PHP show data from Oracle database in a listbox - php

I want to show data from a Oracle database into a listbox.
, but I don't no how to do that.
Now I'm using a textbox and that works good.
This is my HTML code
<form name="form1" method="get" action="Get_opdracht.php"
Opdrachtnummer: <br /> <input id="Password1" type="number" name="nummer1" required="required"/>
<input type="submit" name="submit1" value="Zoeken" />
<hr />
</form>
PHP code (get_opdracht.php)
// database connect
$conn = oci_connect('username', 'password', 'connect');
// variable textbox
$username = $_GET['nummer1'];
// SELECT query
$array = oci_parse($conn, "SELECT * FROM OPD_VW, MDW_VW WHERE OPD_OPDRACHTNUMMER = '$username'");
$query = oci_execute($array);
//show data on page
while (($row = oci_fetch_array($array, OCI_BOTH)) != false) {
echo "<h1>Opdrachtnummer: " . $row['OPD_OPDRACHTNUMMER'] . "</h1><p> <b>Status: </b>" . $row['OPD_STATUS'] . "<p><b>Registratiedatum: </b>" . $row['OPD_REGISTRATIEDATUM'] . "<p><b>Einddatum: </b>" . $row['OPD_EINDDATUM'] . "<p><b>BTW tarief: </b>". $row['OPD_BTW_TARIEF'] . "<p><b>Totale contractsom: €</b>" . $row['OPD_TOTALE_CONTRACTSOM'] . "<p><b>Percentage gerealiseerd: </b>" . $row['OPD_PERCENTAGE_GEREALISEERD'] . "%";
oci_free_statement($array);
oci_close($conn);

To create a listbox you need to use the select multiple as shown below.
<select name="myselect" multiple="multiple">
<option value="value">OPTION</option>
<option value="value">OPTION</option>
</select>

