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));
}
Related
Here is the structure of the database tables in quesiton.
Officials:
CREATE TABLE IF NOT EXISTS `officials` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`last` varchar(250) NOT NULL DEFAULT '',
`first` varchar(250) NOT NULL DEFAULT '',
`middle` varchar(250) NOT NULL DEFAULT '',
`address` varchar(250) NOT NULL DEFAULT '',
`city` varchar(250) NOT NULL DEFAULT '',
`state` varchar(250) NOT NULL DEFAULT '',
`zip` varchar(250) NOT NULL DEFAULT '',
`homeph` varchar(250) NOT NULL DEFAULT '',
`workph` varchar(250) NOT NULL DEFAULT '',
`cellph` varchar(250) NOT NULL DEFAULT '',
`main_phone` varchar(10) DEFAULT NULL,
`email` varchar(250) NOT NULL DEFAULT '',
`notes` text NOT NULL,
`blockreg` varchar(10) NOT NULL,
`inactive` varchar(10) NOT NULL,
`gender` varchar(100) NOT NULL,
`minority` varchar(100) NOT NULL,
`affiliate` tinyint(4) NOT NULL DEFAULT '0' COMMENT 'Is affiliate official? 0 = no. 1 = yes.',
PRIMARY KEY (`id`),
UNIQUE KEY `email` (`email`)
) ENGINE=InnoDB AUTO_INCREMENT=14506 DEFAULT CHARSET=latin1;
Transactions (edited from original post to show foreign key constraints):
CREATE TABLE IF NOT EXISTS `transactions` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`offid` int(11) NOT NULL,
`txn_id` varchar(64) NOT NULL,
`txn_time` varchar(24) NOT NULL,
`approval_code` varchar(24) DEFAULT NULL,
`amount` varchar(8) NOT NULL,
`result_message` varchar(100) NOT NULL,
`ch_phone` varchar(10) NOT NULL COMMENT,
`ch_card_type` varchar(4) NOT NULL,
`ch_address` varchar(50) NOT NULL,
`ch_state` varchar(15) NOT NULL,
`ch_last_name` varchar(50) NOT NULL,
`ch_email` varchar(100) NOT NULL,
`ch_zip` varchar(10) NOT NULL,
`ch_exp_date` varchar(4) NOT NULL,
`ch_city` varchar(50) NOT NULL,
`ch_first_name` varchar(30) NOT NULL,
`ch_transaction_type` varchar(15) NOT NULL,
`ch_card_number` varchar(18) NOT NULL,
PRIMARY KEY (`id`),
UNIQUE KEY `txn_id` (`txn_id`) USING BTREE,
KEY `offid` (`offid`)
) ENGINE=InnoDB AUTO_INCREMENT=72 DEFAULT CHARSET=utf8mb4;
ALTER TABLE `transactions`
ADD CONSTRAINT `transactions_ibfk_1` FOREIGN KEY (`offid`) REFERENCES `officials` (`id`);
Transaction details (edited from original post to show foreign key constraints):
CREATE TABLE IF NOT EXISTS `transaction_details` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`xact_id` varchar(64) NOT NULL,
`detail_amount` float(5,2) NOT NULL,
`sport` varchar(10) NOT NULL,
`regyr` varchar(250) NOT NULL,
PRIMARY KEY (`id`),
KEY `xact_id` (`xact_id`)
) ENGINE=InnoDB AUTO_INCREMENT=63 DEFAULT CHARSET=utf8mb4;
ALTER TABLE `transaction_details`
ADD CONSTRAINT `transaction_details_ibfk_1` FOREIGN KEY (`xact_id`) REFERENCES `transactions` (`txn_id`) ON DELETE CASCADE ON UPDATE CASCADE;
Code that executes upon receiving the data from the credit card company where rows are inserted into the table:
// this creates the main transaction
$sql = "INSERT INTO transactions (offid, txn_id, txn_time, approval_code, amount, result_message, ch_phone,
ch_card_type, ch_address, ch_state, ch_last_name, ch_email, ch_zip, ch_exp_date, ch_city, ch_first_name,
ch_transaction_type, ch_card_number) VALUES ('$offid', '$txn_id', '$txn_time', '$approval_code', '$amount', '$result_message',
'$ch_phone', '$ch_card_type', '$ch_address', '$ch_state', '$ch_last_name', '$ch_email',
'$ch_zip', '$ch_exp_date', '$ch_city', '$ch_first_name', '$ch_transaction_type', '$ch_card_number')";
$result = mysqli_query($conn, $sql);
// this creates the detail transactions
if ($ssl_result == 0) {
$detail_sports = array('fb', 'vb', 'sb', 'bb', 'wr', 'sw', 'so', 'ba', 'tr');
foreach ($detail_sports as $sport) {
$var = $sport . 'total';
// cookie is set for fbtotal, vbtotal, etc. with an amount for that line item
if (isset($_COOKIE[$var])) {
$detail_amount = $_COOKIE[$var];
$sql = "INSERT INTO transaction_details (xact_id, detail_amount, sport, regyr) VALUES ('$txn_id', '$detail_amount', '$sport', '$regyear')";
$result = mysql_query($sql);
}
}
}
The transaction table has id numbers 1, 2, 3, 5, 6, 8, 9, 11, 12, 13, 15, 16, 17, etc. These should be consecutively numbered, but they aren't. It's almost as if they were written to the transaction (parent) table and then deleted. But the code doesn't delete a record.
