postgres update through php - php

I am a complete novice at php and postgres. I have been searching all over the net for an answer but no joy! I have a postgres table called workorders. It has a primary key (ident). I am trying to up date it using a php script/program. Basically the ident is the workorder number. I have written a php script where i type in the workorder number, post it to another php file, which connects to the database, finds the ident that matches and updates it by adding 'paid' to a column called paid. I have made it work using INSERT into, but it will not allow duplicate keys so that is why i am trying to use update. When I do, it says "unexpected T_CONSTANT_ENCAPSED_STRING"
Help would be greatly appreciated.
<?php
include('adodb5/adodb.inc.php');
$DB = &ADONewConnection('postgres');
$DB->PConnect('host=localhost port=5432 dbname=test user=postgres password=secret');
if (!$DB) {
die("Error in connection: " . pg_last_error());
}
$arr = array( array('paid',$_POST['$invoice']), );
$ok = $DB->Execute('insert into workorders (paid, ident) values (?,?)',$arr);
if (!$ok) {
die("Invoice not posted. " . pg_last_error ());
}
?>
If I use this I can insert into my database. If I replace INSERT into with UPDATE or pg_update I get errors.

Have now seen your code, which looks fine.
T_CONSTANT_ENCAPSED_STRING is often the result of not escaping quotation marks when used in strings. But I am not seeing that here. Can you post the example that does not work?
include('adodb5/adodb.inc.php');
$DB = &ADONewConnection('postgres');
$DB->PConnect('host=localhost port=5432 dbname=test user=postgres password=secret');
if (!$DB) {
die("Error in connection: " . pg_last_error());
}
$arr = array( array('paid',$_POST['$invoice']), );
$ok = $DB->Execute('insert into workorders (paid, ident) values (?,?)',$arr);
if (!$ok) {
die("Invoice not posted. " . pg_last_error ());
}

Related

Tutorial issues using INSERT INTO without adding a row to database for certain entries

I am following the last part of the following video tutorial "How to create a database website with PHP and mySQL 07 - Add in input form" :
https://www.youtube.com/watch?v=MGIG00d1Xzc&list=PLhPyEFL5u-i0zEaDF0IPLYvm8zOKnz70r&index=7
At the end here is my code, for the inserting portion to the database for the new_jokes.php script (everything up to this point of the series I have gotten to work fine so far)
Basically I am getting the seemingly classic "INSERT INTO" not working although all my syntax looks correct. Am I missing something obvious here? I get no errors, just the row isn't added.
<?php
include "db_connect.php";
$new_joke_question = $_GET["newjoke"];
$new_joke_answer = $_GET["newanswer"];
// Search the database for the word chicken
echo "<h2>Trying to add a new joke and answer: $new_joke_question
$new_joke_answer </h2>";
$sql = "INSERT INTO Jokes_table (JokeID, Joke_question, Joke_answer) VALUES
(NULL, '$new_joke_question', '$new_joke_answer' )";
$result = $mysqli->query($sql);
include "search_all_jokes.php";
?>
Return to the main page
Here is the db_connect.php code as requested:
<?php
// four variables to connect the database
$host = "localhost";
$username = "root";
$user_pass = "usbw";
$database = "test";
// create a database connection instance
$mysqli = new mysqli($host, $username, $user_pass, $database);
?>
Here is search_all_jokes.php (which has minor error checking):
// if there are any values in the table, select them one at a time
if ($mysqli->connect_errno) {
echo "Connection to MySQL failed: (" . $mysqli->connect_errno . ") " .
$mysqli->connect_error;
}
echo $mysqli->host_info . "<br>";
$sql = "SELECT JokeID, Joke_question, Joke_answer FROM Jokes_table";
$result = $mysqli->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "JokeID: " . $row["JokeID"]. " - Joke_question: " .
$row["Joke_question"]. " " . $row["Joke_answer"]. "<br>";
}
} else {
echo "0 results";
}
?>
Also here is the table structure screenshot viewed in myPHPAdmin:
I added error capturing into new_jokes.php inspired by this Stack Overflow post:
INSERT INTO SYNTAX ERROR
And get the following error:
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 't jump.' )' at line 1localhost via TCP/IP
Thank you everyone for helping out with this! Syntax can really throw a wrench in everything. I also will read up on prepared statements since that also could have prevented the issue. The ultimate help to this I found the solution to by adding the function referenced here for MySQLi real_escape_string to clean the single quote I had within the answer I was submitting to my joke table:
(Can a kangaroo jump higher than the empire state building? Of course, the empire state building can't jump.)
As shown in the documentation #miken32 linked as a comment here it is says: "But if $val1 or $val2 contains single quotes, that will make your SQL be wrong. So you need to escape it before it is used in sql; that is what mysql_real_escape_string is for. (Although a prepared statement is better.)"
But now the code for this part 7 of the tutorial on you tube I found works and adds it into a row on the database table, then displaying the full new table on the next webpage. I spent a good while shooting in the dark on while the answer ended up being fairly simple. Again special thanks to #miken32 for pointing me the right direction.
Here is my completed code that ended up working to at least achieve the goal of the tutorial:
<?php
include "db_connect.php";
$new_joke_question = $_GET["newjoke"];
$new_joke_answer = $_GET["newanswer"];
$new_joke_question = $mysqli->real_escape_string($new_joke_question);
$new_joke_answer = $mysqli->real_escape_string($new_joke_answer);
// Search the database for the word chicken
echo "<h2>Trying to add a new joke and answer: $new_joke_question $new_joke_answer
</h2>";
if ($mysqli->connect_errno) {
echo "Connection to MySQL failed: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
echo $mysqli->host_info . "<br>";
$sql = "INSERT INTO Jokes_table (JokeID, Joke_question, Joke_answer) VALUES (' ',
'$new_joke_question', '$new_joke_answer' )";
$result = $mysqli->query($sql);
if ($mysqli->query($sql) === TRUE) {
echo 'users entry saved successfully';
}
else {
echo 'Error: '. $mysqli->error .'<br>';
}
include "search_all_jokes.php";
?>
Return to the main page

