Mysql query Issue in chrome - php

<?php
session_start();
include("configdb.php");
if(!session_is_registered(username)){
header("location:index.php");
}
?>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Projects</title>
<link href="style.css" rel="stylesheet" type="text/css" />
<script type="text/javascript">
function un_check(){
for (var i = 0; i < document.frmactive.elements.length; i++) {
var e = document.frmactive.elements[i];
if ((e.name != 'allbox') && (e.type == 'checkbox')) {
e.checked = document.frmactive.allbox.checked;
}
}
}
function Confirm(form){
alert("Project has been activated!");
form.submit();
}
function unConfirm(form){
alert("Project has been Deactivated!");
form.submit();
}
</script>
</head>
<body>
<div id="costDiv">
<div id="divErc"></div>
<div id="costBack">
<?php
if(isset($_POST['checkbox'])){$checkbox = $_POST['checkbox'];
if(isset($_POST['activate'])?$activate = $_POST["activate"]:$deactivate = $_POST["deactivate"])
$id = "('" . implode( "','", $checkbox ) . "');" ;
$sql="UPDATE projects SET p_isActive = '".(isset($activate)?'1':'0')."' WHERE p_id IN $id";
$result = mysql_query($sql) or die(mysql_error());
}
?>
<?php include("hor_menu.php"); ?>
<form name="frmactive" method="post" action="">
<table width="350" border="0" cellspacing="1" cellpadding="5" align="center" style="margin-left:150px ; margin-right:auto ; margin-top:20px ; margin-bottom:auto ; position:absolute ; width:400px">
<tr>
<td align="center" ><input type="checkbox" name="allbox" onclick="un_check(this);" title="Select or Deselct ALL" style="background-color:#ccc;"/></td>
<td align="left"><strong>Project</strong></td>
<td align="left"><strong>Country</strong></td>
<td align="left"><strong>Active</strong></td>
</tr>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td align="center"><input name="checkbox[]" type="checkbox" id="checkbox[]" value="<?php echo $rows['p_id']; ?>"/></td>
<td><?php echo $rows['p_name']; ?></td>
<td><?php echo $rows['p_country']; ?></td>
<td><?php if ($rows['p_isActive'] == '1'){ echo'Active';} else{ echo 'Inactive';} ?></td>
</tr>
<?php
}
?>
<tr>
<td colspan="5"><input name="activate" type="submit" id="activate" value="Activate" onClick="Confirm(this.form)" />
<input name="deactivate" type="submit" id="deactivate" value="Deactivate" onClick="unConfirm(this.form)"/></td>
</tr>
</table>
?>
</td>
</tr>
</table>
</form>
</body>
</html>
i have this code.when i check a checkbox project will be either activated or deactivated according to bottom.however this code works perfectly on all browsers except google chrome and safari.can anyone help please.it keeps giving that have an error in my sql syntax espeaciall after the where clause in the update query.thank you

If you get an SQL error depending on browser, then you most likely need to check your inputs. PHP execution does not differ depending on what browser you're using, other than what data it gets from client side controls like textboxes or triggers.
I suggest you log the $sql variable in some way, you could try just using var_dump, to see how the data changes between browsers.

This is not legal: id="checkbox[]" get rid of it.
It's not necessary to assign everything an ID. Only assign an ID if you actually use it in the javascript. Otherwise just leave it out.
INPUT fields do need a name attribute, but that works completely differently from an ID.

