PHP - How Do I Send Data And Return Data From The Server-side Php, Properly?
Here's the code that deals with the client side:
<?php session_start(); if(!isset($_SESSION['Logged_in'])){ header("Location: /page.php?page=login"); } ?> <!DOCTYPE Html> <html> <head> <!--Connections made and head included--> <?php require_once("../INC/head.php"); ?> <?php require_once("../Scripts/DB/connect.php"); ?> <!--Asynchronously Return User Names--> <script> $(document).ready(function(){ function search(){ var textboxvalue = $('input[name=search]').val(); $.ajax( { type: "GET", url: 'search.php', data: {Search: textboxvalue}, success: function(result) { $("#results").html(result); } }); }; </script> </head> <body> <div id="header-wrapper"> <?php include_once("../INC/nav2.php"); ?> </div> <div id="content"> <h1 style="color: red; text-align: center;">Member Directory</h1> <form onsubmit="search()"> <label for="search">Search for User:</label> <input type="text" size="70px" id="search" name="search"> </form> <a href="index.php?do=">Show All Users</a>|<a href="index.php?do=ONLINE">Show All Online Users</a> <div id="results"> <!--Results will be returned HERE!--> </div>search.php <?php //testing if data is sent ok echo "<h1>Hello</h1><br>" . $_GET['search']; ?>This is the link I get after sending foo. http://www.family-li...php?&search=foo Is that mean it was sent, but I'm not processing it correctly? I'm new to the whole AJAX thing. Similar TutorialsHi, For about a month, I have been trying to figure out why my code will not return anything after posting a wwwForm (I have also tried the newer equivalent of this function but I had no luck with that either.) The nameField and passwordField are taken from text boxes within the game and the code used in my login script is copied and pasted from a Register script but I have changed the file location to the login.php file. The register script works fine and I can add new users to my database but the login script only outputs "Form Sent." and not the "present" that should return when the form is returned and it never gets any further than that point meaning that it lets the user through with no consequence if they use an invalid name because the script never returns an answer. What should I do to fix this? Thanks, Unity Code: using System.Collections; using UnityEngine; using UnityEngine.UI; using UnityEngine.Networking; public class Login : MonoBehaviour { public InputField nameField; public InputField passwordField; public Button acceptSubmissionButton; public void CallLogInCoroutine() { StartCoroutine(LogIn()); } IEnumerator LogIn() { WWWForm form = new WWWForm(); form.AddField("username", nameField.text); form.AddField("password", passwordField.text); WWW www = new WWW("http://localhost/sqlconnect/login.php", form); Debug.Log("Form Sent."); yield return www; Debug.Log("Present"); if (www.text[0] == '0') { Debug.Log("Present2"); DatabaseManager.username = nameField.text; DatabaseManager.score = int.Parse(www.text.Split('\t')[1]); Debug.Log("Log In Success."); } else { Debug.Log("User Login Failed. Error #" + www.text); } } public void Validation() { acceptSubmissionButton.interactable = nameField.text.Length >= 7 && passwordField.text.Length >= 8; } } login.php: <?php echo "Test String2"; $con = mysqli_connect('localhost', 'root', 'root', 'computer science coursework'); // check for successful connection. if (mysqli_connect_errno()) { echo "1: Connection failed"; // Error code #1 - connection failed. exit(); } $username = mysqli_escape_string($con, $_POST["username"]); $usernameClean = filter_var($username, FILTER_SANITIZE_STRING, FILTER_FLAG_STRIP_LOW | FILTER_FLAG_STRIP_HIGH); $password = $_POST["password"]; if($username != $usernameClean) { echo "7: Illegal Username, Potential SQL Injection Query. Access Denied."; exit(); } // check for if the name already exists. $namecheckquery = "SELECT username, salt, hash, score FROM players WHERE username='" . $usernameClean . "';"; $namecheck = mysqli_query($con, $namecheckquery) or die("2: Name check query failed"); // Error code # 2 - name check query failed. if (mysqli_num_rows($namecheck) != 1) { echo "5: No User With Your Log In Details Were Found Or More Than One User With Your Log In Details Were Found"; // Error code #5 - other than 1 user found with login details exit(); } // get login info from query $existinginfo = mysqli_fetch_assoc($namecheck); $salt = $existinginfo["salt"]; $hash = $existinginfo["hash"]; $loginhash = crypt($password, $salt); if ($hash != $loginhash) { echo "6: Incorrect Password"; // error code #6 - password does not hash to match table exit; } echo "Test String2"; echo"0\t" . $existinginfo["score"]; ?>
Hello,
I want to grab some data using a script from this site.
