PHP - Insert Labels Of Checked Checkboxes In Database

hi all

I have 3 checkboxes whose values are getting stored in the database with a comma separator.

Code: [Select]
<input name="color[]" type="checkbox" value="red" id="red" />red 
<input name="color[]" type="checkbox" value="blue" id="blue" />blue
<input name="color[]" type="checkbox" value="green" id="green" />green

values getting stored with comma separator

Code: [Select]
red, blue, green

now i want to show the checkboxes as "checked" to show the customer as which colors they chose lasttime when they submitted the form.

How can i make it possible ?

If it is possible without the comma separator then i can remove the comma.

vineet

I have a table made up of time slots, when the user clicks maybe one or two and presses submit, i would like the date selected and the time slots chosen to be saved in to my sql. I am working on the query but a little stuck in regards to the query, i have done the following:

<?php
$username = "root";
$password = "";
$hostname = "localhost";


 $dbhandle = mysql_connect($hostname, $username, $password) or die ("no         
 connection to database");


if(isset($_POST['Submit'])){

$start = mysql_real_escape_string($_POST['start']);
$end = mysql_real_escape_string($_POST['end']);
$booked = mysql_real_escape_string($_POST['booked']);

$selected = mysql_select_db("booking", $dbhandle);

?>

i have set the table as follows

<input data-val='08:30 - 08:45' class='fields' type='checkbox'   
 name="booked[]" />

Hello,   I want to show checkbox is checked when there is entry of that id in a table in my database.    I have 2 tables page and access_level. I am getting data from page table and displays it in <ul><li> tag with checkbox to select all or only few. After selecting the checkbox, i will store only selected checkbox value in access_level table along with table id. Page link and page name details will be stored in page table.    Now if i want to edit , i should display all the pages which is there in page table and i should also mark checked to those which are already stored in access_level table.   I am using LEFT OUT JOIN, It displays all the pages. But it is not displaying the check mark to those which are already selected.   Here is my code

<?php $s1 = mysql_query("SELECT pages.page_id, pages.code, pages.page, pages.href, access_level.aid, access_level.page_id as pgid, access_level.department, access_level.position, access_level.active FROM pages LEFT OUTER JOIN access_level ON pages.page_id=access_level.page_id WHERE access_level.department=".$department." AND access_level.position=".$position." AND pages.code='sn'") or die(mysql_error());
				while($s2 = mysql_fetch_array($s1)) { ?>
				<tr><td><li><?php echo $s2['page']; ?>  </td><td><input type="checkbox" name="sn[]" value="<?php echo $s2['page_id']; ?>"  
                <?php if($row['pgid'] === $s2['page_id']) echo 'checked="checked"';?> />

here is my pages table     pages.JPG   26.55KB   0 downloads   access_level  access_level.JPG   19.09KB   0 downloads   In access_level table i do not have page ids 8 and 9. But i want to display that also from pages table and for 1 to 7 and 10 i should display check mark.    How i can achieve this? Please Help

Hey everyone i having some trouble updating records in my databse using checkbox from a form.

I can update the records, but i can't figure out how to remove records at the same time.

I've posted the code that i'm using to update below

foreach($_POST['tag'] as $k=>$l){
$tag = mysql_real_escape_string($_POST['tag'][$k]);

$check_tags = mysql_query("SELECT * FROM file_cats WHERE category='$tag'") or die(mysql_error());
$num_tags = mysql_num_rows($check_tags);

if($num_tags==0){
$add_tag = mysql_query("INSERT into file_cats (fileID, category) VALUES('$fileID', '$tag')") or die(mysql_error());
}
}


This code inserts new records based on what the user checked. If it exist it won't add a new row, if it doesn't exist then it will add a new row. but like i said how can remove a row if the box is no longer checked?

I've tried using

foreach(empty($_POST['tag'] as $k => $l){
       $tag = empty($_POST['tag']['$k'];
}


But that didn't work basically nothing was deleted.

Has anyone done something like this before? if so i would appreciate any help you could give

Hi people.

I have a form which inputs into a database.

