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Can't Insert Array Into Mysql Query	View Content

I have tried many ways of storing an array of data that comes from a multiple selection form into a mysql table, including serialization. I am currently trying to do it by forming a string, and using the string in the insert query, but keep getting errors, or just nothing happening. I am not sure what I am doing wrong.

Code: [Select]
$categoryString = array();

if ($categoryArray){
foreach ($categoryArray as $category){
$categoryString[] = $category.'<br />';}
}
$categoryquery= "INSERT INTO categoryname VALUES('$categoryString')";

mysql_query($categoryquery);
Thank you for your time as always.

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Mysql Query Doesn't Work. [insert Query]

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If you also have any feedback on my code, please do tell me. I wish to improve my coding base.

Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference.

Code:

Code: [Select]
mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
Full code:

Code: [Select]
<?php
include_once("includes/config.php");

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title><? $title; ?></title>
<meta http-equiv="Content-Language" content="English" />
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<link rel="stylesheet" type="text/css" href="style.css" media="screen" />
</head>
<body>

<div id="wrap">

<div id="header">
<h1><? $title; ?></h1>
<h2><? $description; ?></h2>
</div>

<? include_once("includes/navigation.php"); ?>

<div id="content">
<div id="right">
<h2>Create</h2>
<div id="artlicles">
<?php

if(!$_SESSION['user'])
{

        $username = mysql_real_escape_string($_POST['username']);
        $password = mysql_real_escape_string($_POST['password']);
        $name = mysql_real_escape_string($_POST['name']);
        $server_type = mysql_real_escape_string($_POST['type']);
        $description = mysql_real_escape_string($_POST['description']);

        if(!$username || !$password || !$server_type || !$description || !$name)
        {

        echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>";
        echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'>

        <option value='Any'>Any</option>
        <option value='PvP'>PvP</option>
        <option value='Creative'>Creative</option>
        <option value='Survival'>Survival</option>
        <option value='Roleplay'>RolePlay</option>

        </select></td></tr>

        <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>";
        echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>";
        }
        elseif(strlen($password) < 8)
        {
        echo "Password needs to be higher than 8 characters!";
        }
        elseif(strlen($username) > 13)
        {
                echo "Username can't be greater than 13 characters!";
        }
        else
        {
                $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1");

                if(mysql_num_rows($check1) < 0)
                {
                        echo "Sorry, there is already an account with this username and/or server name!";
                }
                else
                {
                        $ip = $_SERVER['REMOTE_ADDR'];

                        mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
                        echo "Server has been succesfully created!";
                }
        }

}
else
{
        echo "You are currently logged in!";
}

?>
</div>
</div>

<div style="clear: both;"> </div>
</div>

<div id="footer">
<a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop
</div>
</div>

</body>
</html>

Use Insert Ids From An Array In Another Query

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Hi,

I am creating a new menu (food) in my system. This consists of a menu, menu_items and menu_connection table.

I can insert the menu name just fine and return its id just fine.

When inserting the menu items, i need to get each of the menu_item_ids to use in the query that inputs the menu_connection.

This is what i have so far:

if ($_SERVER['REQUEST_METHOD']=="POST") {
	/////////////////////
	//menu name insert //
	/////////////////////
	
	$mname = mysqli_real_escape_string($conn, $_POST['newMenuName']);
	$stmt=$conn->prepare('
	                     INSERT IGNORE INTO ssm_menu (menu_name) VALUES (?);
	                     ');
	$stmt->bind_param('s',$mname);
	$stmt->execute();
	$menuInsId = $stmt->insert_id;
	echo $menuInsId;
	$stmt->close();

	/////////////////////
	//menu item insert //
	/////////////////////
	
	$mitname = $_POST['newMenuItem'];
	$stmt=$conn->prepare('
	                     INSERT IGNORE INTO ssm_menu_items (menu_item_name) VALUES (?);
	                     ');
	foreach ($_POST['newMenuItem'] as $k => $nmItem) {
		$mitname = mysqli_real_escape_string($conn, $nmItem);
		$stmt->bind_param('s',$mitname);
		$stmt->execute();
		$menuItmInsId = $stmt->insert_id;
		echo $menuItmInsId;
	}
	$stmt->close();