PHP won't interface with MySQL in Apache on Raspberry Pi

I installed MySql on my Raspberry Pi 2 Model B+ a few days ago to see if I could use it, PHP, phpmyadmin, and Apache to make an accessible database to organize and catalog books that are around the house. I have a table in a MySQL database set up as a prototype with three columns; Booknumber (set to auto-increment), title, and authorLastName. I'm trying to use a form to insert books into table beta, in database bookProof.
Here's the code for the form:
<html>
<body>
<form action="catalog.php" method="POST">
<p>Book Title: <input type="text" name="title"></p>
<p>Author's Last Name: <input type="text name="authorlastname"></p>
</form>
</body>
</html>
Which links to "catalog.php", which is:
<?php
define('DB_NAME', 'bookProof');
define('DB_USER', 'root');
define('DB_PASSWORD', 'root');
define('DB_HOST', 'localhost');
$conn = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
if ($conn->connect_error) {
die("Could not connect: " . $conn->connect_error);
}
$value = $_POST["title"]
$value2 = $_POST["authorlastname"]
$sql = "INSERT INTO beta ('title', 'authorLastName') VALUES ('".$value."', '".$value2."')"
$query = mysqli_query($conn,$sql);
if ($conn->($sql) === TRUE) {
echo "New entry completed successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
When demoform.php is opened, it functions normally, but when the "Add Books" button is clicked, it goes to catalog.php as intended, but the catalog.php page is blank, the table is unchanged, and Google Chrome's "Inspect" tool gives the error:
POST http://192.168.254.11/Library/catalog.php 500 (Internal Server Error) catalog.php:1
If anyone knows how to get the input to the database, please let me know.
Note: This is just a home system, so security is not a priority (I don't need SQL code injection protection).
Your note, "...security is not a priority (I don't need SQL code injection protection)" - you might think that, but you should do it anyways. Not only does it protect your database should your system be exposed (or made public at a later time), it will handle strings automatically for you, so that your query won't break if your strings have quotes ' in them.
One issue is that you're using singlequotes around column and table names. This should be backticks, or none at all. Then you were missing a semicolon ; after defining your $value, $value2 and $sql strings.
Then you're doing something a bit odd - which is also causing a parse-error (Had you enabled error-reporting and checked your logs, you'd see a "Parse error: syntax error, unexpected (" error in your logs), you're querying the table with mysqli_query(), but then you try to do it again - except you're trying to query on the querystring, and not the query method. Note the comments I've added in the code below.
// Don't use singlequotes ' for columns and table-names
// Use backticks ` - quotes are for strings
$sql = "INSERT INTO beta (`title`, `authorLastName`) VALUES ('".$value."', '".$value2."')"; // You were also missing a semicolon here!
// $query = mysqli_query($conn,$sql); // Remove this line, as you're attempting to query it twice
if ($conn->query($sql) === TRUE) { // You're missing the query() method here
echo "New entry completed successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
Using prepared statements won't be that much of a difference, and you really should do it. There's absolutely no reason to not use prepared statements! Look how little changes that have to be made!
$sql = "INSERT INTO beta (title, authorLastName) VALUES (?, ?)";
if ($stmt = $conn->prepare($sql)) {
$stmt->bind_param("ss", $value, $value2);
$stmt->execute();
$stmt->close();
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
You've also got some invalid HTML which would cause issues - the following line had a missing quote to close off the type attribute.
<input type="text" name="authorlastname">
I suggest you read the following documentation and articles
When to use single quotes, double quotes, and backticks in MySQL
How can I prevent SQL injection in PHP?
PHP manual on mysqli_stmt::bind_param
How to get useful error messages in PHP?
PHP Parse/Syntax Errors; and How to solve them?
As a final note, you should check that the form was submitted and that it has values before inserting into the database. Also, using variable-names like $value and $value2 are not really descriptive - you should avoid it and use proper names for your variables.