Here is the code that inserts a yes no option...
Code: [Select]
<select name = "consent">
        <option value = "Yes" <?php if ($_POST['consent'] == 'Yes') { echo 'selected="selected"'; } ?>>Yes</option>
        <option value = "No" <?php if ($_POST['consent'] == 'No') { echo 'selected="selected"'; } ?>>No</option>
       
     
  </select>
However, I have been asked if I can make it a yes or no checkbox instead.

Please can you tell me how I need to code it so that the "yes" or "no" is recorded in the DB.

At the moment I just have this

Code: [Select]
<input name="consent" type="checkbox" value="Yes" />Yes<br />
    <input name="consent" type="checkbox" value="No" />No<br />

Thanks in advance
VinceG

hello
I want query from one table and insert in another table on another domain .
each database on one domain name.
for example http://www.site.com $con1
and http://www.site1.com $con.

can anyone help me?
my code is :
<?php



$dbuser1 = "insert in this database";
$dbpass1 = "insert in this database";
$dbhost1 = "localhost";
$dbname1 = "insert in this database";

// Connecting, selecting database

$con1 = mysql_connect($dbhost1, $dbuser1, $dbpass1)
   or die('Could not connect: ' . mysql_error());



mysql_select_db($dbname1) or die('Could not select database');


$dbuser = "query from this database";
$dbpass = "query from this database";
$dbhost = "localhost";
$dbname = "query from this database";

// Connecting, selecting database

$con = mysql_connect($dbhost, $dbuser, $dbpass)
   or die('Could not connect: ' . mysql_error());

mysql_select_db($dbname) or die('Could not select database');

//query from database
$query = mysql_query("SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1");
    while($row=mysql_fetch_array($query)){
$result=$row[0];
$text=$row[1]."</br>Size:(".$row[4].")";
$alias=$row[2];
$link = '<a target="_blank" href='.$row[3].'>Download</a>';
echo $result;
}  
//insert into  database
mysql_query("SET NAMES 'utf8'", $con1); 
$query3= " INSERT INTO `jos_content`
 (`id`,
  `title`,
   `alias`,
    `) VALUES 
                        (NULL, 
                        '".$result."', 
                        '".$alias."', 
                       '')";
if (!mysql_query($query3,$con1))
  {
  die('Error: text add' . mysql_error());
  }


            mysql_close($con);
                        mysql_close($con1);
             


?>

I've been trying to simplify this code, let alone make it work.  It's a loop to change the different question checkboxes to "checked" if certain degree_ids are found.  All of the examples i have looked at are very strange and I'm not sure how to go about this, thought it seems to be a common question.  Everyone seems to have their own way.  Any advice would be greatly appreciated.

for($i =1; $i <= $arraycount; $i++){
$query = "SELECT degree_id FROM `user-degree` WHERE `user_id` = '".$user_id."' AND degree_id = '".$i."'";
$result = $conn->query($query) or die(mysql_error());
$row = mysqli_fetch_array($result) ;

echo $row['degree_id'];

if(!$y){
$y = 0;
}

if ($row['degree_id'] == 1){
$q4[0] = 'checked';
}
if ($row['degree_id'] == 2){
$q4[1] = 'checked';
}
if ($row['degree_id'] == 3){
$q4[2] = 'checked';
}
if ($row['degree_id'] == 4){
$q4[3] = 'checked';
}
if ($row['degree_id'] == 5){
$q4[4] = 'checked';
}
if ($row['degree_id'] == 6){
$q4[5] = 'checked';
}
if ($row['degree_id'] == 7){
$q4[6] = 'checked';
}
if ($row['degree_id'] == 8){
$q4[7] = 'checked';
}
if ($row['degree_id'] == 9){
$q4[8] = 'checked';
}
}

how do I make the $image piece so that it inserts "images/$image.php" into the database?

   VALUES('$name', '$id', '$image','$event')";

Hi guys,
I have a table that i want all the entries to be inserted all at once.
Mine is just inserting only one of the item, how can i get it to work? I also attached a sample table.

