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Selecting a specific column in a option box using Javascript

I'm using the following php code to create a select option box:
$TypeLU = mysqli_query($mysqli, "SELECT * FROM LookupList");
while($Row = mysqli_fetch_array($TypeLU)) {
$TypeOptions = $TypeOptions."<option>$Row[1] $Row[2]</option>";
}
In HTML it gets displayed as a list with 2 columns. If I select and item from the list the value would be the concat of both $Row[1] and $Row[2] which is fine for display purposes, but I want to be able to 'extract' for example only $Row[1] as being the 'bound' value which I can then use as refrence.
So in pure Javascript I want to be able to get the value of just $Row[1] for example:
var x = document.getElementById("selectbox").value;
// So x must be only $Row[1] and not the concat of $Row[1] $Row[2]
Thanks

Use:
$TypeOptions = $TypeOptions."<option value='$Row[1]'>$Row[2]</option>";
You may also want to consider escaping the variables in case they contain any < or " or other special HTML characters. See htmlspecialchars.
Full example with escaping:
$TypeLU = mysqli_query($mysqli, "SELECT * FROM LookupList");
while($Row = mysqli_fetch_array($TypeLU))
{
$Row[1] = htmlspecialchars($Row[1]);
$Row[2] = htmlspecialchars($Row[2]);
$TypeOptions = $TypeOptions."<option value='$Row[1]'>$Row[2]</option>";
}

HTML form select to drive PHP/SQLSRV results

I'm working on a very basic web page to facilitate data entry. I have it working with PHP talking to SQL Server to both enter and display data. At the simplest level, they'll select a store from a dropdown and enter some data, then hit submit.
I'm trying to dynamically display the last 7 days worth of data when their store is chosen from the form select input. How do I get the Select value to drive the SQL query? I found how to use onchange to fire a javascript function, which I can alert the value. Do I really need to then use jquery or AJAX to link that back to the PHP select query or is there a simpler way to approach this?
I'm trying to learn this all from scratch and am quite admittedly out of my element. Here are some snippets of what I've got:
...
<script>
function OnSelectChange(obj)
{
alert(obj.options[obj.selectedIndex].value);
}
</script>
...
<td align="right">Store</td><td width="125"><select style="width:100%" name="store" onchange="OnSelectChange(this)">
...
<?php
$tsql = "select * from test where DateStamp >= getdate()-7";
$stmt = sqlsrv_query($conn, $tsql);
//generate table view of data
echo "<table>";
echo "<tr><td>Store ID</td><td>Date</td><td>Data</td></tr>";
while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC))
{
$StoreID = $row['StoreID'];
$DateStamp = $row['DateStamp'];
$Donations = $row['Data'];
echo "<tr><td>".$StoreID."</td><td>".$DateStamp."</td><td>".$Data."</td></tr>";
}
echo "</table>";
?>
I've found a lot of helpful information to get me this far, but am stuck now.
Thank you in advance.

Got it! For anyone else looking, here are relevant snippets of code:
<style>
.hideable {
visibility: collapse;
}
</style>
...
<script language="JavaScript">
function OnSelectChange(cl)
{
var els = document.getElementsByClassName('hideable');
for(var i=0; i<els.length; ++i){
var s = els[i].style;
s.visibility = 'collapse';
};
var els = document.getElementsByClassName(cl);
for(var i=0; i<els.length; ++i){
var s = els[i].style;
s.visibility = 'visible';
};
}
</script>
...
<select style="width:100%" name="store" onchange="OnSelectChange(this.value)">
...
echo "<tr class='hideable ".$StoreID."'><td>".$StoreID."</td><td>".$DateStamp."</td><td>".$Data."</td></tr>";
Leveraging dual classes upon the table (tr) creation and the collapse CSS style property was key.

remain the value of input after hide

I have an edit function where their is a function that has hide/show if i click edit. My problem is how can i remain the value of row if im going to hide it?
for example i have this dialog
and decided to edit sample 1(first row)
and then i decided i realize that i dont want to edit sample 1 then close it (by clicking edit again) then i want to edit sample 5 but i got this error
here is my script
//show and hide update button checker
function update_contain(){
var row = jQuery(".beneficiaries_rows input[type='text']:visible").length;
if(row > 0){
jQuery('.ui-dialog-buttonpane button:contains("Update")').button().show();
}else{
jQuery('.ui-dialog-buttonpane button:contains("Update")').button().hide();
}
}
//beneficiaries edit
jQuery(".edit_beneficiaries").click(function(){
var row = jQuery(this).closest(".beneficiaries_rows");
var show = row.find(".hide");
var hide = row.find(".show");
if(jQuery(show).is(":visible")){
jQuery(show).css({"display":"none"});
jQuery(hide).css({"display":"inline-block"});
}else{
jQuery(hide).css({"display":"none"});
jQuery(show).css({"display":"inline-block"});
}
update_contain();
});
HTML
<table style="border: 2px solid black;margin:auto;">
<tr>
<th style="width:145px;"><center>Name<center></th>
<th><center>Action</center></th>
</tr>
<?php
while($row1 = mysql_fetch_array($result1)){
echo "<tr class='beneficiaries_rows' id='".$row1['id']."' data-id='".$row1['ben_id']."'>";
echo "<td><input class='hide' type='text' name='bename_update' value='".$row1['name']."'></div>";
echo "<div class='show'>".$row1['name']."</td>";
echo "<td>";
echo "<center><button class='del_beneficiaries'>X</button><button class='edit_beneficiaries'>Edit</button></center>";
echo "</td>";
echo "</tr>";
}
?>
</table>
P.S im not good in english thats why im posting a picture to elaborate
my question