Related

php mysql based on user input form

I am having trouble using a checkbox to select one or multiple fields of data for PHP/AJAX to process and display. I have the PHP/AJAX working great on my <select>s but as soon as I try setting up the checkbox all hell breaks lose.
I also am very unsure on how to further prevent SQL injection on the site so if anyone could fill me in a little more about this I would GREATLY appreciate it! I read the link I was provided and just don't understand how bid_param or PDO works exactly.
The ajax script:
(I can't seem to insert the ajax/js so I'll leave a link to the live site)
Link to Agent search page
My php page that displays the data:
<div id="bodyA">
<h1>Find a Local OAHU Agent.</h1>
<!-- This is where the data is placed. -->
</div>
<div id="sideB">
<div class="sideHeader">
<em>Advanced Search</em>
</div>
<form class="formC">
<label for="last">Last Name</label><br />
<select id="last" name="Last_Name" onChange="showUser(this.value)">
<?php
include 'datalogin.php';
$result = mysqli_query($con, "SELECT DISTINCT Last_Name FROM `roster` ORDER BY Last_Name ASC;");
echo '<option value="">' . 'Select an Agent' .'</option>';
while ($row = mysqli_fetch_array($result)) {
echo '<option value="'.$row['Last_Name'].'">'.$row['Last_Name'].'</option>';
}
?>
</select>
<label for="company">Company</label><br />
<select id="company" name="users" onChange="showUser(this.value)">
<?php
include 'datalogin.php';
$result = mysqli_query($con, "SELECT DISTINCT Company FROM `roster` ORDER BY Company ASC;");
echo '<option value="">' . 'Select a Company' .'</option>';
while ($row = mysqli_fetch_array($result)) {
if ($row['Company'] == NULL) {
} else {
echo '<option value="'.$row['Company'].'">'.$row['Company'].'</option>';
}
}
?>
</select>
<label for="WorkCity">City</label><br />
<select id="WorkCity" name="WorkCity" onChange="showUser(this.value)" value="city">
<?php
include 'datalogin.php';
$result = mysqli_query($con, "SELECT DISTINCT WorkCity FROM `roster` ORDER BY WorkCity ASC;");
echo '<option value="">' . 'Select a City' .'</option>';
while ($row = mysqli_fetch_array($result)) {
echo '<option value="'.$row['WorkCity'].'">'.$row['WorkCity'].'</option>';
}
?>
</select>
<label for="WorkZipCode">Zip Code</label><br />
<select id="WorkZipCode" name="WorkZipCode" onChange="showUser(this.value)">
<?php
include 'datalogin.php';
$result = mysqli_query($con, "SELECT DISTINCT WorkZipCode FROM `roster` ORDER BY WorkZipCode + 0 ASC;");
echo '<option value="">' . 'Select a Zip Code' .'</option>';
while ($row = mysqli_fetch_array($result)) {
echo '<option value="'.$row['WorkZipCode'].'">'.$row['WorkZipCode'].'</option>';
}
?>
</select>
<label for="agent">Agent Expertise</label><br />
<label for="ancillary"><input type="checkbox" value="Ancillary" name="Ancillary[]" id="ancillary" />Ancillary</label><br />
<label for="smallgroup"><input type="checkbox" value="Smallgroup" name="Smallgroup[]" id="smallgroup" />Small Group</label><br />
<label for="largegroup"><input type="checkbox" value="LargeGroup" name="LargeGroup[]" id="largegroup" />Large Group</label><br />
<label for="medicare"><input type="checkbox" value="Medicare" name="Medicare[]" id="medicare" />Medicare</label><br />
<label for="longterm"><input type="checkbox" value="LongTerm" name="LongTerm[]" id="longterm" />Long Term Care</label><br />
<label for="individual"><input type="checkbox" value="Individual" name="Individual[]" id="individual" />Individual Plan</label><br />
<label for="tpa"><input type="checkbox" value="TPASelfInsured" name="TPASelfInsured[]" id="tpa" />TPA Self Insured</label><br />
<label for="ppaca"><input type="checkbox" value="CertifiedForPPACA" name="CertifiedForPPACA[]" id="ppaca" />Certified for PPACA</label><br />
</form>
</div>
My php page that pulls the info and places it into a container on the page:
$q = (isset($_GET['q'])) ? $_GET['q'] : false; // Returns results from user input
include 'datalogin.php'; // PHP File to login credentials
$sql="SELECT * FROM `roster` WHERE Company = '".$q."' OR Last_Name = '".$q."' OR WorkCity = '".$q."' OR WorkZipCode = '".$q."' ORDER BY Last_Name ASC";
$result = mysqli_query($con,$sql) // Connects to database or die("Error: ".mysqli_error($con));
echo "<h1>" . "Find a Local OAHU Agent." . "</h1>";
while ($row = mysqli_fetch_array($result)) { // Gets results from the database
echo "<div class='agentcon'>" . "<span class='agentn'>" . "<strong>".$row['First_Name'] . " " .$row['Last_Name'] . "</strong>" . "</span>" . "" . "<span class='email'>".$row['Email'] . "</span>" . "" ."<div class='floathr'></div>";
if ($row['Company'] == NULL) {
echo "<p>";
}
else {
echo "<p>" . "<strong>" .$row['Company'] . "</strong>" . "<br>";
}