	///////////////////////////
	//menu connection insert //
	///////////////////////////

	
	$stmt=$conn->prepare('
	                     INSERT IGNORE INTO ssm_menu_connection (menu_id, menu_item_id) VALUES (?,?)
	                     ');
	foreach ($_POST['newMenuItem'] as $k => $nmItem) {
		$stmt->bind_param('ii',$menuInsId, $menuItmInsId);
		$stmt->execute();
		$connectionInserId = $stmt->insert_id;
		echo $connectionInserId;
	}
	$stmt->close();
}

Currently it is inserting each of the items in the connection table with the same id - i understand why but i dont know how to collect up all of the ids to use later

Mysql Insert Query

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Hello Friends,

Now i m stuck with simple insert query .

here's the code

Code: [Select]
<?php
include_once("conf.php");

$firstName=$_POST['fname'];
$lastName=$_POST['lname'];
$email=$_POST['email'];
$dob=$_POST['dob'];
$password=$_POST['pass'];

$fname= stripslashes($firstName);
$lname=stripslashes($lastName);

$mail= mysql_real_escape_string($email);
$password= mysql_real_escape_string($password);

mysql_select_db('site');
$statement="Insert into Accounts(Name,lastName,emailId,DOB,password) VALUES($fname,$lname,$mail,$dob,$password)";

$query=mysql_real_escape_string($statement);

mysql_query($query) or die("Cannot save data:</br> ".mysql_error());
echo "Data Saved Successfully";

?>

Now please explain me why i m getting the following error

Cannot save data:
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 '@gmail.com,1988/12/20,password)' at line 1

Any help will be highly appreciated!

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Hello every body,, i want to create a new database (auto generate duty assigned using form) in PHP and mysql.. i have four input fields:- Name, Subject, Class & weekly lectures now i want to insert name,subject & class into database, when i insert number of weekly lecture in field four.. insert automaticlly in database multiple time which i write in field four (weekly lecture) database: table structure i have already is::: id-name-subject-class-period-monday-tuesday-wednesday-thursday-friday-saturday anybody please help me to create this database or just insert query ....

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Here is an extract of my while loop

while ($personquery = mysql_fetch_array($personfetch, MYSQL_ASSOC))
          {

          echo "$personfetch[first_name]";
}

This will list each person now I want to do an INSERT to the database on every single member that is being listed I can only get it to INSERT for the one.

Thanks

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I am really just a beginner in PHP but I've got a task to create a simple hotel booking system by using PHP and MySQL. It works in several steps, described in separate .php files. All of them seem to work fine, except the last one. In this last file I want to add all the data into a MySQL database table. I would like to do this by creating a record for each date, a guest is spending in a hotel (so that when booking, another loop checks whether the room is not occupied yet for that date). I decided to use a for-loop for that, but it doesn't seem to be working (the records just don't get inserted into the table). I checked all the variables by echo-ing them and they are all right. I also tried the query without the loop and it works as well. So I guess there is some other problem. This particular .php file looks as follow:
<?php
session_start();
include('db_config.php');
    $name=($_POST['name']);
    $telephone=addslashes($_POST['telephone']);
    $email=addslashes($_POST['email']);
    $code=md5($email.time());
    $checkout_date=$_SESSION['checkout'];
    $checkin_unix=$_SESSION['checkin_unix'];
    $checkout_unix=$_SESSION['checkout_unix'];
    $roomtype=$_SESSION['roomtype'];
    $checkin_date=$_SESSION['checkin'];
    for ($stay_date=$checkin_unix; ; $stay_date=$stay_date+24*60*60) {
      if ($stay_date=$checkout_unix){
      break;
      }
      mysql_query("INSERT INTO reservations VALUES(
        '',
        '$roomtype',
        'pending',
        'date ('d/m/Y', $stay_date)',
        '$stay_date',
        '$checkout_date',
        '$name',
        '$telephone',
        '$email',
        '$code'
        )");

    }
?>

I would appreciate any kind of help, any idea.