Thanks

	if(isset($_POST['send'])){
    
    $category = $_POST['category'];
    $item_type = $_POST['item_type'];
    $item_size = $_POST['item_size'];
    $product_name = $_POST['product_name'];
    $item_qty = $_POST['item_qty'];
    $price = $_POST['price'];
    $total_price = $_POST['total_price'];
    $item_price = $_POST['item_price'];
    
$sql = "
INSERT INTO shopping(
trans_ref,
category,
product_name,
item_type,
item_size,
item_qty,
item_price,
price,
date
)
	VALUES(
:trans_ref,
:category,
:product_name,
:item_type,
:item_size,
:item_qty,
:item_price,
:price,
NOW()
)
";
$stmt = $pdo->prepare($sql);
$stmt->execute(array(
':trans_ref' => $_SESSION['trans_ref'],
':category' => $category,
':product_name' => $product_name,
':item_type' => $item_type,
':item_size' => $item_size,
':item_qty' => $item_qty,
':item_price' => $item_price,
':price' => $price
));
	
if ( $stmt->rowCount()){    
    echo '<div class="alert bg-success text-center">SHOPPING COMPLETED</div>';
}else{
    echo '<div class="alert bg-danger text-center">SHOPPING NOT COMPLETED. A PROBLE OCCURED</div>';
}
	
}
	

I've begun to play around with object oriented php and I was wondering how I would go about writing a method that inserts values into a table.    I have a class, called queries, and I made this very simple function:

class queries {

    public function insert($table, $field, $value){

    global $dbh;
 
    $stmt = $dbh->prepare("INSERT INTO $table ($field) VALUES (:value)");
    $stmt->bindParam(":value", $value);
    $stmt->execute();
    }

}

$queries = new queries();

This topic has been moved to Ajax Help.

http://www.phpfreaks.com/forums/index.php?topic=350408.0

i have this code...
and i want to make it more dynamic....
Code: [Select]
<?php 
$content = $content[1].$content[2].$content[3].$content[4].$content[5].$content[6].$content[7].$content[8].$content[9].$content[10].$content[11].$content[12]
.$content[13].$content[14].$content[15].$content[16] 
.$content[17].$content[18].$content[19].$content[20].$content[21].$content[22].$content[23].$content[24]; 


$sql = mysql_query ("insert into database (content)  values ('$content')") 
               or die(mysql_error()); 
?>

for example... instead  content[1]. content[2]. content[3] ...etc... i want something like this.......

$content = $content[from 1 to  24];
or
$content = $content[from 1 to 777];

any solutions?

Everything on this page works the form loads and i can insert info into the form but when i click submit it just reloads the page(action="this page") and nothing happens.
  thanks in advance
   -Adam
Code: [Select]
<?php
require("connection/connection.php");
include("connection/functions.php");
?>
<HTML>
    <head>
        <meta content="yes" name="apple-mobile-web-app-capable" />
        <meta content="index,follow" name="robots" />
        <meta content="text/html; charset=utf-8" http-equiv="Content-Type" />
        <link href="" rel="apple-touch-icon" />
        <meta content="minimum-scale=1.0, width=device-width, maximum-scale=0.6667, user-scalable=no" name="viewport" />
        <meta name="format-detection" content="telephone=no"/>
        <link href="css/style.css" rel="stylesheet" media="screen" type="text/css" />
        <script src="javascript/functions.js" type="text/javascript"></script>
        <title>.: Bookmarks :.</title>
        <link href="" rel="apple-touch-startup-image" />
       
       
        <meta content="apple-mobile-web-app-status-bar-style" content="default" />
    </head>
    <body class="musiclist">
        <div id="topbar">
           <div id="title">.: Bookmarks :.</div>
            <div id="leftbutton">
                <a href="default.html"><img src="images/home.png"></a>
            </div>
            <div id="rightbutton">
                <a href="">Programs</a>
            </div>
        </div>
        <div id="tributton">
        <div class="links">
        <a id="pressed" href="Bookmarks/Bookmarks.php">Bookmarks</a><a  href="Music/Music.html">Music</a><a  href="">Movies</a>
            </div>
        </div>
        <div id="content" style="top: 36px">
        <form method="post" action="AddBookmark.php">
                <span class="graytitle">Add Bookmark</span>
                <ul class="pageitem">
                    <li class="bigfield"><input type="text" name="BookmarkName" placeholder="Bookmark Name" /></li>
                    <li class="bigfield"><input type="text" name="BookmarkLink" placeholder="Bookmark Link" /></li>
                    <li class="select"><select name="d">
                        <?php 
                            $getbookmarksections = mysql_query("SELECT * FROM bookmarksections");
                            confirm_query($getbookmarksections);
                            