Related

Best Way to Submit Dynamic Repeating Form

I'm trying to teach myself php and mysql and I'm having trouble with a form I'm working on for a football pool I'm trying to put together. I'm looking for a little help on the best way to submit the data from this form. The form gets data from my database and repeats for the appropriate number of rows. Upon submitting the form each row should be inserted or updated to the database. The data that needs to be submitted is: userid, gameid, pick, and tbPoints.
The form is a little rough as I have not finished it yet. I just cant seem to get the form to submit each row as a new entry to the database, it only submits the last game. I know I have a problem with the loop to submit, but I just can't seem to figure out how to get it. Any help is appreciated!
here is my form:
<?php require_once('Connections/t2016.php'); ?>
<?php
mysql_select_db($database_t2016, $t2016);
$query_gamedays = "SELECT DISTINCT schedule.weekDay, schedule.`date` FROM schedule WHERE schedule.weekNum=1";
$gamedays = mysql_query($query_gamedays, $t2016) or die(mysql_error());
$row_gamedays = mysql_fetch_assoc($gamedays);
$totalRows_gamedays = mysql_num_rows($gamedays);
?>
<!doctype html>
<html>
<head>
<meta charset="UTF-8">
<title>Untitled Document</title>
</head>
<body>
<form action="" method="post" name="form1">
<table border="2" cellpadding="3" cellspacing="2">
<tr align="center">
<td>Time</td>
<td colspan="3">Matchup</td>
<td>Current Pick</td>
</tr>
<?php do {
$day = $row_gamedays['weekDay'];
$date = $row_gamedays['date'];
?>
<tr wrap="nowrap">
<td colspan="5"><?php echo '<strong>'.$day.'</strong>, '.$date; ?></td>
</tr>
<?php
mysql_select_db($database_t2016, $t2016);
$query_sched = "SELECT * FROM schedule WHERE schedule.weekNum=1 AND schedule.weekDay='$day'";
$sched = mysql_query($query_sched, $t2016) or die(mysql_error());
$row_sched = mysql_fetch_assoc($sched);
$totalRows_sched = mysql_num_rows($sched);
?>
<?php do {
$gameid = $row_sched['gameID'];
$time = $row_sched['time'];
$vteam = $row_sched['visitorID'];
$hteam = $row_sched['homeID'];
mysql_select_db($database_t2016, $t2016);
$query_picks = "SELECT * FROM picks WHERE picks.gameID=$gameid AND picks.userID=1";
$picks = mysql_query($query_picks, $t2016) or die(mysql_error());
$row_picks = mysql_fetch_assoc($picks);
$totalRows_picks = mysql_num_rows($picks);
if($totalRows_picks > 0) {
$pick = $row_picks['pickID'];
$tbp = $row_picks['tiebreakerPoints'];
} else {
$pick = 'No Pick';
$tbp = '0';
}
$vp = '';
$hp = '';
if($pick == $vteam) {
$vp = 'checked';
} elseif($pick == $hteam) {
$hp = 'checked';
}
?>
<tr align="center">
<td><?php echo $time; ?></td>
<td align="right"><?php echo $vteam.'<input type="radio" name="pickID'.$gameid.'[]" value="'.$vteam.'" '.$vp.'>'; ?></td>
<td align="center"> # </td>
<td align="left"><?php echo '<input type="radio" name="pickID'.$gameid.'[]" value="'.$hteam.'" '.$hp.'>'.$hteam; ?></td>
<td><?php echo $pick; ?></td>
</tr>
<?php if($row_sched['is_tiebreaker'] > 0) { ?>
<tr>
<td colspan="4" align="right" wrap="nowrap"><?php echo 'Enter Tiebreaker Points: <input type="number" name="tbpoints[]" min="0" value="'.$tbp.'">'; ?></td>
<td align="center"><?php echo $tbp; ?></td>
</tr>
<?php } ?>
<input type="hidden" name="gameID[]" value="$gameid">
<input type="hidden" name="userID" value="1">
<?php } while ($row_sched = mysql_fetch_assoc($sched)); ?>
<tr>
<?php } while ($row_gamedays = mysql_fetch_assoc($gamedays)); ?>
<td colspan="5" align="right" wrap="nowrap">
<input type="submit" name"submit" value="Submit Picks">
</td>
</tr>
</table>
</form>
<p> </p>
</body>
</html>
<?php
mysql_free_result($sched);
mysql_free_result($picks);
mysql_free_result($gamedays);
?>
again, the form is really rough so please be kind. Thanks for the help
You just loop through these values like a normal array.
$gameID = $_POST['gameID'];
for($x=0; $x<count($gameID); $x++ ) { .... }
As for the radio buttons/check boxes - you need to give each group a unique name. The only thing to remember with these controls is that if that are not set then nothing is submitted to the server. You need to check if they are part of the post values i.e.
if( isset( $_POST['cbPoints'] ) ) { do something }
Otherwise you'll get errors when trying to access them.

