Mysql result as array - php

I have a php function that interogates a table and gets all the fields in a column if a condition is fulfield. So the function returns a collection of elements.
The problem is that i want this function to return an array that i can parse and display.
The code below:
function get_approved_pictures(){
$con = mysql_connect("localhost","valentinesapp","fBsKAd8RXrfQvBcn");
if (!$con)
{
echo 'eroare de conexiune';
die('Could not connect: ' . mysql_error());
}
mysql_select_db("mynameisbrand_valentineapp", $con);
$all = (mysql_query("SELECT picture FROM users WHERE approved = 1"));
$row=mysql_fetch_assoc($all);
// mysql_close($con);
return $row['picture'];
}
Where am I wrong?

You need to use the loop for traversing all the data fetched by the query:
$pictures=array();
while($row=mysql_fetch_assoc($all))
{
$pictures[]=$row['picture'];
}
return $pictures;

Do it like this
$all = mysql_query("SELECT picture FROM users WHERE approved = 1");
$arr = array(); // Array to hold the datas
while($row = mysql_fetch_array($all)) {
$data = $row['picture'];
array_push($arr,$data);
}
return $arr;
You can now insert it into a function and return the values.
Note : mysql_* functions are being depreciated. Try to avoid them.

For the sake of diversity and to give you some sense of how to use PDO instead of deprecated mysql_*, this is how your function might look like:
function get_approved_pictures(){
$db = new PDO('mysql:host=localhost;dbname=mynameisbrand_valentineapp;charset=UTF-8',
'valentinesapp', 'password');
$query = $db->prepare("SELECT picture FROM users WHERE approved = 1");
$query->execute();
$pictures = $query->fetchAll(PDO::FETCH_ASSOC);
$db = null;
return $pictures;
}
Disclaimer: all error handling intentionally omitted for brevity

For the sake of diversity and to give you some sense of how the things have to be instead of inconvenient and wordy PDO, this is how your function might look like:
function get_approved_pictures(){
global $db;
return $db->getCol("SELECT picture FROM users WHERE approved = 1");
}
Disclaimer: all error handling is up and running but intentionally encapsulated into private methods for invisibility.

Related

how do i create a php function to echo out mysql data, a function that can be reused?

I need to write a PHP function to echo out MySQL rows as I give it the SQL query I want to be executed as the function argument. I have tried out the following code but it is giving me an undefined index error
function runQuery($query) {
$conn = mysqli_connect('localhost', 'root', '', 'mydb');
$result = mysqli_query($conn,$query);
while($row=mysqli_fetch_assoc($result)) {
$resultset[] = $row;
}
if(!empty($resultset))
return $resultset;
the code I am using to call the function is;
runQuery(SELECT * FROM mytable WHERE id='5')
echo $resultset['name'];
this, however, gives me this error, undefined index 'resultset' on line 25. any kind assistance would be appreciated
You dont have a $resultset in the scope of where you call the function. The function creates one, but that is only visible inside the function.
You will also have to put QUOTES around the query, you are passing a string there so it needs to be quoted.
Your errors should have generated quite a few error messages, if you were not getting them I have added 4 lines of code you should add while testing code for example if you are testing on a LIVE server with error reporting turned off.
You should also change the function to ensure you always return something
So amend the call to
ini_set('display_errors', 1);
ini_set('log_errors',1);
error_reporting(E_ALL);
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
function runQuery($conn, $query) {
$resultset = [];
$result = mysqli_query($conn,$query);
while($row=mysqli_fetch_assoc($result)) {
$resultset[] = $row;
}
return $resultset;
}
$resultset = runQuery($conn, "SELECT * FROM mytable WHERE id='5'");
// as result will now be a multidimentional array
// you will need to loop over that to get each returned row
foreach ( $resultset as $row ) {
echo $row['name'];
}
AFTER your edit there is another error
$conn is not created inside the function, so will be invisible in the function code unless passed as a parameter to the function (there is another way but lets not get into the bad habit of using global variables)
First, your code is probably vulnerable to SQL Injection. Please take care of that, by using prepared statements for instance.
https://www.w3schools.com/sql/sql_injection.asp
https://websitebeaver.com/prepared-statements-in-php-mysqli-to-prevent-sql-injection
Other than that, you do not assign the return value of your function to a variable. You cannot use the $resultset defined in the function scope outside the function, as it is a different scope. Try the following:
$resultset = runQuery("SELECT * FROM mytable WHERE id='5'")
echo $resultset['name'];
I built a similar function recently - here is my code
function returnSQL($conn, $nameSql) {
$result = mysqli_query($conn, $nameSql);
if (!$result) {
return 0;
}
while ($res = mysqli_fetch_assoc($result)) {
$data[] = $res;
}
return $data;
}
The connection is setup outside the function and passed in as an argument along with the sql like this...
$conn = mysqli_connect($servername, $username, $password, $DBName);
if (!$conn) {
echo 'Failed to connect to database :- ' . $DBName . '<br>';
die();
}
$sql = "SELECT * FROM table";
$data = returnSQL($conn, $sql);
I'm no expert, but this works for me :)
What I notice from your code is that you are trying to access $resultset outside of the function it is declared in and I think it is not available as a global variable - perhaps it should be something like:
$returnValue = runQuery(SQL statement);
// $returnValue is assigned the array returned from runQuery()
echo $returnValue['name'];

