I'm trying to make a program that takes 5 words from a database randomly and inserts them into an array. The page initially loads as desired, but nothing happens after the button is clicked. None of the alerts are ever triggered, so the function must never be entered, but why is beyond me. Also, I get an error saying name isn't a legitimate index (referencing line 13) the first time I try running it on a browser, so advice about that would be great too.
lingo.php:
<?php
session_start();
if (empty($_POST["name"])):
$_SESSION["error"] = "You did not enter a name.";
header("Location: entername.php");
else:
$name = $_POST["name"];
setcookie("name", "$name", time()+3600);
endif;
?>
<html>
<head>
<b>Welcome to Lingo, <?php echo $_COOKIE["name"]; ?></b><br />
<script src = "http://code.jquery.com/jquery-latest.js"></script>
<script type = "text/javascript" language = "javascript">
var arr = [];
function collectWords() {
$.post("getWord.php",
function(data) {
arr[word1] = $(data).find("Word1").text();
alert("function reached");
alert(arr[word1]);
arr[word2] = $(data).find("Word2").text();
alert(arr[word2]);
arr[word3] = $(data).find("Word3").text();
alert(arr[word3]);
arr[word4] = $(data).find("Word4").text();
alert(arr[word4]);
arr[word5] = $(data).find("Word5").text();
alert(arr[word5]);
});
}
</script>
</head>
<body>
<table id = "theTable" border = "1" class = "thetable"> </table>
<input type = "submit" value = "Start" onclick = "collectWords()">
</body>
</html>
getWord.php
<?php
$db = new mysqli('localhost', 'spj916', "cs4501", 'spj916');
if ($db->connect_error):
die ("Could not connect to db " . $db->connect_error);
endif;
$query = "select word from Words order by rand() limit 1";
$result = $db->query($query);
$rows = $result->num_rows;
if ($rows >= 1):
header('Content-type: text/xml');
echo "<?xml version='1.0' encoding='utf-8'?>";
echo "<Word1>";
$row = $result->fetch_array();
$ans = $row["word"];
echo "<value>$ans</value>";
echo "</Word1>";
else:
die ("DB Error");
endif;
$query = "select word from Words order by rand() limit 1";
$result = $db->query($query);
$rows = $result->num_rows;
if ($rows >= 1):
header('Content-type: text/xml');
echo "<?xml version='1.0' encoding='utf-8'?>";
echo "<Word2>";
$row = $result->fetch_array();
$ans = $row["word"];
echo "<value>$ans</value>";
echo "</Word2>";
else:
die ("DB Error");
endif;
$query = "select word from Words order by rand() limit 1";
$result = $db->query($query);
$rows = $result->num_rows;
if ($rows >= 1):
header('Content-type: text/xml');
echo "<?xml version='1.0' encoding='utf-8'?>";
echo "<Word3>";
$row = $result->fetch_array();
$ans = $row["word"];
echo "<value>$ans</value>";
echo "</Word3>";
else:
die ("DB Error");
endif;
$query = "select word from Words order by rand() limit 1";
$result = $db->query($query);
$rows = $result->num_rows;
if ($rows >= 1):
header('Content-type: text/xml');
echo "<?xml version='1.0' encoding='utf-8'?>";
echo "<Word4>";
$row = $result->fetch_array();
$ans = $row["word"];
echo "<value>$ans</value>";
echo "</Word4>";
else:
die ("DB Error");
endif;
$query = "select word from Words order by rand() limit 1";
$result = $db->query($query);
$rows = $result->num_rows;
if ($rows >= 1):
header('Content-type: text/xml');
echo "<?xml version='1.0' encoding='utf-8'?>";
echo "<Word5>";
$row = $result->fetch_array();
$ans = $row["word"];
echo "<value>$ans</value>";
echo "</Word5>";
else:
die ("DB Error");
endif;
?>
You get the error on $_COOKIE["name"]; because said cookie isn't set until you set it. You don't set the cookie until someone has entered their names, so on the first load it will give an error.
http://www.thesitewizard.com/php/set-cookies.shtml
"Note that you cannot set a cookie in PHP and hope to retrieve the cookie immediately in that same script session. Take the following non-working PHP code as an example:"
Found under the header: "How to get the contents of a cookie.
Fix this with a shorthand if statement, like so:
<b>Welcome to Lingo,
<?php isset($_COOKIE["name"]) ? $_COOKIE["name"] : $_POST["name"]; //Checks if the cookie is set. If not, uses the $_POST name ?>! </b><br />
I have another question:
Why are your words requested 1 at a time? Why not get all 5 words in 1 query? Also, why send them back as XML data? Since you seem to be handling the data yourself, I would personally recommend a simple loop on the PHP side, returning it as a handy pre-made JSON array.
edit: Also important, PHP does not echo any content automatically. An AJAX call can only receive printed data. You need to echo your results at the end of your php script or you will return nothing
Like so:
$query = "select word from Words order by rand() limit 5";
$result = $db->query($query);
$rows = $result->num_rows;
$array = array();
if ($rows >= 1):
$i = 0;//start the wordcount
//While there are results, loop. (Results are limited to 5, so it won't loop more than 5 times)
while($row = $result->fetch_row()){
$i++;//Put this on top so it starts with "1"
$array["word$i"] = $row[0]; //create the array
}
echo json_encode($array); //Turn array into json and echo it.
else:
die ("DB Error");
endif;
Now, you will also need to change your Javascriptside a tiny bit.