checkbox value inserted in mysql

I'm struggling now for a few days to get the value of a checkbox in my code.
Basically I have an admin-page where the customer can select and deselect images that will put online.
You can select and deselect images that will be shown on the homepage, and separate on the gallery-page. Both checked is also possible.
I have another checkbox that can be selected to remove the image from the list(image_deleted).
There is still a database entry and the images are still on file-system but later on I'll create a cleanup-job.
Here is my code:
<?php
ini_set('display_errors', 1);
error_reporting(E_ALL);
ob_start();
require('../../lib/dbconnection.php');
require("../../lib/checklogin.php");
require("includes/upload.inc.php");
$query = 'SELECT * FROM gallery where image_deleted != 1 order by id desc';
$result=$conn->query($query);
$count=$result->num_rows;
?>
<!DOCTYPE html>
<html>
<head>
<title>Classic Nails - CMS</title>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1">
<meta name="description" content="ClassicNails">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" href="../css/screen.css">
<link rel="stylesheet" href="../css/libs/magnific-popup.css">
<script src="../js/libs/min/jquery-min.js" type="text/javascript"></script>
<script src="../js/min/custom-min.js" type="text/javascript"></script>
<script src="js/jquery.magnific-popup.js"></script>
<script>
$(document).ready(function() {
$('.image-link').magnificPopup({
type:'image',
gallery:{
enabled:true
}
});
});
</script>
</head>
<body>
<?php include('includes/header.inc.php'); ?>
<?php include('includes/nav.inc.php'); ?>
<div class="wrapper">
<article class="content">
<h1>Foto gallery</h1>
<?php
if (isset($uploadResult)) {
echo "<p><strong>$uploadResult</strong></p>";
}
?>
<form action="" method="post" enctype="multipart/form-data" name="uploadImage" id="uploadImage">
<p>
<label for="image">Upload image:</label>
<input type="hidden" name="MAX_FILE_SIZE" value="<?php echo MAX_FILE_SIZE; ?>" />
<input type="file" name="images" id="imagesd" />
</p>
<p>
<input type="submit" name="upload" id="upload" value="Upload" />
</p>
</form>
<div id="maincontent">
<h2>Foto informatie</h2>
<form name="FotoInformatie" id="fotoInformatie" method="post" action="">
<table>
<tr>
<td align="center"><strong>Foto<strong></td>
<td align="center"><strong>Titel</strong></td>
<td align="center"><strong>Beschrijving</strong></td>
<td align="center"><strong>Homepage</strong></td>
</tr>
<?php
while ($rows=$result->fetch_assoc()) {
?>
<tr>
<td class="hide" align="center"><?php $id[]=$rows['id']; ?><?php echo $rows['id']; ?></td>
<td><img src="../img/thumbs/<?php echo $rows['filename']; ?>"></td>
<td align="center"><input name="title[]" type="text" id="title" value="<?php echo $rows['title']; ?>"></td>
<td align="center"><input name="caption[]" type="text" id="caption" value="<?php echo $rows['caption']; ?>"></td>
<td><input type="checkbox" name="checkboxHome[]" id="checkBoxHome" value="<?php echo ($rows['home'] == 1) ? 'checked="checked"' : ''; ?>"/></td>
</tr>
<?php
}
?>
<tr>
<td colspan="4" align="center">
<input type="submit" name="submit" value="Submit">
</tr>
</table>
</form>
</div>
</article> <!-- end of content -->
</div> <!-- end of container -->
<?php include('includes/footer.inc.php'); ?>
</body>
</html>
<?php
if(isset($_POST['submit'])) {
$title = $_POST['title'];
$caption = $_POST['caption'];
if ($_POST['checkboxHome'] == "") {
$checkboxHome[] = '0';
} else {
$checkboxHome[] = '1';
}
for($i=0;$i<$count;$i++){
$result1=mysqli_query($conn, "UPDATE gallery SET title='$title[$i]', caption='$caption[$i]', home='$checkboxHome[$i]' WHERE id='$id[$i]'");
header("location:/admin/foto-admin.php");
}
}
?>
The checkbox only works on the first row in my DB. When I select another record, only the first record in my db will be updated.
Another issue is that my checkbox won't be checked so I don't know based on my screen when a image is online or not. in the database I see a 1 of a 0.
I know that sql-injection is possible and I have to prepare the statements, but that is the next step when I get this checkbox-issue working.
Hope someone can help me with my code. It's giving me a headache.
Check these
Attribute name="id[]" for id field is not given. And it should get inside
if(isset($_POST['submit'])) {
$id = $_POST['id'];
}
Incorrect spelling in getting Post value
change
$checkboxHome = $_POST['checkboxHome'];
$checkboxFotoboek= $_POST['checkboxFotoboek'];
$checkboxDelete = $_POST['image_deleted'];
to
$checkboxHome = $_POST['checkBoxHome'];
$checkboxFotoboek= $_POST['checkBoxFotoboek'];
$checkboxDelete = $_POST['checkboxDelete'];
You are trying to get wrong value.
Your check-box name is checkBoxHome and you are trying to get $_POST['checkboxHome'] instead of $_POST['checkBoxHome'] .
Try $_POST['checkBoxHome'] and print it as print_r('checkBoxHome')
Same mistake in checkBoxFotoboek check-box.
try this
if(isset($_POST['submit'])) {
$title = $_POST['title'];
$caption = $_POST['caption'];
$checkboxHome = $_POST['checkBoxHome'];
$checkboxFotoboek= $_POST['checkBoxFotoboek'];
$checkboxDelete = $_POST['checkboxDelete'];
for($i=0;$i<$count;$i++){
$result1=mysqli_query($conn, "UPDATE gallery SET title='$title[$i]', caption='$caption[$i]', home='$checkboxHome[$i]', fotoboek='$checkboxFotoboek[$i]', image_deleted='$checkboxDelete[$i]' WHERE id='$id[$i]'");
header("location:/admin/foto-admin.php");
}
}
?>