The transaction detail table has a many-to-one relationship with transaction and all of its records contain consecutive ids.
Any idea what's going on?
after form submission, the data is stored in database. This is the query i already have:
CREATE TABLE IF NOT EXISTS `registered_users` (
`id` int(8) NOT NULL AUTO_INCREMENT,
`first_name` varchar(255) NOT NULL,
`last_name` varchar(255) NOT NULL,
`gender` varchar(20) NOT NULL,
`date_birth` varchar(255) NOT NULL,
`address` varchar(255) NOT NULL,
`postal_code` varchar(255) NOT NULL,
`place` varchar(255) NOT NULL,
`school_name` varchar(255) NOT NULL,
`email` varchar(55) NOT NULL,
`tel` varchar(55) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;
This is how i insert the data into the DB:
Now i want to insert a new row in the DB which can have only 2 values: "active" or "inactive" for each user. Default on "inactive". That should be inserted automatically for each user after submit. How can i do that?
When you create a table in a mysql db you can specify a default value for attributes. So if you want a boolean "active" / "inactive" flag you might want to try adding an active attribute to your table. See the added last column here:
CREATE TABLE IF NOT EXISTS `registered_users` (
`id` int(8) NOT NULL AUTO_INCREMENT,
`first_name` varchar(255) NOT NULL,
`last_name` varchar(255) NOT NULL,
`gender` varchar(20) NOT NULL,
`date_birth` varchar(255) NOT NULL,
`address` varchar(255) NOT NULL,
`postal_code` varchar(255) NOT NULL,
`place` varchar(255) NOT NULL,
`school_name` varchar(255) NOT NULL,
`email` varchar(55) NOT NULL,
`tel` varchar(55) NOT NULL,
`active` BOOLEAN NOT NULL DEFAULT false,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;
Ref.: https://dev.mysql.com/doc/refman/5.7/en/create-table.html
I am working on creating table in to my database with mysql and here is my code:
mysqli_query($link, "CREATE TABLE IF NOT EXISTS `review` (
`clients_id` int(15) NOT NULL AUTO_INCREMENT,
`client_id` varchar(150) NOT NULL DEFAULT '',
`rating` tinyint(2) NOT NULL DEFAULT '0',
`proj_date` date NOT NULL DEFAULT '0000-00-00',
`proj_desc` text NOT NULL DEFAULT '',
`photoname` text NOT NULL,
`companyname` text NOT NULL,
`feedback` text NOT NULL,
`status` tinyint(1) NOT NULL DEFAULT '0',
`emailid` varchar(100) NOT NULL DEFAULT '',
`customratings` varchar(100) NOT NULL DEFAULT '',
`photo_option` varchar(100) NOT NULL DEFAULT '',
`title` varchar(100) NOT NULL DEFAULT '',
`citation` varchar(100) NOT NULL DEFAULT '',
`date_option` varchar(100) NOT NULL DEFAULT '',
`rating_option` varchar(100) NOT NULL DEFAULT '',
PRIMARY KEY (`clients_id`),
FULLTEXT KEY `feedback` (`feedback`)
) ENGINE=MyISAM AUTO_INCREMENT=1") or mysqli_error($link);
But this is not reflecting into my database ? Why were I may be going wrong ?