                            while($getbm = mysql_fetch_array($getbookmarksections)){
$id = $getbm["ID"];
$sn = $getbm["SectionName"];
                             echo "<option value=\"$id\">$sn</option>";
                            }
                        ?>
                    </select><span class="arrow"></span></li>
                    <li class="button">
                    <input name="AddBookmark" type="submit" value="Add Bookmark" />
                    </li>
                </ul>
            </form>
            <?php
if(isset($_POST["AddBookmark"])){
$SectionID = $id;
$Bookmark = $_POST["BookmarkName"];
$BookmarkLink = $_POST["BookmarkLink"];

$add = "INSERT INTO `bookmarks` (`ID` ,`SectionID` ,`Bookmark` ,`BookmarkLink`)
VALUES ('NULL', '$SectionID', '$Bookmark', '$BookmarkLink')"; 
  if(confirm_query($add)){
  echo "<span class=\"graytitle\">Bookmark '$BookmarkName' in '$sn' Added.</span>";
  }
}
                ?>
        </div>
        <div id="footer">
        </div>
    </body>
</HTML>
<?php 
//Close Connection
if(isset($connection)){
mysql_close($connection);
}
?>
this is the code i suspect is not working
Code: [Select]
<?php
if(isset($_POST["AddBookmark"])){
        $SectionID = $id;
        $Bookmark = $_POST["BookmarkName"];
$BookmarkLink = $_POST["BookmarkLink"];

$add = "INSERT INTO `bookmarks` (`ID` ,`SectionID` ,`Bookmark` ,`BookmarkLink`)
VALUES ('NULL', '$SectionID', '$Bookmark', '$BookmarkLink')"; 
if(confirm_query($add)){
      echo "<span class=\"graytitle\">Bookmark '$BookmarkName' in '$sn' Added.</span>";
        }
}
?>

Hi all,
I've made a page the variables are being recieved and echo'd ok, however they are not being inserted into the database.  below is the file with the insert statement.

any help really appreciated.

MsKazza

Code: [Select]
<?php
// Database connect
$con = mysql_connect("mysql1.myhost.ie","admin_book","root123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("book_test", $con);

$orderid=$_GET['orderid'];



//Parse Values from Coupon.php Form
$orderid  = mysql_real_escape_string(trim($_POST['orderid']));
$design  = mysql_real_escape_string(trim($_POST['design']));
$childname  = mysql_real_escape_string(trim($_POST['childname']));
$address = mysql_real_escape_string(trim($_POST['address']));

echo $orderid;
echo $design;
echo $childname;
echo $address;

$sql="INSERT INTO details (orderid, design, childname, address)
VALUES
('$orderid','$design','$childname','$address')";


?>

<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Digital Scribe Books</title>
<link href="style.css" rel="stylesheet" type="text/css" />

<script type="text/javascript">
function MM_preloadImages() { //v3.0
  var d=document; if(d.images){ if(!d.MM_p) d.MM_p=new Array();
    var i,j=d.MM_p.length,a=MM_preloadImages.arguments; for(i=0; i<a.length; i++)
    if (a[i].indexOf("#")!=0){ d.MM_p[j]=new Image; d.MM_p[j++].src=a[i];}}
}
function MM_swapImgRestore() { //v3.0
  var i,x,a=document.MM_sr; for(i=0;a&&i<a.length&&(x=a[i])&&x.oSrc;i++) x.src=x.oSrc;
}
function MM_findObj(n, d) { //v4.01
  var p,i,x;  if(!d) d=document; if((p=n.indexOf("?"))>0&&parent.frames.length) {
    d=parent.frames[n.substring(p+1)].document; n=n.substring(0,p);}
  if(!(x=d[n])&&d.all) x=d.all[n]; for (i=0;!x&&i<d.forms.length;i++) x=d.forms[i][n];
  for(i=0;!x&&d.layers&&i<d.layers.length;i++) x=MM_findObj(n,d.layers[i].document);
  if(!x && d.getElementById) x=d.getElementById(n); return x;
}

