Develop Reference

How can I access the values of a checkbox outside of the form action that it is declared in?

I successfully created a script that will delete the rows of a table if a checkbox is checked on that row (the checkbox holds the rowID). The checkboxes and button to delete these rows are inside of the same form tags. Now I want to create another button that uses the value of the checkboxes to do a different update statement, but the values of the checkboxes are not appearing in $_POST on this separate page.
Does anyone know how to make the checkbox values accessible outside of the form action it is inside of? Here is my reduced code for the delete that works:
The function below is called on PickTicket.php to display a table.
Function DisplayPickTicket() {
$conn = getDBConnection();
$sql = "SELECT * FROM dbo.BK_NotesRecord WHERE StatusID = 1 ";
$stmt = sqlsrv_query( $conn, $sql );
if ( $stmt === false ) {
die( print_r( sqlsrv_errors(), true) );
}
echo '<form action="updatepickstatus.php" method="post">';
// Delete Checkbox header.
echo '<th class="table-header" style="width:5px;">';
echo 'Delete';
echo '</th>';
// Inventory number header.
echo '<th class="table-header" style="width:90px;">';
echo 'Inventory #';
echo '</th>';
//InventoryID Header
echo '<th class="table-header" style="width:40px;">';
echo 'InventoryID';
echo '</th>';
if (sqlsrv_has_rows($stmt)) {
while($row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC)) {
echo '<tr>';
//Delete checkbox
echo '<td class="cell"><div class="cell">';
echo '<input type = "checkbox" name="chkbox[]" value= "' .$row['InventoryID'].
'">';
echo '</td>';
// Inventory#
echo '<td class="cell"><div class="cell">';
echo $row["InventoryNumber"];
echo '</td>';
// InventoryID.
echo '<td class="cell"><div class="cell">';
echo $row["InventoryID"];
echo '</td>';
}
}
echo "<tr>";
echo "<td>";
echo "<input type='submit' name='submit' Value='Remove'>";
echo '</form>';
echo "</td>";
echo "</tr>";
This is updatepickstatus.php:
<?php
$serverName = "(local)";
$connectionOptions = array("Database"=>"Powerlink");
$conn = sqlsrv_connect( $serverName, $connectionOptions);
if( $conn === false ) {
echo "Connection failed!<br>";
die( print_r( sqlsrv_errors(), true));
}
if (isset($_POST['chkbox'])) {
foreach($_POST['chkbox'] as $Update) {
$sql = "UPDATE BK_NotesRecord set StatusID = '2' WHERE InventoryID LIKE '".$Update."'";
$stmt = sqlsrv_query( $conn, $sql );
//echo '$ids';
}
}
print_r($_POST);
?>
^^I want to accomplish this same basic task, but outside of updatepickstatus.php. When applying similar logic to check the values of the selected checkboxes on a different I get an empty array. Any thoughts?

Try declaring a variable and passing the value you want to use to it then use sessions to move it where ever you want.

How to send data and run a link with that data with one click

I want to send latitude and longitude data so I can search it up on Google Map. Those data has to be selected from a database but I can't figure how to do it in one click.
Here is my code.
<?php
$con = mysqli_connect("localhost","root","","project");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM table1");
echo "<table border='1'>
<tr>
<th>ID</th>
<th>Latitude</th>
<th>Longitude</th>
<th>Date</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['latitude'] . "</td>";
echo "<td>" . $row['longitude'] . "</td>";
echo "<td>" . $row['date'] . "</td>";
echo "<td><a href='./clicktest.php?id=".$row['id']."'>View Location</a></td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
And as you can see, it sends the id value of a row that I want to another page which is
<?php
$con=mysqli_connect("localhost","root","","project");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(isset($_GET['id'])) {
$result = "SELECT * FROM table1 WHERE id='".$_GET['id']."'" ;
$show = mysqli_query($con, $result);
while($row = mysqli_fetch_array($show))
{
$latitude = $row['latitude'];
$longitude = $row['longitude'];
echo "<a href='http://www.google.com/maps/place/".$latitude.",".$longitude."'>click here</a>";
}
mysqli_close($con);
}
?>
And finally can run a link to Google Map on this page. But what I want is to click on the 'view location' on the first page and jump to the google map page with data on that row without having to go through another page and another click.
I did some research and maybe this is about AJAX? And are there any ways to do it without using AJAX?(since I never use it before)
Thanks