PHP MySQL Query Insert consistent error

//--------------------------------------------------------------------------
// php script for adding data from mysql database
//--------------------------------------------------------------------------
$ip = $_GET['ip']; //for debugging sake, will be POST from Ajax
$key = substr(md5(microtime()),rand(0,26),5); //random referral ID - will implement exist analysis
echo $ip; //debugging
$dbhost = 'localhost';
$dbuser = user;
$dbpass = 'pass';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn ) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("database", $conn);
$tbl_name = "refs";
$sql="INSERT INTO $tbl_name(ip, key)VALUES('frfr', 'grgr')";
if (!mysql_query($sql,$con)) {
die('Error: ' . mysql_error());
}
echo "1 record added";
I'm not sure if it's my Digital Ocean server or what, but the only syntax my PhpMyAdmin will accept as a query is as INSERT INTOrefs(ip,key) VALUES ("insert","432")
with the double quoted values. I cannot seem to get this implemented in the PHP without getting a flat out error or an Unknown column in 'field list' error.
Similar questions suggest junk non-printable characters from copy-paste, however I've retyped my code within the editor. Thanks for all the help
I'm creating a basic referral system by the way, storing requested IP's in 'refs' table with a key, or id.
key is a reserve word and thus needs to be escaped using backtique. Along with that you have spacing issue as well. Your query should looks like below
INSERT INTO refs(ip,`key`) VALUES ('insert','432')
Never use a reserve word as column or table name. if in doubt, then escape all the columns present in query.
Start referring MySQL Documentation for more inforamtion. It's way easier than posting it as question in stackoverflow.