function MM_swapImage() { //v3.0
  var i,j=0,x,a=MM_swapImage.arguments; document.MM_sr=new Array; for(i=0;i<(a.length-2);i+=3)
   if ((x=MM_findObj(a[i]))!=null){document.MM_sr[j++]=x; if(!x.oSrc) x.oSrc=x.src; x.src=a[i+2];}
}
</script>


</head>

<body onload="MM_preloadImages('images/buttons/home_over.png','images/buttons/books_over.png','images/buttons/cards_over.png','images/buttons/letters_over.png')">


<div id="snow">
<div id="wrapper">


<div id="header">

<div id="logo"><img src="images/digital_scripe.png" width="218" height="91" /></div>


<div id="menu"><a href="index.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Home','','images/buttons/home_over.png',1)"><img src="images/buttons/home_act.png" name="Home" width="131" height="132" border="0" id="Home" /></a><a href="books.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Books','','images/buttons/books_over.png',1)"><img src="images/buttons/books_act.png" name="Books" width="131" height="132" border="0" id="Books" /></a><a href="cards.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Cards','','images/buttons/cards_over.png',1)"><img src="images/buttons/cards_act.png" name="Cards" width="131" height="132" border="0" id="Cards" /></a><a href="letters.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Letters','','images/buttons/letters_over.png',1)"><img src="images/buttons/letters_act.png" name="Letters" width="131" height="132" border="0" id="Letters" /></a></div>
 </div>
 
 
 
 
<div id="content">
 <div id="info_bar"><br /><br /><br />
 <p>Your personal order number is:</p><br />
 
<br />
<p>Please fill out the details of where you would like your order to be shipped to.</p>
 </div>


<form action="card_paynow.php" method="POST" name="custdetails" id="custdetails">
  <table width="60%"  border="0" cellspacing="0" cellpadding="0">
    <tr>
      <td width="19%">orderid</td>
      <td width="81%"><input name="orderid" type="hidden" value="<? echo $orderid; ?>"></td>
    </tr>
    <tr>
      <td>name</td>
      <td><input name="name" type="text" id="name"></td>
    </tr>
    <tr>
      <td>surname</td>
      <td><input name="surname" type="text" id="surname"></td>
    </tr>
    <tr>
      <td>address 1 </td>
      <td><input name="add1" type="text" id="add1"></td>
    </tr>
    <tr>
      <td>address 2 </td>
      <td><input name="add2" type="text" id="add2"></td>
    </tr>
    <tr>
      <td>town</td>
      <td><input name="town" type="text" id="town"></td>
    </tr>
    <tr>
      <td>county</td>
      <td><input name="county" type="text" id="county"></td>
    </tr>
    <tr>
      <td>postcode</td>
      <td><input name="postcode" type="text" id="postcode"></td>
    </tr>
    <tr>
      <td>number</td>
      <td><input name="phone" type="text" id="phone"></td>
    </tr>
    <tr>
      <td>email</td>
      <td><input name="email" type="text" id="email"></td>
    </tr>
    <tr>
      <td>&nbsp;</td>
      <td>&nbsp;</td>
    </tr>
    <tr>
      <td>&nbsp;</td>
      <td><input type="submit" name="Submit" value="Continue..."></td>
    </tr>
    <tr>
      <td>&nbsp;</td>
      <td>&nbsp;</td>
    </tr>
  </table>
 
 
</form>

</div>


<div id="footer" class="clear"><div id="sign"><div id="sign_text">Personalised<br />
Books</div>
</div></div>

</div></div>






</body>
</html>

Hi guys,

Is it possible to insert into two tables in a database?

Thanks