But I am stuck right at the beginning. If you make a selection with the drop down boxes you get some output. For example you select:
Selecteer competitienaam: Najaarscompetitie 2014 Selecteer competitiegroep: Eredivisie dames Selecteer weergave: Programma (incl. uitslagen en stand) Optioneel poule filter: De Treffers R Selecteer poule(s): Eredivisie - Poule A I want to grab this output. When I look at the page source it is inside an iframe: <iframe src="http://www.nttb-competitie.nl/" width="100%" height="1200" scrolling="yes" frameborder="0" name="NTTB_Competitie"></iframe>I figured out: The script on the site first it gets url: http://www.nttb-competitie.nl/selectie.php?anr=0And after last selection it gets url: http://www.nttb-competitie.nl/web_programma.php?reset=0&pidString=1009267&sc=0&vastgesteldeAfdelingsnr=0&cnid=10085&cid=10704&view=programma&pf=1269&pid=1009267My problem is when I copy those URL's in a webbroser I get a page with only the words: Ongeldige aanroep!Which means "Invalid Call!" So my question is: How can I grab the data instead of this stupid message. Is it even possible or is it somehow protected? Please help! All good with sending single dimensional form data to the server using application/x-www-form-urlencoded, and having PHP convert it into an array. Also, good with sending simple arrays using []. But then I find myself needing to send a deeper object to the server . For example, I have a form with three text inputs simpleFormName 1, 2, and 3 plus some deeper object. Array ( [simpleFormName1] => bla [simpleFormName2] => bla [simpleFormName3] => bla [deeperObject] => Array ( [0] => Array ( [prop1] => bla [prop2] => true [data] => Array ( [0] => Array ( [p1] => 321 [p2] => 123 ) [1] => Array ( [p1] => 121 [p2] => 423 ) [2] => Array ( [p1] => 221 [p2] => 133 ) ) ) [1] => Array ( [prop1] => blabla [prop2] => false [data] => Array ( [0] => Array ( [p1] => 222 [p2] => 443 ) [1] => Array ( [p1] => 321 [p2] => 213 ) [2] => Array ( [p1] => 111 [p2] => 421 ) ) ) ) )
I see at least four options: Come up with naming structure which "flattens" the data. I have used this approach in the past, but it quickly becomes difficult to manage and I don't want to do so. Not use PHP's POST and and instead use Content-Type: application/json and file_get_contents(php://input) and json_decode the entire request server side. Use PHP's POST and urlencoding but make deeperObject a string using JSON.stringify and json_decode this one field server side. Urlencode the entire object.simpleFormName1=bla&simpleFormName2=bla&simpleFormName3=bla&deeperObject%5B0%5D%5Bprop1%5D=bla&deeperObject%5B0%5D%5Bprop2%5D=true&deeperObject%5B0%5D%5Bdata%5D%5B0%5D%5Bp1%5D=321&deeperObject%5B0%5D%5Bdata%5D%5B0%5D%5Bp2%5D=123&deeperObject%5B0%5D%5Bdata%5D%5B1%5D%5Bp1%5D=121&deeperObject%5B0%5D%5Bdata%5D%5B1%5D%5Bp2%5D=423&deeperObject%5B0%5D%5Bdata%5D%5B2%5D%5Bp1%5D=221&deeperObject%5B0%5D%5Bdata%5D%5B2%5D%5Bp2%5D=133&deeperObject%5B1%5D%5Bprop1%5D=blabla&deeperObject%5B1%5D%5Bprop2%5D=false&deeperObject%5B1%5D%5Bdata%5D%5B0%5D%5Bp1%5D=222&deeperObject%5B1%5D%5Bdata%5D%5B0%5D%5Bp2%5D=443&deeperObject%5B1%5D%5Bdata%5D%5B1%5D%5Bp1%5D=321&deeperObject%5B1%5D%5Bdata%5D%5B1%5D%5Bp2%5D=213&deeperObject%5B1%5D%5Bdata%5D%5B2%5D%5Bp1%5D=111&deeperObject%5B1%5D%5Bdata%5D%5B2%5D%5Bp2%5D=421 Any recommendations how to best implement? If you recommend using urlencoding for simple forms, but some other approach for more complex data, what criteria do you use to transition from one approach to another? Thanks I wish to create validation rules once which are used both on the client and on the server.