You'll want to set the text inside your label with something like this
jQuery("[selector for your label]").html(jQuery("[selector for input]").val());
I suggest doing this on click, just before you hide the input.
edit; since you're not really following:
jQuery(".edit_beneficiaries").click(function(){
var row_display = jQuery(this).parents("tr").find(".show");
var row_edit = jQuery(this).parents("tr").find(".hide");
jQuery(row_display).html(jQuery(row_edit).val());
});

Text not marking up

I am writing a page which allows a tournament director (TD) to add players to an event. When the TD goes to search for a player I want the search results to appear on keyup below the search input (without needing to refresh the page).
At this stage I have 2 php pages and 1 js page and have the live search happening, but the output is not getting marked up. I have written a piece of code to convert a PHP $_SESSION[''] array to a javascript array, but my brain is frying and I'm not sure where to go from here.
I effectively want the processed version of this code as output (presuming the SQL returns only 1 player named John Smith)
<tr></td> 123 </td><td> John Smith </td></tr>
My main php page has:
<table border='1px'>
<tr><th colspan='6'> <?php echo ($eName . " - " . $vn); ?></th></tr>
<tr><th>Player ID</th>
<th>Player Name</th>
<th>Place</th>
<th>Points</th>
<th>Cash</th>
<th>Ticket?</th></tr>
<tr><td colspan='3'>Search <input type='text' id='newsearch'</input>
</table>
<br>
<div id="newsearch-data"> </div>
<script>
var players = [];
</script>
<?php
//Makes php array into js array
foreach ($_SESSION['players'] as $id){
echo "<script> players.push('" . $id . "') </script>";
}
?>
<script src="http://code.jquery.com/jquery-1.8.0.min.js"> </script>
<script src="global.js"> </script>
js page:
$('input#newsearch').on('keyup', function(){
var newsearch = $('input#newsearch').val();
if ($.trim(newsearch)!= "") {
$.post('newsearch.php', {newsearch: newsearch}, function(data) {
$('div#newsearch-data').text(data);
});
}
});
And 2nd php page
<?php
session_start();
unset($_SESSION['players']);
if (isset($_POST['newsearch']) === true && empty($_POST['newsearch'] === false)){
require 'dbconnect.php';
$term = $_POST['newsearch'];
$terms = "%" . $term . "%";
$_SESSION['players'] = array();
$query = ("SELECT * FROM players WHERE Username LIKE '$terms'");
$run_query = mysqli_query($dbcon, $query);
while ($dbsearch = mysqli_fetch_assoc($run_query))
{
array_push ($_SESSION['players'],$dbsearch['PlayerID']);
echo "<tr><td>" . $dbsearch['PlayerID'] . "</tr></td>";
}}?
How can I output a new row on a html table rather than raw html code?

replace $('div#newsearch-data').text(data);
with $('div#newsearch-data').html(data);
$.text() escapes the input string and you should use it when you know the input is a string, and that you want to preset it "as is". If the input includes actual HTML that you want to inject into the DOM, $.html() is your friend :-)

You need .html() to render output as a HTML rather than .text(), because .text() will output the result as a raw data, then try this solution instead :
$('div#newsearch-data').html(data);
Live example - credited to #Peter Bailey for live example.
References
http://api.jquery.com/html/ - .html()
http://api.jquery.com/text/ - .text()

how to display only 4 result in each page using html & sqlite

This is me once again as I got some error in previous question in the code. So asking this question once again.
This is code from phonegap app from index html page. I don't know how to get only 4 result from database at each page when a sqlite query processed?
Also I want to add next page button. When clicking on this button next 4 result from database should come. This is code.
function querySuccess(tx, results){
var len = results.rows.length;
var output = '';
for (var i=0; i<len; i++){
output = output + '<li id="' + results.rows.item(i).id + '">' + results.rows.item(i).list_action + '</li>';
}
messageElement.html('<p>There are ' + len + ' items in your list:</p>');
listElement.html('<ul>' + output + '</ul>');
}

Dividng your solution into 3 phases
Phase 1: Using LIMIT Clause you can limit number of rows to be displayed
SELECT expressions FROM tables WHERE conditions ORDER BY expression [ ASC | DESC ] LIMIT number_rows OFFSET offset_value;
For Example:
SELECT employee_id, last_name, first_name
FROM employees
WHERE favorite_website = 'divyashah.in'
ORDER BY employee_id DESC
LIMIT 4;
Phase 2:
As the code provided by you is not efficient to explain but still... Now for your Next button
1) Fetch number of rows from your database table (mysql_num_rows).
2) Store that number in a Variable say 'a'.
3) Divide it(a) with 4 and store it in variable 'b'.
4) Use if > if(b!=0) {display next button} else {no need to display}
Phase 3:
This will fetch your first four rows i.e. from 0 to 4.
<?PHP
$fetch = mysql_query("SELECT * FROM table LIMIT 0, 4")or
die(mysql_error());
?>
Now how can you make the next page show the next 4 records?
you simply have to store the value of the starting row in a variable and pass it in the URL as a GET variable. Also have to check if there was a value already passed or not so we can set a default value in case it wasn't (zero to start from first row):
<?PHP
//check if the starting row variable was passed in URL or not
if (!isset($_GET['startrow']) or !is_numeric($_GET['startrow'])) {
//we give the value of the starting row to 0 because nothing was found in URL
$startrow = 0;
//otherwise we take the value from the URL
} else {
$startrow = (int)$_GET['startrow'];
}
?>
Now your query should have this new variable ($startrow) in the LIMIT clause
<?PHP
//this part goes after the checking of the $_GET var
$fetch = mysql_query("SELECT * FROM table LIMIT $startrow, 4")or
die(mysql_error());
?>
Now to see the next 4 records you should have a link which will add 4 to $startrow so you can view the next 4 records
<?PHP
//now this is the link..
echo 'Next';
?>