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$qry='insert into tablename values('a','b');';

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I have created a registration page to access my website. After the user registrate himself should appear an alert saying that the registration was OK and a redirect to main.php page... however for some reason if I create an insert statement the alert and the redirect don't appear... If I remove the insert the alert and the redirect works... why?

This is part of the code of my 3 files:

registration.php (ajax call)

$('#submit').click(function() {
var username2 = $('#uname2').val();
var password2 = $('#psw2').val();
$.ajax({
url: 'ajax/response.php',
type: 'GET',
data: {username2 : username2, password2: password2},
success: function(data) {
if(data === 'correct') {
alert("Username and Password have been created!"); //don' work with the insert
location.replace("main.php"); //don' work with the insert
} else {
alert("Username or password are not correct... please register yourself!");
}
}
});
});

response.php (answer to ajax call)

if(isset($_GET['username2']) && isset($_GET['password2'])) {
$username2 = $_GET['username2'];
$password2 = $_GET['password2'];
if (checkUser($pdo, $username2) === true) {
echo 'duplicate';
} else {
insertUserPwd($pdo, $username2, $password2); //including this line the redirect and the alert doesn't work... the insert is OK
echo 'correct';
}
}

data_access.php (the function works but doesn't permit alert and redirect to appear)

function insertUserPwd(PDO $pdo, $usr, $pwd)
{
$data = [
'id' => '',
'user' => $usr,
'password' => $pwd
];
$sql = "INSERT INTO users (id, user, password) VALUES (:id, :user, :password)";
$stmt= $pdo->prepare($sql);
$stmt->execute($data);
}

Can someone help me to fix the code?

Insert Array In Mysql

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I'm trying to insert array of posts categories into MySQL. I'm getting category IDs from drop down list..

ERR: Notice: Notice: Array to string conversion in C:\laragon\*********\includes\functions.php on line 477

Array

Array
(
    [0] => Array
        (
            [0] => 4
            [1] => 5
        )

)

PHP

try {
        $sql = "INSERT INTO posts_category_ids (post_id, category_id) VALUES";

        $insertQuery = array();
        $insertData = array();
        foreach ($filter_category as $row) {
            $insertQuery[] = '(?, ?)';
            $insertData[] = $id;
            $insertData[] = $row;
        }

        if (!empty($insertQuery)) {
            $stmt = $this->conn->prepare($sql);
            $stmt->execute($insertData); //477
        }


         }catch (Exception $e){

            echo $e->getMessage();
}
}

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Hi,

I have an 2 arrays ( $match[1] and $match[2] ) with 50+ fields that i need to insert into a database, but im not getting how i could do it. Until now im trying with with the following code:

Code: [Select]
$sql="INSERT INTO is_trigger_reference (triggerName) VALUES ('$match[2][0]')";
mysql_query($sql) or die(mysql_error());

But when i check into the db table i only got:

Code: [Select]
triggername ( field name )
Array[0] ( data inserted)

I know i can set a var to each array ( 40+ ) but i belive there's something more clean than that

Thanks!

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hi guys so i have this add contacts page and the form is divided into 3 different froms 1) primary contact 2)spouse 3)child and the child form data is inserted as array into database because in the primary contact part of the form there is a "Children ?" with yes and no radio button and if yes a drop down list is enabled where if user chooses say 2 then there would be 2 child form that appears. and since theres 2 children then in the database a new row and data will be added accordingly. image attached to be clearer. from2.jpg 24.15KB 0 downloads

i got it inserted into database but in the specified field it says array: |child_name|dob|house_add1|mobile|office|email|
inserted:
|array|array-array-array|array|array|array|array| query:

"INSERT INTO child VALUES('','".$childsalutations." $childfname $childlname',' ".$cday."-".$cmonth."-$cyear ','$childline1','$childline2','$childm','$childoff','$childemail')"

in a stackoverflow question(not my own question) someone says: information stating arrays need to be split, before inserting into the table. does that mean something like this?:

$cday = ($_POST['cday']);
$cmonth = ($_POST['cmonth']);
$cyear = ($_POST['cyear']);
$childsalutations = ($_POST['child-salutations']);
$childfname = ($_POST['child-fname']);
$childlname = ($_POST['child-lname']);
$childline1 = ($_POST['child-line1']);
$childemail = ($_POST['child-email']);
$childm = ($_POST['child-mobile']);
$childoff = ($_POST['child-office']);

$info = array('c_name' => $childsalutation $childfname $childlname, 'c_dob' => $cday-$cmonth-$cyear, 'c_line1' => $childline1, 'c_mobile' => $childm, 'c_office' => $childoff, 'c_email' => $childemail)

just in case u wanted to c the html child form(warning its abit long,very!):

<table class="prime">
	<tbody>
	<br>
	<tr><td style="font-size:20px;font-weight:bold">Child <span id="number"></span></td></tr>
	<tr>
		<td>Salutation :</td> 
		<td><select name="child-salutations[]" id="child-salutations">
				<option value="" disabled selected>Salutations</option>
				<option value="Datin">Datin</option>
				<option value="Datin Paduka">Datin Paduka</option>
				<option value="Dato Paduka">Dato Paduka</option>
				<option value="Dato'">Dato'</option>
				<option value="Dato' Seri">Dato' Seri</option>
				<option value="Datuk">Datuk</option>
				<option value="Datuk Seri">Datuk Seri</option>
				<option value="Dr.">Dr.</option>
				<option value="Haji">Haji</option>
				<option value="Hajjah">Hajjah</option>
				<option value="HM">HM</option>
				<option value="HRH">HRH</option>
				<option value="Miss">Miss</option>
				<option value="Mrs.">Mrs.</option>
				<option value="Mr.">Mr.</option>
				<option value="Pehin">Pehin</option>
				<option value="Professor">Professor</option>
				<option value="Raja">Raja</option>
				<option value="Tan Sri">Tan Sri</option>
				<option value="Tengku">Tengku</option>
				<option value="Tuanku">Tuanku</option>
				<option value="Tun">Tun</option>
				<option value="Tunku">Tunku</option>
				<option value="Ungku">Ungku</option>
				</select>
		</td>
	</tr>
	<tr><td colspan="2"><label class="label" style="color:Red">*If a person has many salutations, choose the highest form of salutation</label></td></tr>
	<tr><td>First Name :</td><td><input type="text" name="child-fname[]" id="child-fname" class="style" /></td>
			<td>Last Name :</td><td><input type="text" name="child-lname[]" id="child-lname" class="style" /></td></tr>
	<tr>
		<td>Date of Birth : </td>
		<td>
			<select name="cday[]">
			<option value=""selected disabled>Day</option>
			<option value="1">1</option>
			<option value="2">2</option>
			<option value="3">3</option>
			<option value="4">4</option>
			<option value="5">5</option>
			<option value="6">6</option>
			<option value="7">7</option>
			<option value="8">8</option>
			<option value="9">9</option>
			<option value="10">10</option>
			<option value="11">11</option>
			<option value="12">12</option>
			<option value="13">13</option>
			<option value="14">14</option>
			<option value="15">15</option>
			<option value="16">16</option>
			<option value="17">17</option>
			<option value="18">18</option>
			<option value="19">19</option>
			<option value="20">20</option>
			<option value="21">21</option>
			<option value="22">22</option>
			<option value="23">23</option>
			<option value="24">24</option>
			<option value="25">25</option>
			<option value="26">26</option>
			<option value="27">27</option>
			<option value="28">28</option>
			<option value="29">29</option>
			<option value="30">30</option>
			<option value="31">31</option>
		</select>
		<select name="cmonth[]">
			<option value="" selected disabled>Month</option>
			<option value="1">January</option>
			<option value="2">February</option>
			<option value="3">March</option>
			<option value="4">April</option>
			<option value="5">May</option>
			<option value="6">June</option>
			<option value="7">July</option>
			<option value="8">August</option>
			<option value="9">September</option>
			<option value="10">October</option>
			<option value="11">November</option>
			<option value="12">December</option>
		</select>
		Year : <input type="text" name="cyear[]" maxlength="4" size="4" class="year">
		</td>
	</tr>
		<tr><td>Where do they live ?</td><td colspan="3"><input type="radio" name="living[]" id="living-me" class="living-me"/>With Me<input type="radio" name="living[]" id="living-other" class="living-other"/>With Other Parent<input type="radio" name="living[]" id="living-own" class="living-own"/>Own</td></tr>
		<tr><td>House Address</td></tr>
		<tr><td>Line 1  :</td><td><input type="text" name="child-line1[]" id="child-line1" size="20" class="style" /></td>
		<td>Mobile No :</td><td><input type="text" name="child-mobile[]" id="child-mobile" class="style" /></td></tr>
		<tr><td>Office No :</td><td><input type="text" name="child-office[]" id="child-office" class="style" /></td>
		<td>Email Address : </td><td><input type="email" name="child-email[]" id="email" class="style" /></td></tr>
	</tbody>
</table>