Dynamically enable and disable DIV via MySQL

Last night I was trying to figure out how I can how I can dynamically enable and disable span#txtCaptchaDiv on my contact form at the very bottom, above the submit button.
So I added a new field to MySQL, called captcha where I wanted to 1 to show and 0 to hide
So if I add 1 to field captcha the following code will show on my form.php
<label for="code">Write code below > <span id="txtCaptchaDiv" style="color:#F00"></span><!-- this is where the script will place the generated code -->
<input type="hidden" id="txtCaptcha" /></label><!-- this is where the script will place a copy of the code for validation: this is a hidden field -->
<input type="text" name="txtInput" id="txtInput" size="30" />
If I add 0 to field captcha the captcha area will be blank on my form.php.
Can you guy help me out please?
here is my index.php code I currently have:
<?php
require_once("/config/database.php");
$con = mysql_connect($config["db_server"],$config["db_user"],$config["db_pass"]);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Email FORM</title>
</head>
<body>
<div style="width: 550px; text-align: center;">
<span style="filter:alpha(opacity=60); opacity:.6; padding-left: 10px;"><br />
<?php
$data = mysql_query("SELECT * FROM formrelated")
or die(mysql_error());
while($info = mysql_fetch_array( $data ))
Print " ".$info['welcomemsg'] . "";
?>
</span></div>
<form id="form1" name="form1" method="post" action="submit.php" onsubmit="return checkform(this);">
<table width="454" border="1" align="center" cellpadding="0" cellspacing="0">
<tr>
<td width="123">Name</td>
<td width="325">
<input name="name" type="text" />
</td>
</tr>
<tr>
<td height="21">Address</td>
<td><input name="adress" type="text" /></td>
</tr>
<tr>
<td height="21"> </td>
<td><input name="address2" type="text" /></td>
</tr>
<tr>
<td height="21">Email</td>
<td><input name="email" type="text" /></td>
</tr>
<tr>
<td height="21">Tel</td>
<td><input name="email" type="text" /></td>
</tr>
</table>
<!--- captcha code here--->
<center>
<table width="454" height="122" border="0" cellspacing="0" cellpadding="0" background="reCAPbg.png">
<tr>
<td height="73" colspan="2" align="center" valign="middle"><label for="code"><span id="txtCaptchaDiv" style="color:#333; font-size:18px;"></span><!-- this is where the script will place the generated code -->
<input type="hidden" id="txtCaptcha" /></label></td>
<td width="136" rowspan="2"> </td>
</tr>
<tr>
<td width="145"> type the code here:</td>
<td width="173" height="47" align="center"><input type="text" name="txtInput" id="txtInput" size="20" /></td>
</tr>
</table>
</center>
<!--- captcha code ends here--->
<input name="Submit" type="button" value="submit" />
</form>
<script type="text/javascript">
//Generates the captcha function
var a = Math.ceil(Math.random() * 9)+ '';
var b = Math.ceil(Math.random() * 9)+ '';
var c = Math.ceil(Math.random() * 9)+ '';
var d = Math.ceil(Math.random() * 9)+ '';
var e = Math.ceil(Math.random() * 9)+ '';
var code = a + b + c + d + e;
document.getElementById("txtCaptcha").value = code;
document.getElementById("txtCaptchaDiv").innerHTML = code;
</script>
<script type="text/javascript">
function checkform(theform){
var why = "";
if(theform.txtInput.value == ""){