I think you need to directly link to google maps page and use "target=_blank" attribute on it. for example:
echo 'View Location';

You can add as many parameter to the anchor tag as you like, so add the lat and long like this for example
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['latitude'] . "</td>";
echo "<td>" . $row['longitude'] . "</td>";
echo "<td>" . $row['date'] . "</td>";
echo "<td><a href='./clicktest.php?id=".$row['id']
.'&lat=' . $row['latitude']
. '&long' . $row['longitude']"'>View Location</a></td>";
echo "</tr>";
}

value of $_POST is always the last value of my <tr>

I have a problem in which I cannot solve. I have a simple page where I query all Users and list them in a table. When a user clicks on one of the table rows, it should be taken to another page where the user can edit information of the that they picked. The problem is that in my script, the $_POST value is always the value of the last
CODE
<?php
include "conn.php";
$pquery = "SELECT * FROM Patient NATURAL JOIN User ORDER BY LastName;";
$patientQuery = $conn->query($pquery);
if (mysqli_num_rows($patientQuery) == 0)
echo "<p>No patients found.</p>";
else{
while($assoc = $patientQuery->fetch_assoc()){
echo "<tr onclick = 'sub();'>";
echo "<td>";
echo $assoc['UserID'];
echo "<input type = 'hidden' name = 'UserID' value = '". $assoc['UserID'] ."' />";
echo "</td>";
echo "<td>";
echo $assoc['FirstName'];
echo "</td>";
echo "<td>";
echo $assoc['LastName'];
echo "</td>";
echo "</tr>";
}
}
?>
<script>
function sub(){
document.getElementById("edit").submit();
return false;
}
</script>

I've slightly modified your code - this should work:
<?php
include "conn.php";
$pquery = "SELECT * FROM Patient NATURAL JOIN User ORDER BY LastName;";
$patientQuery = $conn->query($pquery);
if (mysqli_num_rows($patientQuery) == 0)
echo "<p>No patients found.</p>";
else{
while($assoc = $patientQuery->fetch_assoc()){
echo "<tr onclick = 'sub(". $assoc['UserID'] .");'>";
echo "<td>";
echo $assoc['UserID'];
echo "</td>";
echo "<td>";
echo $assoc['FirstName'];
echo "</td>";
echo "<td>";
echo $assoc['LastName'];
echo "</td>";
echo "</tr>";
}
}
?>
<script>
function sub(UserID){
document.location.href = 'http://www.yourdomain.com/something.php?UserID='+UserID;
return false;
}
</script>

