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.
Related
So I'm running a PDO update working, and for some reason it won't update the table...
$business_id = 9874128;
$hidden = 1;
$query = "UPDATE business_property_overrides SET hidden=? WHERE business_id=?";
try {
$stmt = $pdo->prepare($query);
$stmt->execute(array($business_id, $hidden));
}
For some reason this won't update, even though I get no errors. The existing tables schema looks like this, and the data is:
There is an existing data set with business_id = 9874128 and hidden set to 0, but it won't update when I run the above code.
CREATE TABLE `business_property_overrides` (
`business_id` int(11) NOT NULL,
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(512) NOT NULL,
`apt_type` varchar(25) DEFAULT NULL,
`apt_num` varchar(9) DEFAULT NULL,
`street_address` varchar(255) DEFAULT NULL,
`city` varchar(255) DEFAULT NULL,
`state` varchar(255) DEFAULT NULL,
`zip` varchar(25) DEFAULT NULL,
`phone` varchar(11) DEFAULT NULL,
`url` varchar(512) DEFAULT NULL,
`hours` varchar(100) DEFAULT NULL,
`openhours` varchar(100) DEFAULT NULL,
`location` point DEFAULT NULL,
`yelp` varchar(512) DEFAULT '0',
`twitter` varchar(512) DEFAULT '0',
`hidden` tinyint(1) DEFAULT '0',
`merged` int(11) DEFAULT NULL,
`closed` tinyint(1) DEFAULT '0',
PRIMARY KEY (`id`),
UNIQUE KEY `business_id` (`business_id`),
UNIQUE KEY `id` (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=9874134 DEFAULT CHARSET=utf8;
The hidden is TINYINT 1 characters long, you are assigning it business_id which is 7 characters long, that is the error.
Change
$stmt->execute(array($business_id, $hidden));
To:
$stmt->execute(array($hidden,$business_id))
As I've already commented over here, or you can simply use the placeholders of taking no care about the occurence like as
$query = "UPDATE business_property_overrides SET hidden = :hidden WHERE business_id = :business_id";
try {
$stmt = $pdo->prepare($query);
$stmt->execute(array(":business_id" => $business_id, ":hidden" => $hidden));
}
Here's my info table:
CREATE TABLE `info` (
`id_info` int(10) NOT NULL auto_increment,
`judul_info` varchar(50) collate latin1_general_ci NOT NULL,
`konten` varchar(255) collate latin1_general_ci NOT NULL,
`diubah_oleh` varchar(20) collate latin1_general_ci NOT NULL,
`id_kategori` int(10) NOT NULL,
`tgl_buat` timestamp NOT NULL default '0000-00-00 00:00:00',
`tgl_ubah` timestamp NOT NULL default CURRENT_TIMESTAMP on update CURRENT_TIMESTAMP,
`dibuat_oleh` varchar(20) collate latin1_general_ci NOT NULL,
`id` int(10) NOT NULL,
PRIMARY KEY (`id_info`),
KEY `id_kategori` (`id_kategori`),
KEY `id` (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci AUTO_INCREMENT=62 ;
Here's my upload table
CREATE TABLE `upload` (
`id` int(10) unsigned NOT NULL auto_increment,
`deskripsi` text,
`filetype` varchar(200) default NULL,
`filedata` longblob,
`filename` varchar(200) default NULL,
`filesize` bigint(20) default NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=34 ;
I'm using this query :
$sql1="INSERT INTO info VALUES('','$judul', '$konten','$diubah_oleh','$kategori',now(),'$tgl_ubah','$dibuat_oleh','')";
$sql2="insert into upload values ('','$keterangan','$tipe','$filedata','$nama_file',$ukuran)";
$sql3="UPDATE info SET id=last_insert_id()";
$result=mysql_query($sql1);
$result=mysql_query($sql2);
$result=mysql_query($sql3);
I want info.id has the same value as upload.id but with this query all of the value i get in info.id is the same as value i last inserted in upload.id.
CREATE TABLE `upload` (
`id` int(10) unsigned NOT NULL,
`deskripsi` text,
`filetype` varchar(200) default NULL,
`filedata` longblob,
`filename` varchar(200) default NULL,
`filesize` bigint(20) default NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=34 ;
$sql1="INSERT INTO info VALUES('','$judul', '$konten','$diubah_oleh','$kategori',now(),'$tgl_ubah','$dibuat_oleh','')";
$result=mysql_query($sql1);
$lastId = mysql_insert_id();
$sql2="insert into upload values ('$lastId','$keterangan','$tipe','$filedata','$nama_file',$ukuran)";
$result=mysql_query($sql2);
Your last update statement below is updating all the rows in your info tables with the same id because there is no where statement.
Since you need the upload table id information inside the info table.
Follow these steps:
Run the $sql2 first.
Then run the $sql1 inserting the last_insert_id() in info.id.
This way you don't need to use update statement as well.