why += "- Security code should not be empty.\n";
}
if(theform.txtInput.value != ""){
if(ValidCaptcha(theform.txtInput.value) == false){
why += "- Security code did not match.\n";
}
}
if(why != ""){
alert(why);
return false;
}
}
// Validate the Entered input aganist the generated security code function
function ValidCaptcha(){
var str1 = removeSpaces(document.getElementById('txtCaptcha').value);
var str2 = removeSpaces(document.getElementById('txtInput').value);
if (str1 == str2){
return true;
}else{
return false;
}
}
// Remove the spaces from the entered and generated code
function removeSpaces(string){
return string.split(' ').join('');
}
</script>
</body>
</html>
This will work for you... enjoy!
<?PHP
$query = mysql_query("SELECT captcha FROM formrelated WHERE id = '1'");
while ($row = mysql_fetch_assoc($query)) {
$captchathis = $row['captcha'];
if ($captchathis == "1") {
echo "YOUR HTML CODE HERE";
}
else {
echo "BLANK";
}
}
?>
Try it like this
<?PHP
if($mysqlResult['captcha'] === 1)
{
echo $myHtml;
}
?>
Where $mysqlResult is an array with the result from the query, $mysqlResult['captcha']is the value of the row captcha from your query and $myHtml is that HTML code you just showed on your answer.
Good luck! ;)
Reffer to
http://php.net/manual/en/
EDIT:
http://www.php.net/manual/en/language.types.array.php ( Array type on the manual )
http://www.php.net/manual/en/control-structures.if.php ( If control structure on the manual )
http://www.php.net/manual/en/ref.mysql.php ( MySQL native functions. deprecated. Preffer MySQLi )
http://www.php.net/manual/en/book.mysqli.php ( MySQLi extension )
http://www.php.net/manual/en/book.pdo.php ( PDO native php class )
Another answer to explain the basic construct of IF logic.
Suppose i have some condition i want to meet to do something; in this case, the following logic
SHOW my form with the basic inputs
IF condition 'captcha = 1' is met, SHOW input2 (captcha)
SHOW rest of the HTML
it would be like this in PHP
<?PHP
echo $myFormWithBasicInputs;
if($captcha === 1)
{
echo $input2;
}
echo $restOfHTML;
?>
In your case, $myFormWithBasicInput and $restOfHTML is already outputed as HTML. All you want to do is inject an PHP code in it to check if some condition is matched. It will be like this
<html>
<!-- MY FORM WITH BASIC INPUTS -->
<?PHP
$captcha = $mySQLresult['captchaRow'];
if($captcha === 1)
{
?>
<!-- CAPTCHA INPUT HERE -->
<?PHP
}
?>
<!-- REST OF HTML -->
</html>
be aware that this is an workaround with example code.
<?PHP
$mysql_query = "SELECT captcha FROM formrelated";
$captcha = $mySQLresult['captchaRow'];
if($captcha === 1)
{
?>
<!--- CODE---->
<table width="454" height="122" border="0" cellspacing="0" cellpadding="0" background="reCAPbg.png">
<tr>
<td height="73" colspan="2" align="center" valign="middle"><label for="code"><span id="txtCaptchaDiv" style="color:#333; font-size:18px;"></span><!-- this is where the script will place the generated code -->
<input type="hidden" id="txtCaptcha" /></label></td>
<td width="136" rowspan="2"> </td>
</tr>
<tr>
<td width="145"> type the code here:</td>
<td width="173" height="47" align="center"><input type="text" name="txtInput" id="txtInput" size="20" /></td>
</tr>
</table>
<?PHP
}
?>
<!-- REST OF HTML -->