Javascript not working in AJAX response

Below code is the AJAX response to another page. I have added onclick event to table row and written javascript code to handle it. But javascript code doesn't work.Is this wrong way of coding or there is any problem in code. Suggest me a simple solution
<?php
echo '<script type=\"text/javascript\">
function clicked(){
alert("I am an alert box!");
}
</script>';
$q = $_GET['q'];
include 'db_connect.php';
$sql="SELECT name,address,mobile,email,pan,tan FROM client WHERE name = '$q'";
$sql_bill="SELECT clientname,financialyear,receiptno,amount,ddate,type,chequeno,category FROM billing WHERE clientname = '$q'";
$sql_total="SELECT SUM(amount) AS TotalAmount FROM billing";
$result = mysql_query($sql);
$result_bill = mysql_query($sql_bill);
$result_total = mysql_query($sql_total);
$total= mysql_fetch_array($result_total);
echo "<h4><b>Client details</b></h4><table align='center' border='2'>
<tr>
<th>Name</th>
<th>Address</th>
<th>Mobile</th>
<th>Email</th>
<th>PAN</th>
<th>VAT TIN</th>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['address'] . "</td>";
echo "<td>" . $row['mobile'] . "</td>";
echo "<td>" . $row['email'] . "</td>";
echo "<td>" . $row['pan'] . "</td>";
echo "<td>" . $row['tan'] . "</td>";
echo "</tr>";
}
echo "</table>";
echo "<h4><b>Payment received details</b></h4><table align='center' border='2'>
<tr>
<th>Client Name</th>
<th>Financial Year</th>
<th>Receipt No</th>
<th>Date</th>
<th>Type</th>
<th>Chequeno</th>
<th>Category</th>
<th>Amount</th>
</tr>";
while($row = mysql_fetch_array($result_bill))
{
echo "<tr onclick=\"clicked()\">";
echo "<td>" . $row['clientname'] . "</td>";
echo "<td>" . $row['financialyear'] . "</td>";
echo "<td>" . $row['receiptno'] . "</td>";
echo "<td>" . $row['ddate'] . "</td>";
echo "<td>" . $row['type'] . "</td>";
echo "<td>" . $row['chequeno'] . "</td>";
echo "<td>" . $row['category'] . "</td>";
echo "<td>" . $row['amount'] . "</td>";
echo "</tr>";
}
echo "<tr>";
echo "<td colspan=7>Total</td>";
echo "<td>".$total['TotalAmount']. "</td>";
echo "</tr>";
echo "</table>";
?>

Try this instead of your script code:
echo <<<EOD
<script type="text/javascript">
function clicked(){
alert("I am an alert box!");
}
</script>
EOD;

Edit 1:
Look in the console (F12), What happens when you click on a tr?
The following fiddle replicates your code and works:
http://jsfiddle.net/Yaj44/
Edit 2:
I think it has something to do with the way you call the data.
Put the function in your main page,
then call the AJAX.
Set innerHTML of specific div with responseData.
If you do it in that order, it might work.
Your way to adding this functionality is very dirty..
If you use JQuery, you can add on the end:
$('tr').click(function() {
alert('Do something on click!');
});
mysql_fetch_array is outdated, use mysqli (or PDO) instead.
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");
$query = "SELECT Name, CountryCode FROM City ORDER by ID DESC LIMIT 50,5";
if ($result = $mysqli->query($query)) {
/* fetch associative array */
while ($row = $result->fetch_assoc()) {
printf ("%s (%s)\n", $row["Name"], $row["CountryCode"]);
}
/* free result set */
$result->free();
}
/* close connection */
$mysqli->close();

Clickable data displayed from mysql using PHP

I'm not very good at JavaScript but here is my problem.
I have three pages: page1.php, page2.php, page3.php
On page1.php, I have a form for users to select grade level they want to view, then the action is performed on page2.php -- displaying the list of all students in that grade.
This is the code for page2.php
<?php
//database variables
require_once('admin_settings.php');
//these variables are from a form used to display the current data
$level = $_POST['level_group'];
$room = $_POST['room_group'];
$con=mysqli_connect("$host","$dbuser","$dbpass","$dbname");
mysqli_set_charset($con, "utf8");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT std_id, std_name FROM students WHERE std_level LIKE '$level%' AND std_room LIKE '$room';");
//table
echo"
<table border='1' id='mytable'>
<tr bgcolor = #99CCFF>
<th><b>Student ID</b></th>
<th><b>Name</b></th>
<th><b>Action</b></th>
</tr>";
//loop through the database
while($row = mysqli_fetch_array($result))
{
echo"<form action='view_one_student.php' method='post'>";
echo "<tr bgcolor = '#c0eae4n' id = 'listings'>";
echo "<td name= 'stdid'>" . $row['std_id'] . "</td>";
echo "<td>" . $row['std_name'] . "</td>";
echo "<td>" . '<input type="submit" value="view"> <input type="submit" value="sdq">' . "</td>";
echo "</tr>";
echo "</form>";
}
echo "</table>";
mysqli_close($con);
?>
The question...now is How can I write the code on page3.php so that when users clicks on view or sdq button next to each column, the student ID should be captured, send a request to the database, and query other data related to this particular student such as age, address, phone..etc. and display them on that page3.php