You can do this by using mysql_insert_id(). This function returns the AUTO_INCREMENT ID generated from the previous INSERT operation. Your code should look like this
$sql1="INSERT INTO info VALUES('','$judul', '$konten','$diubah_oleh','$kategori',now(),'$tgl_ubah','$dibuat_oleh','')";
$result=mysql_query($sql1);
$lastinsertedid= mysql_insert_id();
$sql2="insert into upload values ('$lastinsertedid','$keterangan','$tipe','$filedata','$nama_file',$ukuran)";
$result=mysql_query($sql2);
Hope this helps you
There is an alternate php function to that mysql_insert_id() . You can use this to generate the ID inserted in the last executed query.
can you try to do this:
$sql1="INSERT INTO info VALUES('','$judul', '$konten','$diubah_oleh','$kategori',now(),'$tgl_ubah','$dibuat_oleh','')";
$sql2="insert into upload values ('','$keterangan','$tipe','$filedata','$nama_file',$ukuran)";
$last_id = last_insert_id();
$sql3="UPDATE info SET id=".$last_id;
$result=mysql_query($sql1);
$result=mysql_query($sql2);
$result=mysql_query($sql3);
Info table query:
CREATE TABLE `info` (
`id_info` int(10) NOT NULL auto_increment,
`judul_info` varchar(50) collate latin1_general_ci NOT NULL,
`konten` varchar(255) collate latin1_general_ci NOT NULL,
`diubah_oleh` varchar(20) collate latin1_general_ci NOT NULL,
`id_kategori` int(10) NOT NULL,
`tgl_buat` timestamp NOT NULL default '0000-00-00 00:00:00',
`tgl_ubah` timestamp NOT NULL default CURRENT_TIMESTAMP on update CURRENT_TIMESTAMP,
`dibuat_oleh` varchar(20) collate latin1_general_ci NOT NULL,
`id` int(10) NOT NULL,
PRIMARY KEY (`id_info`),
KEY `id_kategori` (`id_kategori`),
KEY `id` (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci AUTO_INCREMENT=62 ;
Upload table query :
CREATE TABLE `upload` (
`id` int(10) unsigned NOT NULL auto_increment,
`deskripsi` text,
`filetype` varchar(200) default NULL,
`filedata` longblob,
`filename` varchar(200) default NULL,
`filesize` bigint(20) default NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=34 ;
i want to insert the upload.id same as info.id when i post a new info. please help me.
here's my source code(sorry for the query i can't upload an image cause of my reputation still low):
<?php
include "config.php";
$id_info=$_POST['id_info'];
$judul=$_POST['judul_info'];
$konten=$_POST['konten'];
$tgl_ubah=$_POST['tgl_ubah'];
$diubah_oleh=$_POST['diubah_oleh'];
$id_kategori=$_POST['id_kategori'];
// update data in mysql database
$sql="UPDATE info SET judul_info='$judul', konten='$konten', diubah_oleh='$diubah_oleh', id_kategori='$id_kategori' WHERE id_info='$id_info'";
$result=mysql_query($sql);
// if successfully updated.
if($result){
echo "Telah diupdate";
echo "<BR>";
echo "<a href='list_info.php'>View result</a>";
}
else {
echo "ERROR";
}
?>
You need to use the last_insert_id() or mysql_insert_id() after your insert into the info table to get the new id inserted and then then use that in your next query while inserting record into upload table. Refer to links below on how to use them.
http://dev.mysql.com/doc/apis-php/en/apis-php-function.mysql-insert-id.html
http://php.net/manual/en/function.mysql-insert-id.php
http://dev.mysql.com/doc/refman/5.0/en/information-functions.html#function_last-insert-id
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
Here is my USER table
CREATE TABLE IF NOT EXISTS `users` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`username` varchar(100) NOT NULL,
`expiry` varchar(6) NOT NULL,
`contact_id` int(11) NOT NULL,
`email` varchar(255) NOT NULL,
`password` varchar(100) NOT NULL,
`level` int(3) NOT NULL,
`active` tinyint(4) NOT NULL DEFAULT '1',
PRIMARY KEY (`id`,`email`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;
And here is my contact_info table
CREATE TABLE IF NOT EXISTS `contact_info` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(255) NOT NULL,
`email_address` varchar(255) NOT NULL,
`company_name` varchar(255) NOT NULL,
`license_number` varchar(255) NOT NULL,
`phone` varchar(30) NOT NULL,
`fax` varchar(30) NOT NULL,
`mobile` varchar(30) NOT NULL,
`category` varchar(100) NOT NULL,
`country` varchar(20) NOT NULL,
`state` varchar(20) NOT NULL,
`city` varchar(100) NOT NULL,
`postcode` varchar(50) NOT NULL,
PRIMARY KEY (`id`,`email_address`),
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;
The system uses username to login users. I want to modify it in such a way that it uses email for login. But there is no email_address in users table.
I have added foreign key - email in user table(which is email_address in contact_info).
How should I query database?
No, no, no, no no. Seriously, no. Don't make me come over there :-)
You're breaking third normal form by storing the email address twice.
The relationship need only be a short one, that of id. Assuming you're not guaranteeing the IDs will be identical in the two tables (i.e., my users.id isn't necessarily equal to my contact_info.id), just add a ci_id to the users table to act as a foreign key to the contact_info table.
Then the query to get a user's username and email will be something like:
select u.username, ci.email
from users u, contact_info ci
where u.username = 'paxdiablo'
and u.ci_id = ci.id;