Firefox does not submit my form

I have a php application that fetches the requests from mysql database and displays them for further approval. The form is fetched from send_req.php and is displayed inside the div on showrequests.php. This is the code for send_req.php
<table style="border:0;border-color:transparent">
<tr style="background-color:lightblue">
<td>Product ID</td>
<td>Name</td>
<td>Quantity</td>
<td><input type="checkbox" name="selectAll" /></td>
<td>Authorized Quantity</td>
</tr>
<form method="post" action="send_req.php">
<?php
$reqNum = $_POST['rId'];
echo "<h3>Request # $reqNum</h3>";
$showReqs = mysql_query("Select * from request where request_number='".$reqNum."' and status=0");
while($resultA = mysql_fetch_array($showReqs))
{
$rBy = $resultA['requested_by'];
$rTime = $resultA['request_time'];
$rId = $resultA['id'];
$pId = $resultA['product_id'];
$getPrName = mysql_query("select name from products where id='$pId'");
$prN = mysql_fetch_array($getPrName);
$prName = $prN['name'];
$rQuantity = $resultA['requested_quantity'];
$status = $resultA['status'];
?>
<tr>
<input type="hidden" name="rId[]" value="<?php echo $rId; ?>"/>
<td style="background-color:orange"><input type="text" name="prId[]" value="<?php echo $pId; ?>" readonly="readonly" style="border:0px"/></td>
<td style="background-color:orange"><input type="text" name="prName[]" value="<?php echo $prName; ?>" readonly="readonly" style="border:0px"/></td>
<td style="background-color:orange"><input type="text" name="quantity[]" value="<?php echo $rQuantity; ?>" readonly="readonly" style="border:0px"/></td>
<td style="background-color:orange"></td>
<td><input type="text" name="pQuantity[]" /></td>
</tr>
<?php }
?>
<tr>
<td></td>
<td></td>
<td></td>
<input type="hidden" name="rNum" value="<?php echo $reqNum; ?>" />
<td></td>
<td><input type="submit" name="submitReq" value="Send" id="submit_req" style="backgroundColor:Transparent;border:0;color:blue;width:100;"/></td>
</tr>
</form>
</table>
<?php
echo "Requested By:$rBy at ".substr($rTime,11,18)." ".substr($rTime,0,10);
?>
This is the showrequests.php page
<html>
<head>
<script type="text/javascript">
function getRequest(ob)
{
var id = ob.id;
if(window.XMLHttpRequest)
{
ajaxOb = new XMLHttpRequest();
}
else if(window.ActiveXObject)
{
ajaxOb = new ActiveXObject("Microsoft.XMLHTTP");
}
ajaxOb.open("POST", "send_req.php");
ajaxOb.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
ajaxOb.send("rId=" + id);
ajaxOb.onreadystatechange = function()
{
if(ajaxOb.readyState == 4)
{
if(ajaxOb.status == 200)
{
document.getElementById("showTable").innerHTML = ajaxOb.responseText;
}
}
}
}
</script>
</head>
<body>
<?php
$mysql_con = mysql_connect("localhost","root","") or die("Could not connect ".mysql_error());
$mysql_db = mysql_select_db("cart",$mysql_con) or die("Unable to select db ".mysql_error());
echo "<h2 align='center'>Pending Requests</h2>";
$showReq = mysql_query("Select distinct(request_number) as rNums from request where status=0");
?>
<div style="float:left;margin-right:15px;">
<br/>
<?php
while($result = mysql_fetch_array($showReq))
{
$rNum = $result['rNums'];
?>
<input type="button" name="fetchReq" id="<?php echo $rNum; ?>" value="<?php echo "Request # $rNum"; ?>" style="margin-bottom:5px;backgroundColor:Transparent;border:0;color:blue;width:100;text-Decoration:underline" onclick="getRequest(this)"/>
<?php
echo "<br/>";
}
?>
</div>
<div id="showTable" style="float: left">
</div>
</body>
</html>
My problem now is that everything works fine in chrome and IE but the form is not submitted when i click the submit button in firefox. I am using firefox 20.0.1. Update: I have removed the html,head and body tags from send_req.php
still not working
form is not allowed inside table. Please see also
Form inside a table
Regards,
Michael
Reminder : the structure of an HTML document is :
<!-- No div before html tag -->
<!DOCTYPE html> <!-- Doctype for HTML5 ; use whatever doctype you need -->
<html>
<head>
</head>
<!-- No div before body tag -->
<body>
<!-- Divs only belongs here -->
</body>
</html>
<!-- No div after html tag -->
If you don't follow this basic structure, you're forcing the browser to interpret your invalid code (+ quirks mode when you don't provide a doctype).
Some browser guess well what you tried to do, others don't, as Firefox might.
Please use a HTML validator as W3's validator to check your syntax.

Where do I set the $id variable in this PHP code?