PHP MySQL Query Where x = $variable from function

Why is this not working:
function listOrderComments ($factnr){
global $connection;
//$factnr = 123; //or $factnr = "123"; (Both work)
$query = "SELECT * FROM orderstatus WHERE factuurnummer = '$factnr'";
$result = mysqli_query($connection, $query);
When I echo $factnr I get "123" back.
When I uncommented //$factnr = 123; my function is working.
Looked everywhere for a solution. check the type $factnr is (string).
Well if you're using a variable in your query you're opening yourself up to an injection attack for one.
If you're going to be using that variable I would recommend you use bind_param for your query
Read the PHP manual link below and you will be able to figure out the issue
http://php.net/manual/en/mysqli-stmt.bind-param.php
If you're passing in a variable to your function it should already be set so I don't understand why you're setting it to 123 anyway. So execute the sql statement and bind the parameter following the first example on the PHP docs page.
public function listOrderComments ($factnr)
{
global $connection;
$query = "SELECT * FROM orderstatus WHERE factuurnummer = ?";
$sql->prepare($query);
$sql->bind_param("s", $factnr);
$sql->execute();
$result = $sql->get_result();
$data = mysqli_fetch_all($result, MYSQLI_ASSOC);
foreach ($data as $row) {
print_r($row);
}
}
Then do what you want with the result
You can go with:
$query = "SELECT * FROM orderstatus WHERE factuurnummer = ". $factnr;
Concatenating your code is not good practise. Your best solution is to use PDO statements. It means that your code is easier to look at and this prevents SQL injection from occuring if malice code slipped through your validation.
Here is an example of the code you would use.
<?php
// START ESTABLISHING CONNECTION...
$dsn = 'mysql:host=host_name_here;dbname=db_name_here';
//DB username
$uname = 'username_here';
//DB password
$pass = 'password_here';
try
{
$db = new PDO($dsn, $uname, $pass);
$db->setAttribute(PDO::ERRMODE_SILENT, PDO::ATTR_EMULATE_PREPARES);
error_reporting(0);
} catch (PDOException $ex)
{
echo "Database error:" . $ex->getMessage();
}
// END ESTABLISHING CONNECTION - CONNECTION IS MADE.
$factnr = "123" // or where-ever you get your input from.
$query = "SELECT * FROM orderstatus WHERE factuurnummer = :factnr";
$statement = $db->prepare($query);
// The values you wish to put in.
$statementInputs = array("factnr" => $factnr);
$statement->execute($statementInputs);
//Returns results as an associative array.
$result = $statement->fetchAll(PDO::FETCH_ASSOC);
$statement->closeCursor();
//Shows array of results.
print_r($result);
?>
Use it correctly over "doted" concat. Following will just work fine:
$factnr = 123;
$query = "SELECT * FROM orderstatus WHERE factuurnummer = " . $factnr;
UPDATE:
here is $factnr is passing as argument that supposed to be integer. Safe code way is DO NOT use havvy functions even going over more complicated PDO, but just verify, is this variable integer or not before any operation with it, and return some error code by function if not integer. Here is no danger of code injection into SQL query then.
function listOrderComments ($factnr){
global $connection;
if (!is_int($factnr)) return -1
//$factnr = 123; //or $factnr = "123"; (Both work)
$query = "SELECT * FROM orderstatus WHERE factuurnummer = " . $factnr;
$result = mysqli_query($connection, $query);

PHP Using a while loop to fetch items, using a dynamic select function?

Okay, I've never ever used dynamic functions, not sure why, I've never liked using explode(), implode(), etc.
but I've tried it out, and something went wrong.
public function fetch($table, array $criteria = null)
{
// The query base
$query = "SELECT * FROM $table";
// Start checking
if ($criteria) {
$query .= ' WHERE ' . implode(' AND ', array_map(function($column) {
return "$column = ?";
}, array_keys($criteria)));
}
$check = $this->pdo->prepare($query) or die('An error has occurred with the following message:' . $query);
$check->execute(array_values($criteria));
$fetch = $check->fetch(PDO::FETCH_ASSOC);
return $fetch;
}
This is my query.
Basically I will return the variable $fetch which holds the fetch method.
and then somewhere, where I want to use the while loop to fetch data, I will use that:
$r = new Database();
while ($row = $r->fetch("argonite_servers", array("server_map" => "Wilderness")))
{
echo $row['server_map'];
}
Now, I am not getting any errors, but the browser is loading and loading forever, and eventually will get stuck due to lack of memory.
That's because the loop is running and running without stopping.
Why is it doing this? How can I get this dynamic query to work?
EDIT:
$r = new Database();
$q = $r->fetch("argonite_servers", array("server_map" => "Wilderness"));
while ($row = $q->fetch(PDO::FETCH_ASSOC))
{
echo $row['server_map'];
}
One nice feature of PDO is that the PDOStatement implements the Traversable. This means you can iterate it directly:
// `$check` is a `PDOStatement` object
$check = $this->pdo->prepare($query) or die('An error has occurred with the following message:' . $query);
$check->execute(array_values($criteria));
$check->setFetchMode(PDO::FETCH_ASSOC);
return $check;
Use it:
$statement = $r->fetch("argonite_servers", array("server_map" => "Wilderness"));
foreach ($statement as $row) {
}
this is because you call your fetch function in a loop and it re-starts the query every time. You need to call the $check->fetch() in loop instead.
or in other words, if your fetch function (which should probably have a different name) would return $check, then on the returned object you should call fetch() in a loop:
$r = new Database();
$q = $r->fetch(...);
while($q->fetch()){...}
you also need to edit your fetch function to end like this:
$check->execute(array_values($criteria));
return $check;
}