echo $row['WorkAddress1'] . " " .$row['WorkCity'] . "," . " " .$row['WorkStateProvince'] . " " .$row['WorkZipCode'] . "<br>";
if ($row['Work_Phone'] !== NULL) {
echo "<strong>" . "Work" . " " . "</strong>" .$row['Work_Phone'] . "<br>";
}
if ($row['Fax'] !== NULL) {
echo "<strong>" . "Fax" . " " . "</strong>" .$row['Fax'] . "<br>";
}
echo "<strong>" . "Agent Expertise:" . "</strong>";
if ($row['Ancillary'] == 1) {
echo " " . "Ancillary" . "/";
}
if ($row['SmallGroup'] == 1) {
echo " " . "Small Group" . "/";
}
if ($row['IndividualPlans'] == 1) {
echo " " . "Individual Plans" . "/";
}
if ($row['LongTermCare'] == 1) {
echo " " . "Long Term Care" . "/";
}
if ($row['Medicare'] == 1) {
echo " " . "Medicare" . "/";
}
if ($row['LargeGroup'] == 1) {
echo " " . "LargeGroup" . "/";
}
if ($row['TPASelfInsured'] == 1) {
echo " " . "TPA Self Insured" . "/";
}
if ($row['CertifiedForPPACA'] == 1) {
echo " " . "Certified For PPACA";
}
echo "</p>" . "</div>";
}
mysqli_close($con);
?>
I appreciate any and all help on this topic! Any time I add the checkbox values to my php file it ends up displaying everyone in the database for all fields in the form.
I am also trying to prevent sql injection on this but how can a user do this if I don't have a field the user can input text into?
EDIT As of today I gave a try with using jQuery to activate the checkboxes and then call some AJAX.
Here is the script I wrote and it is pulling an agent, just not everyone that has that "expertise".
$('input').click(function() {
$.ajax({
url: "process.php",
data: { value: 1},
success: function (data) {
$('#bodyA').html(data);
}
});
});
Here's a quick example of something I recently worked on in which I needed to loop through multiple checkboxes and pass those values into a SQL statement. Although this example happens on a button click, hopefully its something along the lines of what you are trying to accomplish, or at least at start... :)
<?php
$array = array();
if (isset($_POST['medicare'])) {
foreach ($_POST['medicare'] as $value) {
array_push($array, $value);
}
}
// this will return the value of each selected checkbox, separating each with a comma
$result = implode(",", $array);
// if you want to loop through each individually (for example pass each into a SQL statement)
foreach ($_POST['medicare'] as $value) {
// Do your SQL here
// $value will be the value of each selected checkbox (Smallgroup, Largegroup, etc.)
$sql = "insert into tablename(fieldname) values ('$value')"; // just an example
}
?>
<input type="checkbox" name="medicare[]" id="smallgroup" value="Smallgroup" />
<label for="smallgroup">Small Group</label>
<br />
<input type="checkbox" name="medicare[]" id="largegroup" value="Largegroup" />
<label for="largegroup">Large Group</label>
<br />
<input type="checkbox" name="medicare[]" id="medicare" value="Medicare" />
<label for="medicare">Medicare</label>
<br />
<input type="checkbox" name="medicare[]" id="individualplan" value="IndividualPlan" />
<label for="individualplan">Individual Plan</label>
<br />
<input type="submit" value="Submit" id="btnSubmit" name="btnSubmit" />
UPDATE
Instead of setting one variable, try setting a variable for each select control and putting your SQL statement in a foreach loop. I just tested this with some dummy data and didn't have any issues with it.
<?php
$lastname = (isset($_GET['Last_Name'])) ? $_GET['Last_Name'] : false;
$users = (isset($_GET['users'])) ? $_GET['users'] : false;
$workCity = (isset($_GET['WorkCity'])) ? $_GET['WorkCity'] : false;
$WorkZipCode = (isset($_GET['WorkZipCode'])) ? $_GET['WorkZipCode'] : false;
foreach ($_GET['medicare'] as $value) {
//echo $value;
$sql="SELECT * FROM roster WHERE Company = '$users' OR Last_Name = '$lastname' OR WorkCity = '$workCity' OR WorkZipCode = '$WorkZipCode' OR Ancillary = '$value' ORDER BY Last_Name ASC";
}
...continue as you were...
?>
I DID IT!! Wohoo! I ended up just making a separate php page called expertise.php to process the checkboxs using jquery/ajax.
The jQuery that achieved this: (Thank god I went onto the jQuery website to look up functions!)
$('input').click(function() {
$.ajax({
url: "expertise.php",
data: { value: 1},
success: function (data) {
$('#bodyA').html(data);
}
});
});
The PHP page is the same as my process.php page except for the sql:
$sql="SELECT * FROM `roster` WHERE Ancillary = '1' AND SmallGroup = '1' AND CertifiedForPPACA = '1' ORDER BY Last_Name ASC";
If anyone would enlighten me more on making this better protected against sql injections, feel free to!
Agent Search Page
Well I at least got both parts of the search working but a new problem has arose :p
Now in the sql I can use AND or OR, with AND it pulls only agents that have everyone of those expertise and with OR it seems to pull everyone. Any ideas?