This is how you will access the new array (created by php)
<script type = "text/javascript" language = "javascript">
function collectWords() {
$.post("getWord.php",
function(data) {
alert(data); // show whether you get any data back in the first place. thanks #jDo
var arr = $.parseJSON(data);
alert(arr.word1);
alert(arr.word2);
alert(arr.word3);
alert(arr.word4);
alert(arr.word5);
});
}
</script>
As you can see, this way saves you quite a lot of code and saves you a lot of word replacements
Related
How do I echo values that are submitted from a form on another page (which they are submitted into the database) into an HTML tag as options?
This is my latest code and I'm still stuck here. When I clicked on the drop-down list, it still shows nothing.
<select name="comName" id="comName" class="form-control" required>
<?php include('db_company.php');
$query_option = "SELECT * FROM company";
$result = mysqli_query($query_option);
while($row = mysqli_fetch_array($result)) {
echo "<option value='{$row['comName']}'>{$row['comName']}</option>";
}
?>
</select>
There are some mistakes that you have done in here.First of all I would like to suggest you to start using error reporting using,
ini_set('display_errors', '1');
ini_set('display_startup_errors', '1');
error_reporting(E_ALL);
If you have used error reporting from the beginning I believe you would have noticed all the issues that you are facing now.
1.You need to use the DB connection when you are using mysqli_query
$link = mysqli_connect($DB_HOST, $DB_USER, $DB_PASS, $DATABASE);
$result = mysqli_query($link,$query_option);
2.You are using the wrong value inside the while loop.
while($row = mysqli_fetch_array($result)) {
echo "<option>".$row{'company'}."</option>";
}
You are getting comName from the query so you have to use comName instead of company.
while($row = mysqli_fetch_array($result)) {
echo "<option value=".$row{'comName'.">".$row{'comName'}."</option>";
}
Your echo in row is wrong
<?php include('db_company.php');
$query_option = "SELECT * FROM company";
$result = mysqli_query($query_option);
while($row = mysqli_fetch_array($result)) {
echo "<option value='{$row['id']}'>{$row['company']}</option>";
}
?>
I want to build an array in PHP from SQL query and send it back via ajax to my JS file.
$id = clear(filter_input(INPUT_POST, 'id'));
$sql = 'SELECT * FROM `counties` WHERE `id`="'.$id.'"';
$query = mysqli_query($con, $sql);
$array = array();
while($result = mysqli_fetch_array($query)) {
$id = $result['id'];
$name = $result['name'];
$array[] = array('id' => $id, 'name' => $name);
}
echo json_encode($array);
This is my code. In response I have always just one element. There's a lot of more. How could i do that correctly? I was browsing whole Internet and I didn't find anything useful... :(
$id = $_POST['id'];
$query = mysqli_query($con, "SELECT id,name FROM `counties` WHERE `id`='$id'");
$array = mysqli_fetch_all($query,MYSQLI_ASSOC);
echo json_encode($array);
this may simplified code
I made a server that checks if people are online if they logged in, and want to put that information into a html table.
<?php
session_start();
include_once '../php\connect.php';
function removeRefresh(){
$query = connection()->prepare("UPDATE `online` SET `time` = `time`+1");
$query->execute();
$query = connection()->prepare("DELETE FROM `online` WHERE `time` > 5");
$query->execute();
}
function addOrRefresh($ID){
$query = connection()->prepare("SELECT COUNT('id') FROM `online` WHERE `ID` = :ID");
$query->bindParam(':ID', $ID);
$query->execute();
$count = $query->fetchColumn();
if($count == 0){
$query = connection()->prepare("INSERT INTO `online`(`ID`, `time`) VALUES(:ID, 0)");
$query->bindParam(':ID', $ID);
$query->execute();
}
else{
$query = connection()->prepare("UPDATE `online` SET `time` = 0 WHERE `ID` = :ID");
$query->bindParam(':ID', $ID);
$query->execute();
}
}
$action = filter_input(INPUT_GET, 'action');
if($action == 'heartbeat'){
removeRefresh();
$ID = $_GET['id'];
addOrRefresh($ID);
$query = connection()->prepare("SELECT u.username FROM `nusers` u INNER JOIN `online` o ON o.ID = u.ID");
$query->execute();
$onlineUsers = $query->fetchAll(PDO::FETCH_ASSOC);
$resultaat = "";
foreach($onlineUsers as $user){
$resultaat .= "<p>{$user['username']} <br></p>";
}
echo "$resultaat";
}
?>
You can register, and when you log in into the lobby, you are added into an 'online' table, that checks every second with the function 'heartbeat'.