I can't get result from mysqli query with PHP (converted from mysql_result)

I have a PHP function that I am converting from using the mysql extension to the mysqli extension.
Everything is going okay, until here. I previously used a mysql_result to get a single piece of data. There is no direct equivalent in mysqli, so I have tried the following but it still doesn't work.
function getdbvalue($table,$value,$idfield,$id) {
$qrytext = "SELECT $value FROM $table WHERE $idfield LIKE '$id'";
$valueqry = mysqli_query($dbh,$qrytext);
if (FALSE === $valueqry) die("Select failed: ".mysqli_error);
$result = mysqli_fetch_row($valueqry);
$returnvalue = $result[0];
return $returnvalue;
}
I have verified that the variables are passing to the function okay, and the function is actually getting triggered. If I return $id I see the ID numbers.
I don't get an error for the query.
SOLVED:
I needed to add the database connection variable as a global in the function:
Working code:
function getdbvalue($table,$value,$idfield,$id) {
global $dbh; // This was missing!
$qrytext = "SELECT $value FROM $table WHERE $idfield LIKE '$id'";
$valueqry = mysqli_query($dbh,$qrytext);
if (FALSE === $valueqry) die("Select failed: ".mysqli_error);
$result = mysqli_fetch_row($valueqry);
$returnvalue = $result[0];
return $returnvalue;
}
Thanks to everyone for their help. :)
Although it's good idea to automate simple selects, the implementation is highly insecure, and should never be used.
Make it accept SQL query and parameters. It will make it secure.
And also you have to use PDO instead of mysqli
function getdbvalue() {
global $pdo;
$args = func_get_args();
$sql = array_shift($args);
$stm = $pdo->prepare($sql);
$stm->execute($args);
return $stm->fetchColumn();
}
have to be used like this (you have to connect to PDO first):
$name = getdbvalue("SELECT name FROM users WHERE id=?", $is);
this is the only proper way

Using PHP to place database rows into an array?

I was just wondering how i would be able to code perform an SQL query and then place each row into a new array, for example, lets say a table looked like the following:
$people= mysql_query("SELECT * FROM friends")
Output:
| ID | Name | Age |
--1----tom----32
--2----dan----22
--3----pat----52
--4----nik----32
--5----dre----65
How could i create a multidimensional array that works in the following way, the first rows second column data could be accessed using $people[0][1] and fifth rows third column could be accessed using $people[4][2].
How would i go about constructing this type of array?
Sorry if this is a strange question, its just that i am new to PHP+SQL and would like to know how to directly access data. Performance and speed is not a issue as i am just writing small test scripts to get to grips with the language.
$rows = array();
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
$rows[] = $row;
}
Are you open to using a DB module, like the PEAR::DB module? If so, check out this article by Paul Dubois on Writing Scripts with PHP's Pear DB Module. The Module has been superseded, but it will show you the basics of some more advanced (and more commonplace) DB practices.
As for your actual question, you could iterate over all the rows and populate an array...
$dsn = "mysqli://testuser:testpass#localhost/test";
$conn =& DB::connect ($dsn);
if (DB::isError ($conn)) { /* ... */ }
$result =& $conn->query ("SELECT * FROM friends");
if (DB::isError ($result)){ /* ... */ }
while ($row =& $result->fetchRow()) {
$people[] = $row;
}
$result->free ();
Or you could write an object which implements the ArrayAccess interface, requesting a particular row when you refer to that index. (This code could be completely wrong but here's my try)
class FriendsTable implements ArrayAccess {
function offsetGet($key) {
$result =& $conn->query ("SELECT * FROM friends LIMIT $key, 1",); // careful; this is vulnerable to injection...
if (DB::isError ($result)){ die ("SELECT failed: " . $result->getMessage () . "\n"); }
$people = null;
if ($row =& $result->fetchRow ()) {
$people = $row;
}
$result->free ();
return $people;
}
function offsetSet($key, $value) {
/*...*/
}
function offsetUnset($key) {
/*...*/
}
function offsetExists($offset) {
/*...*/
}
}
$people = new FriendsTable();
$person = $people[2]; // will theoretically return row #2, as an array
... or something.
$array = array();
$sql = "SELECT * FROM friends";
$res = mysql_query($sql) or trigger_error(mysql_error().$sql);
while($row = mysql_fetch_assoc($res)) $array[]=$row;

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