advice on php post operation not running

I have donation page which when the user clicks donate it posts the data to a php file named test.php I am trying this out my first trying to echo the first name and last name but this is not working ultimately I want this php page to run a MySQL query to update the total_Donation row within a database, here is my main php page first.
Database code which sits at top of file
<?php
$con = mysql_connect("localhost","root","null");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("snr", $con);
$names_sql = "SELECT first_Name, last_Name FROM donate WHERE user_ID > 0";
$names_query = mysql_query($names_sql)or die(mysql_error());
$rsNames= mysql_fetch_assoc($names_query);
if(isset($_POST['donation']) && $_POST['donation'] != '')
{
$donation = mysql_real_escape_string($_GET['donation']);
$fname = mysql_real_escape_string($_GET['first_Name']);
$lname = mysql_real_escape_string($_GET['last_Name']);
$donate_sql = "UPDATE `donate` SET donate_Total = donate_Total + '{$donation}' WHERE first_Name = '{$fname}' AND last_Name = '{$lname}'";
}
mysql_close($con);
?>
Here is my form section of html
form method ="post" action="test.php">
<table>
<tr><td><label>Runner:</label></td>
<td>
<select>
<?php do{?>
<option> <?php echo $rsNames['first_Name'];?> <?php echo $rsNames['last_Name'];?></option>
<?php } while ( $rsNames= mysql_fetch_assoc($names_query))?>
</select>
</td>
</tr>
<tr><td><label>Donation £</label></td><td><input type="text" maxlength="9" value="0.00" name="donation"/></td></tr>
<tr><td><input id="submit" type="submit" value="DONATE"/></td></tr>
</table>
</form>
the option gets all the first names and last names fine when the user hits donate I want it to run the $donation_sql but all i get are errors saying unidentified index, I'm even trying the below in the test.php to simply just echo the first_Name this is giving the same error.
<?php
echo $_POST['first_Name'];
?>
Can someone please help me with this, thanks.
index.php
<?php
$con = mysql_connect("localhost","root","null");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("snr", $con);
$names_sql = "SELECT first_Name, last_Name FROM donate WHERE user_ID > 0";
$names_query = mysql_query($names_sql)or die(mysql_error());
?>
<form method ="post" action="test.php">
<table>
<tr><td><label>Runner:</label></td>
<td>
<select name="name">
<?php
while($list = mysql_fetch_array($names_query))
{
?>
<option value="<?php echo $list['first_Name'] . ' ' . $list['last_Name']; ?>">
<?php echo $list['first_Name'] . ' ' . $list['last_Name']; ?>
</option>
<?php
}
?>
</select>
</td>
</tr>
<tr><td><label>Donation £</label></td><td><input type="text" maxlength="9" value="0.00" name="donation" /></td></tr>
<tr><td><input id="submit" type="submit" name="send" value="DONATE"/></td></tr>
</table>
</form>
test.php
<?php
$con = mysql_connect("localhost","root","null");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("snr", $con);
if(isset($_POST['donation']) && $_POST['donation'] != '')
{
$names = explode(' ',$_POST['name']);
$first_name= $names[0];
$last_name= $names[1];
$donation = mysql_real_escape_string($_POST['donation']);
$fname = mysql_real_escape_string($first_name);
$lname = mysql_real_escape_string($last_name);
$donate_sql = "UPDATE `donate` SET donate_Total = donate_Total + '" .$donation. "' WHERE first_Name = '" .$fname. "' AND last_Name = '" .$lname. "'";
echo 'DEBUG (remove after OK): <br>' .$donate_sql. '<br>';
$res = mysql_query($donate_sql);
echo 'Thanks ' .$first_name. ' ' .$last_name. '<br>';
}
mysql_close($con);
?>
That´s it!
\make sure you set name for select and you have valua attr in option tag
<select name="first_Name">
<otpion value="<?php echo $rsNames['first_Name'];?>"><?php echo $rsNames['first_Name'];?>
<?php echo $rsNames['last_Name'];?>
</option>
</select>
YOu need to give a name attribute to the select:
<select name="first_Name">
<?php while ( $rsNames= mysql_fetch_assoc($names_query)):?>
<option value="<?php echo htmlspecialchars($rsNames['first_Name']).' '.htmlspecialchars($rsNames['last_Name']);?>"> [option displayed to the user here]</option>
<?php endwhile;?>
</select>
And of course use the $_POST array, not the $_GET, since you're using the POST method.