How To Insert An Array Into A Mysql Table?

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Insert Multi Array Into Mysql

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Hello, I have been at this for awhile now and cant seem to figure it out. I have 2 separate arrays like shown below and I am trying to insert each value into my database using one update query. I have tried a for loop inside each array, but only one column populates. I have tried breaking the values outside their for loop, etc, etc.

Does anyone know a way this can be done?

Code: [Select]

//Cars
$cars = array('ford', 'toyota', 'bmw');

//Places
$places =arrray('Orlando','California','Texas');

foreach ($cars as $car){
foreach ($places as $place){
$place .=$place;
}
mysql_query("INSERT INTO vehicles (cars, places) VALUES ('" . $car. "', '" . $place . "'')");
}

Mysql Insert: Whatever's Inside Json Array?

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An affiliate marketer refers some products which are stored in a simple array.
My website creates a JSON variable of all products sold.

$all_items_sold = '{"7777":{"item":"hammer","price":"4.99"},"8888":{"item":"nail","price":"1.99"},"9999":{"item":"apple","price":"2.00"}}';
$referred_by_Affiliate = array('1234','8888','7777');

So, out of all the 3 items that sold, only 2 of them were referred by the affiliate marketer.
DESIRED EFFECT: insert this product 8888: $1.99 and insert this product 7777: $4.99 into mysql.

Currently I do this:
1.) foreach loop,
2.) use strpos to see if it's in the raw JSON variable.
3.) If there, I use preg_match to find the price.
4.) Do a mySQL insert while still in the foreach loop.

Is there a "best practices" way to accomplish this?
I'm guessing there is a "one liner" so I don't have to do a foreach loop and using strpos.
And I'm guessing there is a way to do multiple mySQL inserts all at once with just one line of code, instead of from inside a foreach loop.
This probably can all be accomplished with just one line of code total, error trapping included 😁

Thank you.

Edited July 28 by ChenXiu

Adding A Value From Mysql Query Array Into A Php Array

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So I'm querying my database to add the results (mapID's) into a PHP array.
The MySQL query I used in the below code would usually return 10 values (only 10 mapID's in the database)

Code: [Select]
while($data = mysql_fetch_array(mysql_query("SELECT mapID FROM maps"))){
$sqlsearchdata[] = $data['mapID'];
}
Instead the page takes ages to load then gives this error:
Quote

Fatal error: Allowed memory size of 8388608 bytes exhausted (tried to allocate 16 bytes)

It says the error begins on the first line of the above code. I'm assuming this is not the right way to add the value from the MySQL array into a normal PHP array.

Can anyone help me?

Insert In Mysql Array Value From Drop Select Box Value - Cannot Pass The Var

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I am simply trying to insert a value generated from an array in a while loop, but it seems the value is not global and I can't pass it as I like. I did some research on this, but could not find an answer that solved my issue...

