerid = '".$customerid."'";
or you can get away with:
$tSQL = "delete from events where customerid = '$customerid'";
but I have always hated that for some reason.
if all of that fails troubleshoot by spitting out the $customerid (or even the whole $tSQL) variable and then trying the query manually in phpmyadmin or toad or whatever db client you use, and see what it tells you. If it just says 0 rows affected, then run it like a select instead. Tailor to fit.
This is my table:
CREATE TABLE `users` (
`id` int(10) unsigned NOT NULL auto_increment,
`name` varchar(20) NOT NULL default '',
`pass` varchar(32) NOT NULL default '',
`lang` varchar(2) default NULL,
`locale` varchar(2) default NULL,
`pic` varchar(255) default NULL,
`sex` char(1) default NULL,
`birthday` date default NULL,
`mail` varchar(64) default NULL,
`created` timestamp NOT NULL default CURRENT_TIMESTAMP,
PRIMARY KEY (`id`),
UNIQUE KEY `mail` (`mail`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=27 ;
And this is my query:
$query = "INSERT IGNORE INTO `users` (`name`, `mail`, `birthday`, `lang`, `locale`, `sex`, `pic`) VALUES ('".$name."', '".$email."', '".date_format($birthdaynew, 'Y-m-d H:i:s')."', '".substr($locale, 0, 2)."', '".substr($locale, -2, 2)."', '".$sex."', 'pic/".$uid.".jpg')";
$rows = mysql_query($query) or die("Failed: " . mysql_error());
$_SESSION['id'] = mysql_insert_id(); // I have tryed also mysql_insert_id($db_con) where $db_con is the link to db.
$_SESSION['name'] = $name;
$_SESSION['name'] contains correctly the name but $_SESSION['id'] contains 0.
Why ?
I'm going crazy!
Is there a particular reason why you are using INSERT IGNORE?
If you use INSERT IGNORE, then the row won't actually get inserted if there is a duplicate key (PRIMARY or UNIQUE), or inserting a NULL into a column with a NOT NULL constraint.
Referring to the pass column, as you have not defined anything to insert into it, and it has NOT NULL constraint.
EDIT:
Referring also to the mail column, as you have a UNIQUE constraint on it.
When registering my first user in table 'users' the id is the same value as user_id in the linking table,1, (language). However, when I register another user (id2 in users) the user_id in language is still 1. See SQL:
CREATE TABLE `language` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`user_id` int(11) NOT NULL,
`native` varchar(30) NOT NULL,
`other` varchar(30) NOT NULL,
`other_list` varchar(9) NOT NULL,
`other_read` varchar(9) NOT NULL,
`other_spokint` varchar(9) NOT NULL,
`other_spokprod` varchar(9) NOT NULL,
`other_writ` varchar(9) NOT NULL,
PRIMARY KEY (`id`),
KEY `user_id` (`user_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=9 ;
CREATE TABLE IF NOT EXISTS `users` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`md5_id` varchar(200) NOT NULL,
`full_name` tinytext CHARACTER SET latin1 COLLATE latin1_general_ci NOT NULL,
`user_name` varchar(10) NOT NULL,
`user_email` varchar(30) NOT NULL,
`user_level` tinyint(4) NOT NULL DEFAULT '1',
`pwd` varchar(220) NOT NULL,
`nationality` varchar(30) NOT NULL,
`department` varchar(20) NOT NULL,
`birthday` date NOT NULL,
`date` date NOT NULL DEFAULT '0000-00-00',
`users_ip` varchar(200) NOT NULL,
`activation_code` int(10) NOT NULL DEFAULT '0',
`banned` int(1) NOT NULL,
`ckey` varchar(200) NOT NULL,
`ctime` varchar(220) NOT NULL,
`approved` int(1) NOT NULL DEFAULT '1',
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;
This is my PHP code:
if(empty($_SESSION['$user_id'])) { // user not logged in; redirect to somewhere else }
if (!empty($_POST['doLanguage']) && $_POST['doLanguage'] == 'Submit')
{
$result = mysql_query("SELECT `id` FROM users WHERE `banned` = '0'") or
die (mysql_error());
list($id) = mysql_fetch_row($result);
session_start();
$_SESSION['user_id']= $id;
sql_insert = "INSERT into `language`
(`user_id`,`native`,`other`,`other_list`,`other_read`, `other_spokint`
,`other_spokprod`,`other_writ` )
VALUES
('$id','$native','$other','$other_list','$other_read','$other_spokint',
'$other_spokprod','$other_writ') ";
mysql_query($sql_insert,$link) or die("Insertion Failed:" . mysql_error());
}
header("Location: myaccount.php?id=' . $_SESSION[user_id] .'");
exit();
Would appreciate any help!
I think this is because your initial select query is selecting all users that are not banned
So need to change the query to filter by the new user:
"SELECT `id` FROM users WHERE `banned` = '0' and id=" . $_SESSION['$user_id'];
The reason why the user_id field kept getting populated as 1 was because mysql_fetch_row was getting the first record which was user id 1.
So if you filter it by the new user, mysql_fetch_row should get the id of the new user.
EDIT
I'm just looking at the code again, and it looks like you do not store the user id in php after you insert it, so your user_id for $_SESSION is null.
So my above example will not work. Instead of the above query, use the following function to get the id of your newly created user. You call this function after you run your insert query. That function will get the new of your newly created user.
mysql_insert_id()
http://php.net/manual/en/function.mysql-insert-id.php
EDIT 2
Ok, since mysql_insert_id() didn't work for you, maybe you can try the following. I just changed your select query to order by id desc.
$result = mysql_query("SELECT `id` FROM users WHERE `banned` = '0' order by id desc")
Just a note, this is probably not the best solution. But on a low traffic website it should be fine. To make the query a bit more accurate, you'd want to search for the user id based on a unique field like a username or email. This would make sure you get the correct id back.
EDIT 3
Here is an example of checking for the user's email. This is just the mysql query, you'll have to adjust this for php. Sorry I'm in class right now.
"SELECT `id` FROM users WHERE `banned` = '0' and email='user#email.com' limit 1