retieve data from mysql and display in form

I have a radio box and 2 drop down menus which when submitted save to mysql. The radio box is either yes or no and the 2 dropdowns where created in html. It is currently fully working and saves all the data.
What I now wish to do is when a user logs back in, it will show what they have previously selected (if they have).
PHP SCRIPT:
<?php
session_start();
require_once("config.php");
if(!isset($_SESSION['username'])){
header('Location: login.php');
exit;
}else{
$sql = "SELECT attendance1 FROM user WHERE username = '".mysql_real_escape_string($_SESSION['username'])."'";
$res = mysql_query($sql);
$row = mysql_fetch_array($res);
if(($row[0] == "Yes") || ($row[0] == "No")){
header("Location: errorsubmit.html");
exit;
}
}
if(isset($_POST['submit'])){
$sql = "UPDATE user SET attendance1 = '" . mysql_real_escape_string($_POST['attendance1']) . "' WHERE username = '" . mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());
$sql = "UPDATE user SET colour1= '" . mysql_real_escape_string($_POST['colour1']) . "' WHERE username = '" . mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());
$sql = "UPDATE user SET shade1= '" . mysql_real_escape_string($_POST['shade1']) . "' WHERE username = '" . mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());
header("Location: thanks.html", true, 303);
}
?>
FORM:
<form>
<input name="attendance1" type="radio" id="Yes" value="Yes" checked="checked"/>Yes
<br />
<input name="attendance1" type="radio" id="No" value="No" />No
</h3></td>
<td>
<select name="colour1" id="colour1" >
<option selected="selected">Please Select</option>
<option>Red</option>
<option>White</option>
<option>Green</option>
</select>
</td>
<td><h3>
<select name="shade1" id="shade1" >
<option selected="selected">Please Select</option>
<option>light</option>
<option>heavy</option>
</select>
<td> </td>
<td><label>
<input type="submit" name="submit" id="button" value="Submit" />
</label></td>
</tr>
</table>
Try with below:
you need to fetch values from database and match them with select box values to show them selected.
<select name="shade1" id="shade1" >
<option>Please Select</option>
<option value="light" <?php if($val=='light') echo 'selected'; ?>>light</option>
<option value="heavy"<?php if($val=='heavy') echo 'selected'; ?>>heavy</option>
</select>
Here $val is variable having value retrieved from database.
while adding you should have :
<option value="light" >light</option>
<option value="heavy">heavy</option>
You just have to check if the value in the database has the value of the option field, and if so you echo a "selected='true'" to your option tag. Like
<option <?php if($row["column_name"] == "light") echo "selected=\"true\""; ?>>light</option>

assign database value as selected value in a drop down box [duplicate]

This question already exists:
Closed 11 years ago.
Possible Duplicate:
Populating a dynamic drop down menu from a MySQL database
my form successfully updates mysql database but when the user logs back in I want to assign database value as selected value in a drop down box to what the user has selected before which does not work.
PHP:
<?php
session_start();
require_once("config.php");
if(!isset($_SESSION['username'])){
header('Location: login.php'); /
}
}
if(isset($_POST['submit'])){
$sql = "UPDATE user SET attendance1 = '" . mysql_real_escape_string($_POST['attendance1']) . "' WHERE username = '" . mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());
$sql = "UPDATE user SET food1 = '" . mysql_real_escape_string($_POST['food1']) . "' WHERE username = '" . mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());
$sql = "UPDATE user SET drink1 = '" . mysql_real_escape_string($_POST['drink1']) . "' WHERE username = '" . mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());
header("Location: thanks.html", true, 303);
}
$row2 = "SELECT * FROM user WHERE username = '" . mysql_real_escape_string($_SESSION['username']) . "'";
$result = mysql_query($row2) or die("Error in SQL: " . mysql_error());
$row3 = mysql_fetch_array($result);
echo $row3['shade1'];
?>
FORM:
<form>
<input name="attendance1" type="radio" id="Yes" value="Yes" checked="checked"/>Yes
<br />
<input name="attendance1" type="radio" id="No" value="No" />No
</h3></td>
<td>
<select name="colour1" id="colour1" >
<option selected="selected">Please Select</option>
<option>Red</option>
<option>White</option>
<option>Green</option>
</select>
</td>
<td><h3>
<select name="shade1" id="shade1" >
<option selected="selected">Please Select</option>
<option value="Light" <?php if($row2['shade1']=="Light") { echo "selected"; }?>>Light</option>
<option value="Heavy" <?php if($row2['shade1']=="Heavy") { echo "selected"; }?>>Heavy</option>
</select>
<td> </td>
<td><label>
<input type="submit" name="submit" id="button" value="Submit" />
</label></td>
</tr>
</table>
</form>
You can use JQuery:
1) Use the $.ajax call in $(document).ready() to get the values you want
2) Set them like this:
Say your <select></select> element has an id of "options", set it's currently selected value like this:
$("#options").val(optionToBeSelected);
where optionToBeSelected is the value attribute of the <option></option> element you want to become selected.
This code should work for you.
If anyone sees a major/minor flaw in my answer, please let me know!