Here is my select box code which is working perfect:

<select name="city">
<?php
$sql = "SELECT id, city_name FROM cities ".
"ORDER BY city_name";
$results_set = (mysqli_query($cxn, $sql)) or die("Was not able to produce the result set!");
while($row = mysqli_fetch_array($results_set)) {
echo "<option value=$row[id]>$row[city_name]</option>";
}
?>
</select>

Are any variables defined in a while loop global to the while loop only?

Here is my SQL which you can see my $row[id] being passed thru field city_id... The value is being generated as supposed to based on value like: value="1", value="2" etc.. for the select options for each city name. So the values are there... But I CANNOT get that numerical id to pass to the database when submitting my form.

Any ideas for a workaround to get this value passing as normal?

if (isset($_POST['addPosting'])) {
$query = "INSERT INTO Postings (id, city_id, title, description) VALUES
('','$row[id]','$_POST[title]','$_POST[description]')";

Mysql Insert Into Putting Array On Seperate Table Rows

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Hi again all,

Why does the foreach loop im doing, put the array values from collection, into seperate table rows rather than 1 row per whole array?

    <?php
        class registration{
            public $fields = array("username", "email", "password");
            public $data = array();
            public $table = "users";
            public $dateTime = "";
            public $datePos = 0;
            public $dateEntryName = "date";
            public $connection;

            function timeStamp(){
                return($this->dateTime = date("Y-m-d H:i:s"));
            }

            function insertRow($collection){
                    //HERE
                    foreach($this->fields as $row => $value){
                        mysql_query("INSERT INTO $this->table ($value)
                        VALUES ('$collection[$row]')");
                    }

                    mysql_close($this->connection->connectData);
            }

            function validateFields(){

                $this->connection = new connection();
                $this->connection->connect();

                foreach($this->fields as $key => $value){
                    array_push($this->data, $_POST[$this->fields[$key]]);
                }
                $this->dateTime = $this->timeStamp();
                array_unshift($this->data, $this->dateTime);
                array_unshift($this->fields, $this->dateEntryName);

                foreach($this->data as $value){
                    echo "$value";
                }

                $this->insertRow($this->data);
            }
        }

        $registration = new registration();
        $registration->validateFields();
    ?>

I end up with 3 rows

row 1: username
row 2: email
row 3 : password

rather than row 1:username email password.

Array In Mysql Query

Similar Tutorials

Hello guys,

Got a question here..

If I have a database table called features with the following columns with example

f_id | f_name | f_status
1 | deck | 1
2 | Fireplace | 1
3 | Alarm | 1

I have another table another table called ads that has a column called features
and it has the following
1,2,3 (for example)

I want to be able to query the db and display all the f_name.

Here is what I have so far but it's not working and need some help

Code: [Select]

// Query the ads table
$result3 = mysql_query("SELECT * FROM ads WHERE ad_id='$id2'")
or die(mysql_error());
$row3 = mysql_fetch_array( $result3 );
$features = $row3['features'];

echo'This is the features ' . $features .'';
$feature2 = explode(",", $features);

print_r($feature2);

echo "test" . $feature2[0]; // piece1
echo $feature2[1]; // piece2

$result = mysql_query("SELECT * FROM features WHERE f_id=$feature2[1]") or die("Sql error : " . mysql_error());

while($row = mysql_fetch_assoc ($result)){
$f_name=$row["f_name"];
echo $f_name;
}

Php Array Error In Mysql Query

Similar Tutorials

when i run the below code i get an error... Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in... but when i take out the h>0 then there is no errors in the code. in table users, i need all columns in the array. How to get h>0 working?

$query = "SELECT * FROM users WHERE h>0 ORDER BY id";
$result = mysqli_query($link, $query);

$data = array();
while($row1 = mysqli_fetch_array($result))
{
	$data[] = $row1;
} 

print_r($data);

Edited by kalster, 27 October 2014 - 07:38 PM.