page2.php
You have to add a hidden input with the id of the student in page2.
<input type="hidden" value="'.$row["std_id"].'" name="std_id">
Change your action page to page3 (if that's where you want to display the student info: name, age...)
<form action='page3.php' method='post'>page2.php
The code for page2:
<?php
//database variables
require_once('admin_settings.php');
//these variables are from a form used to display the current data
$level = $_POST['level_group'];
$room = $_POST['room_group'];
$con=mysqli_connect("$host","$dbuser","$dbpass","$dbname");
mysqli_set_charset($con, "utf8");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT std_id, std_name FROM students WHERE std_level LIKE '$level%' AND std_room LIKE '$room';");
//table
echo"
<table border='1' id='mytable'>
<tr bgcolor = #99CCFF>
<th><b>Student ID</b></th>
<th><b>Name</b></th>
<th><b>Action</b></th>
</tr>";
//loop through the database
while($row = mysqli_fetch_array($result))
{
echo"<form action='page3.php' method='post'>";
echo "<tr bgcolor = '#c0eae4n' id = 'listings'>";
echo "<td name= 'stdid'>" . $row['std_id'] . "</td>";
echo "<td>" . $row['std_name'] . "</td>";
echo "<td> <input type='hidden' value='" . $row["std_id"] . "' name='std_id'>";
echo "<input type='submit' value='view'>";
echo "<input type='submit' value='sdq'></td>";
echo "</tr>";
echo "</form>";
}
echo "</table>";
mysqli_close($con);
?>
page3.php
A posible solution for page3 can be: (add your columns you need: age, address, phone..)
<?php
//database variables
require_once('admin_settings.php');
//these variables are from a form used to display the current data
$id = $_POST['std_id'];
$con=mysqli_connect("$host","$dbuser","$dbpass","$dbname");
mysqli_set_charset($con, "utf8");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM students WHERE std_id ='$id';");
echo '<h3>Student detail</h3>';
//table
echo"
<table border='1' id='mytable'>
<tr bgcolor = #99CCFF>
<th><b>Student ID</b></th>
<th><b>Name</b></th>
<th><b>Age</b></th>
</tr>";
//loop through the database
while($row = mysqli_fetch_array($result))
{
echo "<tr bgcolor = '#c0eae4n' id = 'listings'>";
echo "<td name= 'stdid'>" . $row['std_id'] . "</td>";
echo "<td>" . $row['std_name'] . "</td>";
echo "<td>" . $row['std_age'] . "</td>";
echo "</tr>";
echo "</form>";
}
echo "</table>";
mysqli_close($con);
?>

You can modify your code like this. A hidden field will send the student id to the page of view_one_student.php.
while($row = mysqli_fetch_array($result))
{
echo"<form action='view_one_student.php' method='post'>";
echo "<tr bgcolor = '#c0eae4n' id = 'listings'>";
echo "<td name= 'stdid'>" . $row['std_id'] . "</td>";
echo "<td>" . $row['std_name'] . "</td>";
echo '<input type="hidden" value="'.$row["std_id"].'" name="std_id">';
echo "<td>" . '<input type="submit" value="view"> <input type="submit" value="sdq">' . "</td>";
echo "</tr>";
echo "</form>";
}
echo "</table>";
mysqli_close($con);
On view_one_student.php you have to catch that with:
$_POST["std_id"]
And eventually input that into a sql sentense in view_one_student.php:
$sql = "SELECT * FROM <table> WHERE id=".$_POST["std_id"];
Thats the general idea.