but I tried creating other table with the following code
mysqli_query($link, "CREATE TABLE IF NOT EXISTS `setting` (
`id` int(11) NOT NULL auto_increment,
`script_url` text NOT NULL,
`date` varchar(4) NOT NULL,
`rateing` varchar(4) NOT NULL,
`photo` varchar(4) NOT NULL,
`dateformat` varchar(4) NOT NULL,
`page_limit` int(4) NOT NULL,
`proj_desc` varchar(4) NOT NULL,
`companyname` varchar(4) NOT NULL,
`text_color` varchar(255) NOT NULL,
`citation_color` varchar(255) NOT NULL,
`bg_color` varchar(255) NOT NULL,
`border_color` varchar(255) NOT NULL,
`ratingsformat` varchar(250) NOT NULL,
`rating` varchar(250) NOT NULL,
`customratings` varchar(250) NOT NULL,
`speed` varchar(250) NOT NULL,
`pagination` varchar(250) NOT NULL,
`version` varchar(250) NOT NULL,
`global_option` varchar(250) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=0;
") or mysqli_error($link);
it is being created correctly and both the tables are in the same file
No you cannot create table using an insert statement.
There are four types of SQL statements:
DML (DATA MANIPULATION LANGUAGE)
DDL (DATA DEFINITION LANGUAGE)
DCL (DATA CONTROL LANGUAGE)
TCL (TRANSACTION CONTROL LANGUAGE)
DML is your SELECT, INSERT, UPDATE and DELETE statements...
DDL is you your CREATE, ALTER and DROP statements.
See more info about types of SQL statements
In order to insert data in your table, first you need to create it.
How to create sql table from php
No. "INSERT INTO" used to insert the data to existing table only.You should create the table with respective column before try to insert the values.
Or you can check the table exist or not before going insert."If exists" u can insert the value else just create and insert the values.
This worked for me but I don't know how?
I just removed DEFAULT '' present while creating review table and table just got created
Here is edited code
mysqli_query($link, "CREATE TABLE IF NOT EXISTS `review` (
`clients_id` int(15) NOT NULL AUTO_INCREMENT,
`client_id` varchar(150) NOT NULL,
`rating` tinyint(2) NOT NULL,
`proj_date` date NOT NULL,
`proj_desc` text NOT NULL,
`photoname` text NOT NULL,
`companyname` text NOT NULL,
`feedback` text NOT NULL,
`status` tinyint(1) NOT NULL,
`emailid` varchar(100) NOT NULL,
`customratings` varchar(100) NOT NULL,
`photo_option` varchar(100) NOT NULL,
`title` varchar(100) NOT NULL,
`citation` varchar(100) NOT NULL,
`date_option` varchar(100) NOT NULL,
`rating_option` varchar(100) NOT NULL,
PRIMARY KEY (`clients_id`),
FULLTEXT KEY `feedback` (`feedback`)
) ENGINE=MyISAM AUTO_INCREMENT=1") or mysqli_error($link);
i have a problem in my examination system. In the admin panel, in the results page displays the correct score but the firstname and the last name not. I dont know what to do.