Display MySQL Database as an array

I have a MySQL database full of user information, like their username, password, email, etc.
I want a PHP script that allows me to pull JUST their username and display it like so:
"username1","username2","username3"
Literally exactly like that, the quotes and all.
EDIT: Sorry for not supplying enough information.
The table is named "users" the field I want to pull off it is "username" I can get it to pull and display all the information, my only problem is imploding it.
OK dude, read the comments
<?php // open a php tag
$dbc = mysql_connect("host", "username", "password"); // connect to database
mysql_select_db("db_name", $dbc) // select the database
$sql = "SELECT `username` FROM `users_table`"; // select only the username field from the table "users_table"
$result = mysql_query($sql); // process the query
$username_array = array(); // start an array
while($row = mysql_fetch_array($result)){ // cycle through each record returned
$username_array[] = "\"".$row['username']."\""; // get the username field and add to the array above with surrounding quotes
}
$username_string = implode(",", $username_array); // implode the array to "stick together" all the usernames with a comma inbetween each
echo $username_string; // output the string to the display
?>
I've seen all the other answers, however have you considered using PDO instead of mysql_query functions? It's a much nicer way to work with the database.
Here's what you want to achieve in a few lines of code (using lamba functions):
$dbh = new PDO("mysql:host=localhost;dbname=test", "yourusername", "yourpassword");
$results = $dbh->prepare("SELECT u.username FROM users u");
$results->execute();
$results = $results->fetchAll();
echo implode(", ", array_map(function(&$r) { return $r['username']; }, $results));
Output: Jamie, Bob, Chris
Nice and clean. Also, you should check if you have any results that have been returned and if the query was successful.
Just another approach.
EDIT: I've just realised you're a beginner so my answer may be a bit too advanced. However, i'll leave it for others to see as a solution, and perhaps you might look into using PDO an lamba functions when you learn a bit more. Best of luck.
Let's assume that you have a 'mydb' database and 'users' table in it.
SQL needed:
USE mydb;
SELECT username from users;
Short version:
Wrap it in PHP calls to mysql PHP library
Get result as an array then implode it with comma symbol.
Long version:
First we need to connect to database:
$db = mysql_connect('DATABASE_HOST', 'USER', 'PASSWORD');
if (!$db) {
die('Not connected : ' . mysql_error());
}
$db_selected = mysql_select_db('mydb', $db);
if (!$db_selected) {
die ('Can\'t use mydb: ' . mysql_error());
}
Remember to always check the return values of functions.
Then we query the database:
$result = mysql_query('select username from users', $db);
...and fetch results in flat array (we need only usernames):
while ($row = mysql_fetch_array($result, MYSQLI_ASSOC))
{
$data[] = $row['login'];
}
Then we format the returned data according to your specs:
$string_result = '"'. implode('", "', $data) . '"';
You can do with $string_result anything you want, just close the database connection immediately after use:
mysql_close($db);
Good luck with learning PHP, BTW. ;)
You could using PHP's implode, but it's probably easier just do it in SQL assuming that the list won't be too long:
SELECT GROUP_CONCAT(CONCAT('"', username, '"')) AS usernames
FROM your_table

INSERT IGNORE In mySQL

I wonder whether someone can help me please.
I'm trying to put together a PHP script that takes data from an xml file and places the data in a mySQL data. I've been working on this for a few days and I'm still can't seem to get this right.
This is the code that I've managed to put together:
<?
$objDOM = new DOMDocument();
$objDOM->load("xmlfile.xml");
$Details = $objDOM->getElementsByTagName("Details");
foreach( $Details as $value )
{
$listentry = $value->getElementsByTagName("listentry");
$listentrys = $listentry->item(0)->nodeValue;
$sitetype = $value->getElementsByTagName("sitetype");
$sitetypes = $sitetype->item(0)->nodeValue;
$sitedescription = $value->getElementsByTagName("sitedescription");
$sitedescriptions = $sitedescription->item(0)->nodeValue;
$siteosgb36lat = $value->getElementsByTagName("siteosgb36lat");
$siteosgb36lats = $siteosgb36lat->item(0)->nodeValue;
$siteosgb36lon = $value->getElementsByTagName("siteosgb36lon");
$siteosgb36lons = $siteosgb36lon->item(0)->nodeValue;
//echo "$listentrys :: $sitetypes :: $sitedescriptions :: $siteosgb36lats :: $siteosgb36lons <br>";
}
require("phpfile.php");
//Opens a connection to a MySQL server
$connection = mysql_connect ("hostname", $username, $password);
if (!$connection) {
die('Not connected : ' . mysql_error());
}
// Set the active MySQL database
$db_selected = mysql_select_db($database, $connection);
if (!$db_selected) {
die ('Can\'t use db : ' . mysql_error());
}
mysql_query("INSERT IGNORE INTO scheduledsites (listentry, sitetype, sitedescription, siteosgb36lat, siteosgb36lon) VALUES('$listentrys','$sitetypes','$sitedescriptions','$siteosgb36lats','$siteosgb36lons') ")
or die(mysql_error());
echo "Data Inserted!";
?>
I can pull the data from the xml file, but it's the part of the script that sends the data to my database table that I'm having trouble with.
The script runs but only the last record is saved to the database.
I can parse the fields from the xml file without any problems and the check I'm trying to put in place is, if there is a 'listentry' number in the new data that is matched to one already in the table then I don't want that record to be added to the table, i.e. ignore it.
I just wondered whether someone could perhaps take a look at this please and let me know where I'm going wrong.
Many thanks
You are only calling mysql_query once. So it will only insert one row.
The sql needs to be inside the loop.

Categories