Troubleshooting HTML and PHP / MySQL

Long time reader, first time poster. I am a novice PHP enthusiast, and I have a page that I have been working. Right now I have the DB connection working well and my SELECT statement is giving me the info needed. My problems are two fold (maybe more after this post; set your phasers to cringe):
At one point, I had the INSERT working, but it suddenly stopped and no amount of tweaking seems to bring it back. I have verified that the INSERT statement works in a seperate PHP file without variables.
When I did have the INSERT working, every refresh of the page would duplicate the last entry. I have tried tried several ways to clear out the $_POST array, but I think some of my experimenting lead back to problem #1.
<?php
$dbhost = "REDACTED";
$dbuser = "REDACTED";
$dbpass = "REDACTED";
$dbname = "guest_list";
// Create a database connection
$connection = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);
// Test if connection succeeded
if(mysqli_connect_errno()) {
die("DB's not here, man: " .
mysqli_connect_error() .
" (" . mysqli_connect_errno() . ")"
);
}
// replacement for mysql_real_escape_string()
function html_escape($html_escape) {
$html_escape = htmlspecialchars($html_escape, ENT_QUOTES | ENT_HTML5, 'UTF-8');
return $html_escape;
}
// Posting new data into the DB
if (isset($_POST['submit'])) {
$first = html_escape($_POST['first']);
$last = html_escape($_POST['last']);
$contact = html_escape($_POST['contact']);
$associate = html_escape($_POST['associate']);
$insert = "INSERT INTO g_list (";
$insert .= "g_fname, g_lname, g_phone, g_association) ";
$insert .= "VALUES ('{$first}', '{$last}', '{$contact}', '{$associate}')";
$insert .= "LIMIT 1";
$i_result = mysqli_query($connection, $insert);
// I have verified that the above works by setting the varialble
// in the VALUES area to strings and seeing it update
}
$query = "SELECT * ";
$query .= "FROM g_list ";
$query .= "ORDER BY g_id DESC";
$q_result = mysqli_query($connection, $query);
?>
<!DOCTYPE html>
<html lang="en">
<head>
<title>Guest List</title>
<link href="guest.css" media="all" rel="stylesheet" type="text/css" />
</head>
<body>
<header>
<h1>REDACTED</h1>
<h2>Guest Registry</h2>
</header>
<div class="container">
<div class="registry">
<form name="formup" id="main_form" method="post">
<fieldset>
<legend>Please enter your name into the registry</legend>
<p class="first">First Name:
<input type="text" name="first" value="" placeholder="One or more first names" size="64"></p>
<p class="last">Last Name:
<input type="text" name="last" value="" placeholder="Last name" size="64"></p>
<p class="contact">Phone Number or Email:
<input type="text" name="contact" value="" placeholder="" size="32"></p>
<p class="associate">Your relation?
<input type="text" name="associate" value="" placeholder="" size="128"></p>
<p class="submit">
<input type="submit" name="submit" title="add" value="submit" placeholder=""></p>
</fieldset>
</form>
</div>
</div>
<h3>Guest List:</h3>
<table>
<tr>
<th>Firstname(s)</th><th>Lastname</th>
<th>Phone or Email</th><th>Association</th>
</tr>
<?php while($guest = mysqli_fetch_assoc($q_result)) {
echo "<tr>" . "<td>" . $guest["g_fname"] . "</td>"
. "<td>" . $guest["g_lname"] . "</td>"
. "<td>" . $guest["g_phone"] . "</td>"
. "<td>" . $guest["g_association"] . "</td>" . "</tr>";
} ?>
</table>
<footer>
<div>Copyright <?php echo date("Y"); ?>, REDACTED, LLC.</div>
<?php
if (isset($connection)) {
mysqli_close($connection);
}
?>
</footer>
</body>
</html>
These two lines will fail:
$insert .= "VALUES ('{$first}', '{$last}', '{$contact}', '{$associate}')";
$insert .= "LIMIT 1";
Two problems here, all with the second line:
No SPACE between ) and LIMIT: )LIMIT 1 is your code;
LIMIT 1 in an INSERT is not allowed....

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