I'm troubleshooting this code, I am very new to PHP, so any help would be appreciated.
I think the error is coming from not having the $id variable set anywhere, so where do I set it?
Here is the code:
// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name WHERE depot = 'plainview'";
$result=mysql_query($sql);
// Count table rows
$count=mysql_num_rows($result);
//error reporting
error_reporting(E_ALL); ini_set('display_errors', '1');
$result1 = false;
//update
if(isset($_POST['Submit'])){
for($i=0;$i<$count;$i++){
$sql1 = "UPDATE $tbl_name SET
available='".mysql_real_escape_string($_POST['available'][$i])."',
rent='".mysql_real_escape_string($_POST['rent'][$i])."',
corp_ready='".mysql_real_escape_string($_POST['corp_ready'][$i])."',
down='".mysql_real_escape_string($_POST['down'][$i])."',
gfs='".mysql_real_escape_string($_POST['gfs'][$i])."',
dateTime = NOW()
WHERE id='".$id[$i]."'";
$result1 = mysql_query($sql1) or die(mysql_error());
}
}
//redirect
if($result1){
header("location: success.php");
}
else
header("location: fail.php");
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<script language="JavaScript1.1" type="text/javascript">
<!--
function mm_jumpmenu(targ,selObj,restore){ //v3.0
eval(targ+".location='"+selObj.options[selObj.selectedIndex].value+"'");
if (restore) selObj.selectedIndex=0;
}
//-->
</script>
<title>Untitled Document</title>
</head>
<body>
<div>
<p>Plainview, North East Region</p>
<p>Select a different region: <select onchange="mm_jumpmenu('parent',this,0)" name="lostlist">
<option value="" selected="selected">Choose Your Depot</option>
<option value="plainview.php">Plainview</option>
<option value="worcrester.php">Worcrester</option>
</select></p>
</div><Br />
<table width="500" border="0" cellspacing="1" cellpadding="0">
<form name="form1" method="post" action="">
<tr>
<td>
<table width="700" border="0" cellspacing="1" cellpadding="0">
<tr>
<td>ID</td>
<td align="center"><strong>Product Name</strong></td>
<td align="center"><strong>Available</strong></td>
<td align="center"><strong>Rent</strong></td>
<td align="center"><strong>Corp Ready</strong></td>
<td align="center"><strong>Down</strong></td>
<td align="center"><strong>GFS</strong></td>
</tr>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td align="left"><?php $id[]=$rows['id']; ?><?php echo $rows['id']; ?></td>
<td align="left"><?php echo $rows['product']; ?></td>
<td align="center"><input name="available[]" type="text" id="available" value="<?php echo $rows['available']; ?>" size="5"></td>
<td align="center"><input name="rent[]" type="text" id="rent" value="<?php echo $rows['rent']; ?>" size="5"></td>
<td align="center"><input name="corp_ready[]" type="text" id="corp_ready" value="<?php echo $rows['corp_ready']; ?>" size="5"></td>
<td align="center"><input name="down[]" type="text" id="down" value="<?php echo $rows['down']; ?>" size="5" /></td>
<td align="center"><input name="gfs[]" type="text" id="gfs" value="<?php echo $rows['gfs']; ?>" size="5"></td>
</tr>
<?php
}
?>
<tr>
<td colspan="4" align="center"><input type="submit" name="Submit" value="Submit"></td>
</tr>
</table>
</td>
</tr>
</form>
</table>
<?php
mysql_close();
?>
</body>
</html>
Per the statement:
if($result1){
header("location: success.php");
}
else
header("location: fail.php");
Every time I try to access the page it redirects to fail.php, so why does it fail?
Either $_POST['Submit'] is not set, or $count is 0.
if($result1 === true){
header("location: success.php");
} else {
header("location: fail.php");
}
Maybe I missed a line, But I don't see where you put a value to $id... The first time you mention it is where you expect to receive a value from it.
If you don't post anything, $result1 will remain false.
If you post something, the test will only be done on the last iteration of your loop, so every other UPDATE could fail but the last one, it would be considered as a success.
Btw, why do you have some code after your redirection? It will obviously never be displayed.
You also need to exit() after any header('Location') to avoid unwanted code execution (code that would remain after the header call).
I think you might want to include the header()instructions INSIDE the if(isset($_POST['Submit'])) test to avoid being redirected just by loading the page without posting anything.
Location should also be a complete URL with protocol and domain, even if most browser will accept a partial own, this is violating some RFCs.
You also need to loop through results to defined $id:
while ($row = mysql_fetch_assoc($result)) $id[] = $row['primary_column_name'];

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