Please excuse my poor English.
EDIT: I have systems working. I merely want to change the 'echo "$resultaat"' so that it puts the $resultaat into a tablerow.
adding after a '}' won't work, I've tried working around that, not certain if I tried all posibilities in that retrospect. After request, I've posted the entire document.
the other javascript part is integrated into another document; containing:
<?php
session_start();
$ID = $_SESSION['ID'];
if (isset($_POST['logout'])) {
session_destroy();
$query = connection()->prepare("DELETE FROM `online` WHERE `ID` = :ID");
$query->bindParam(':ID', $ID);
$query->execute();
}
?>
<html>
<head>
<title>Stratego</title>
<Link href="../layout/style.css" type="text/css" rel="stylesheet" />
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>
<script src="./js/HeartBeat.js"></script>
<script>
$(document).ready(function () {
heartbeat.setSpelerId( <?php echo $ID; ?> );
heartbeat.polling();
});
</script>
</head>
<body>
<form method='post' name='logout' action="../setup/logout.php">
<input type='submit' name='loggout' value='Log me out'>
</form>
Singleplayer
<div id="online-list">
</div>
</body>
<?php echo "<script>
function heartbeat(){
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (xhttp.readyState == 4 && xhttp.status == 200) {
console.log(xhttp.responseText);
document.getElementById(\"online-list\").innerHTML = xhttp.responseText;
}
};
xhttp.open(\"GET\", \"./wachtruimterInterface.php?action=heartbeat&id=" . $ID . "\", true);
console.log('keeping in touch ༼つ ◕_◕ ༽つ');
xhttp.send();
}
setInterval(heartbeat, 1000);
</script>"; ?>
</html>
The console log is there to check if the connection is being maintained. People use their ID to login, no passwords.
This is how it looks now, with 1 person being online; named Luuk.
http://www.filedropper.com/screenproof
My goal is to get all the people that are online into a table.
I've tried
$resultaat .= "<tr><td>{$user['username']} <br></td></tr>";
}
echo "<table>$resultaat</table>";
Just now, doesn't seem to work, any tips on how to progress?
I've fixed the issue;
$resultaat = "";
foreach($onlineUsers as $user){
$naam= $user['username']
$resultaat .= "<tr><td>$naam</td></tr>";
}
echo "<table border=1>$resultaat</table>";
I am making a array from values from databse in PHP then storing it into javascript array. When I alert the array the single quotes are not visible, although when PHP is seen it shows the quotes around values. What’s wrong in my code? It should be ['1','2','3','4'] but it shows [1,2,3,4].
<script>
var values = [
<?php
$conn = mysql_connect("localhost","root","") or die ("we couldn't connect!");
mysql_select_db("test_value");
$rs = mysql_query("SELECT * FROM test") or die(mysql_error());
while($row = mysql_fetch_array($rs))
{
echo "'".$row['var1']."',";
}
?>
];
window.alert(values);
</script>
<script>
var values=[
<?php
$conn =mysql_connect("localhost","root","") or die ("we couldn't connect!");
mysql_select_db("test_value");
$rs = mysql_query("SELECT * FROM test") or die(mysql_error());
while($row = mysql_fetch_array($rs)){
echo "\'".$row['var1']."\',";
}?>
];
window.alert(values);
</script>
Use this , this may help you
I have the following in my page header
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.5.1/jquery.min.js"></script>
<script type="text/javascript" src="javascript/autoSuggest.js"></script>
<script type="text/javascript" src="javascript/suggest.js"></script>
suggest.js is made of:
$(function(){
$("#idName input").autoSuggest("../Test.php", {minChars: 2, matchCase: true});
});
and autoSuggest.js is a plugin by Drew Wilson (http://code.drewwilson.com/entry/autosuggest-jquery-plugin)
Test.php is
<?php
include('database_info.inc');
$input = $_POST["idName"];
$data = array();
var_dump($data);
// query database to see which entries match the input
$query = mysql_query("SELECT * FROM test WHERE title LIKE '%$input%'");
while ($row = mysql_fetch_assoc($query)) {
$json = array();
$json['value'] = $row['id'];
$json['name'] = $row['title'];
$data[] = $json;
}
header("Content-type: application/json");
echo json_encode($data);
?>
My var_dump() doesn't do anything and no items are suggested ...what could I be doing wrong? seems like there's no communication with Test.php
A quick look at the plug-in suggests it expects a parameter called q passed as a GET string, not as a POST.
<?
$input = $_GET["q"];
$data = array();
// query your DataBase here looking for a match to $input
$query = mysql_query("SELECT * FROM my_table WHERE my_field LIKE '%$input%'");
while ($row = mysql_fetch_assoc($query)) {
$json = array();
$json['value'] = $row['id'];
$json['name'] = $row['username'];
$json['image'] = $row['user_photo'];
$data[] = $json;
}
header("Content-type: application/json");
echo json_encode($data);
?>