Here is the code:
$result1=executeQuery("select s.nume_student,s.prenume_student,s.stdname,s.stdid, IFNULL((select sum(q.marks) from studentquestion as sq,question as q where q.qnid=sq.qnid and sq.testid=".$_REQUEST['testid']." and sq.stdid=st.stdid and sq.stdanswer=q.correctanswer),0) as om from studenttest as st, student as s where s.stdid=st.stdid and st.testid=".$_REQUEST['testid'].";" );
$result3=executeQuery("SELECT sq.stdid, sq.testid, COUNT(*) AS correctAnswers, SUM(q.marks) AS studentScore,(SELECT SUM(marks) FROM question WHERE testid=".$_REQUEST['testid'].") AS totalScore FROM question q, studentquestion sq WHERE sq.testid=".$_REQUEST['testid']." AND q.testid = sq.testid AND q.qnid = sq.qnid AND sq.answered = 'answered' AND q.correctanswer = sq.stdanswer GROUP BY sq.stdid");
without $result1 doesen't display anything
And here are the tables from database:
question table:
CREATE TABLE IF NOT EXISTS `question` (
`testid` bigint(20) NOT NULL DEFAULT '0',
`qnid` int(11) NOT NULL DEFAULT '0',
`question` varchar(500) DEFAULT NULL,
`optiona` varchar(100) DEFAULT NULL,
`optionb` varchar(100) DEFAULT NULL,
`optionc` varchar(100) DEFAULT NULL,
`optiond` varchar(100) DEFAULT NULL,
`correctanswer` enum('optiona','optionb','optionc','optiond') DEFAULT NULL,
`marks` float(3,2) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
studentquestion table:
CREATE TABLE IF NOT EXISTS `studentquestion` (
`stdid` bigint(20) NOT NULL DEFAULT '0',
`testid` bigint(20) NOT NULL DEFAULT '0',
`qnid` int(11) NOT NULL DEFAULT '0',
`answered` enum('answered','unanswered','review') DEFAULT NULL,
`stdanswer` enum('optiona','optionb','optionc','optiond') DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
student table:
CREATE TABLE IF NOT EXISTS `student` (
`stdid` bigint(20) NOT NULL,
`stdname` varchar(40) DEFAULT NULL,
`stdpassword` varchar(40) DEFAULT NULL,
`emailid` varchar(40) DEFAULT NULL,
`contactno` varchar(20) DEFAULT NULL,
`fname` varchar(40) DEFAULT NULL,
`lname` varchar(40) DEFAULT NULL,
`pincode` varchar(20) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
student test table:
CREATE TABLE IF NOT EXISTS `studenttest` (
`stdid` bigint(20) NOT NULL DEFAULT '0',
`testid` bigint(20) NOT NULL DEFAULT '0',
`starttime` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`endtime` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
`correctlyanswered` int(11) DEFAULT NULL,
`status` enum('over','inprogress') DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
and test table:
CREATE TABLE IF NOT EXISTS `test` (
`testid` bigint(20) NOT NULL,
`testname` varchar(30) NOT NULL,
`testdesc` varchar(100) DEFAULT NULL,
`testdate` date DEFAULT NULL,
`testtime` time DEFAULT NULL,
`subid` int(11) DEFAULT NULL,
`testfrom` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`testto` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
`duration` int(11) DEFAULT NULL,
`totalquestions` int(11) DEFAULT NULL,
`attemptedstudents` bigint(20) DEFAULT NULL,
`testcode` varchar(40) NOT NULL,
`tcid` bigint(20) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
I resolved the same problem of here:
How to get sql WHERE statement auto update from localhost link
but i'm stucked here with fname and last name.
If your second query returns the correct student ids, you can just change it to add student names:
SELECT s.lastname, s.firstname, e.*
from
students s join (
... your second query ... ) e
on s.stdid=e.stdid
you can change your join type depends on what you want.
Thanks, but i resolved it like this:
SELECT ***s.fname, s.lname***, sq.stdid, sq.testid, COUNT(*) AS correctAnswers,
SUM(q.marks) AS studentScore,(SELECT SUM(marks) FROM question
WHERE testid=1) AS totalScore
FROM question q, studentquestion sq, ***student s***
WHERE sq.testid=1 AND q.testid = sq.testid
AND q.qnid = sq.qnid AND sq.answered = 'answered' ***and sq.stdid=s.stdid***
AND q.correctanswer = sq.stdanswer GROUP BY sq.stdid
I have this table structure
CREATE TABLE IF NOT EXISTS `users` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`username` varchar(50) DEFAULT NULL,
`realname` varchar(50) NOT NULL,
`password` varchar(50) DEFAULT NULL,
`email` varchar(50) DEFAULT NULL,
`bos1` varchar(30) DEFAULT NULL,
`emailbos1` varchar(50) DEFAULT NULL,
`role` varchar(20) DEFAULT NULL,
`created` datetime DEFAULT NULL,
`modified` datetime DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=21 ;
I got an issue here :
Two persons (eg: username=steve, and balmer) has bos1 name (eg: bill)
When I login as steve and/or balmer I can see my own detail / data,
but when I login as bill, I want to show steve's and balmer's detail as well.
How do i suppose to do it ?
Thank you.