For instance, I will start off with the following PHP object:
stdClass Object ( [rules] => stdClass Object ( [email] => stdClass Object ( [required] => 1 [email] => 1 [remote] => stdClass Object ( [url] => check-email.php [type] => post [data] => stdClass Object ( [username] => function() {return $( '#username' ).val();} ) ) ) ) [messages] => stdClass Object ( [email] => stdClass Object ( [required] => an email is required ) ) )When the edit page is downloaded to the client, I will include this object in some format suitable to the client. The client will then use the jQuery Validation plugin (http://jqueryvalidation.org/) along with the validation object, and client side validate the page. When the form passes client side validation and is uploaded, PHP will use the same validation object to serverside validate the form (I have this part working as desired). My question is how should I pass this data to the client? Originally, I would just use PHP to write some JavaScript. exit('var myObj='.json_encode($myObj));Note that when I json_encode the object, the value of $myObj->rules->email->remote->data->username is a string with quotes around it, however, I can easily use PHP to strip these tags before sending it to the client. As Jacques1 pointed out in http://forums.phpfre...ascript-client/, I should never ever use PHP to generate JavaScript, and should use AJAX to download the JSON directly. I tried doing the later, but found that a callback function could not be included in the JSON. Please advise on the best way to accomplish this. Thank you Hello to all, I have problem figuring out how to properly display data fetched from MySQL database in a HTML table. In the below example I am using two while loops, where the second one is nested inside first one, that check two different expressions fetching data from tables found in a MySQL database. The second expression compares the two tables IDs and after their match it displays the email of the account holder in each column in the HTML table. The main problem is that the 'email' row is displayed properly while its while expression is not nested and alone(meaning the other data is omitted or commented out), but either nested or neighbored to the first while loop, it is displayed horizontally and the other data ('validity', 'valid_from', 'valid_to') is not displayed.'
Can someone help me on this, I guess the problem lies in the while loop? <thead> <tr> <th data-column-id="id" data-type="numeric">ID</th> <th data-column-id="email">Subscriber's Email</th> <th data-column-id="validity">Validity</th> <th data-column-id="valid_from">Valid From</th> <th data-column-id="valid_to">Valid To</th> </tr> </thead> Here is part of the PHP code:
<?php while($row = $stmt->fetch(PDO::FETCH_ASSOC)) { echo ' <tr> <td>'.$row["id"].'</td> '; while ($row1 = $stmt1->fetch(PDO::FETCH_ASSOC)) { echo ' <td>'.$row1["email"].'</td> '; } if($row["validity"] == 1) { echo '<td>'.$row["validity"].' month</td>'; }else{ echo '<td>'.$row["validity"].' months</td>'; } echo ' <td>'.$row["valid_from"].'</td> <td>'.$row["valid_to"].'</td> </tr>'; } ?>
Thank you. the second database found on the cloud
i try to get JSON data but how to insert and update them to another online database with the same table my php script to return json data <?php include_once('db.php'); $users = array(); $users_data = $db -> prepare('SELECT id, username FROM users'); $users_data -> execute(); while($fetched = $users_data->fetch()) { $users[$fetchedt['id']] = array ( 'id' => $fetched['id'], 'username' => $fetched['name'] ); } echo json_encode($leaders);
i get
{"1":{"id":1,"username":"jeremia"},"2":{"id":2,"username":"Ernest"}} Edited March 24 by mahenda Hi guys, im trying to connect to a database and get the value for the user in the row called 'user_credit', if it equals 1 or more then i want to show the ''You have £ ....'' bit in the script. Problem is nothing shows at all, even without the if statement. I have changed the value for me in the database so in user_credit the value is 100, which is more than 1 so it should appear. I have probably done something wrong. Any ideas? Code: [Select] <? include '../admin/database/membership_dbc.php'; $r = mysql_query("SELECT * FROM users WHERE user_name='".safe($_SESSION['user_name'])."'") or die ("Cannot find table"); while( $cred = mysql_fetch_array($r) ) { if ($cred >= '1' ) { ?> <p>You have £<? echo $cred['user_credit']; ?> available on you account, would you like to use it on this order?<br> <label for="credit"></label> <select name="credit" id="credit"> <option value="Y" selected>Yes, use credit</option> <option value="N">No, save credit</option> </select> </p> <? } } ?> I have a problem with my rest service as below: ..and its not working (i replaced <textarea> with <div> to apply some html tags inside it ) and this is my form : Code: [Select] <form action="proc.php" method="post"> <div id="text" name="question_text" class="text" contenteditable="true"></div> </form> but when i submit some text the $_POST['question_text'] is not set ! Hey, I'm trying to send a form input to a page, and then get the response (as in, the body) from that page without leaving my PHP script. So instead of when you normally submit a form and it will take you to the 'action="/somepage"'. I want it to get that information and bring it back to the original page. No navigation should happen. I'm almost certain this is possible, but the only thing I've found at the moment requires a custom addon - which, isn't on most hosting sites. Sorry if I haven't described this very well, but any help would be appreciated Thanks tree Hello, Curious to know if someone could point me in the right direction, been struggling with this for a bit now. I have a HTML page with a search field, I can enter a search term and hit the submit button and I am directed to my search.php page with the appropriate results. What I am looking to accomplish is having the search results from the search.php page displayed in a text area below my search field in my HTML page. I have included an image to better describe what I am looking to accomplish: Additionally below is the source from my HTML page and search.php page: page.html <form name="search" action='search.php' method="post"> <input type="text" class="myinputstyle" name="search" value="search" onClick="this.value=''"/><br> <input type="submit" value="submit" class="myinputstyle"> </form> search.php <?php $search = "%" . $_POST["search"] . "%"; mysql_connect ("localhost", "game_over", "Ge7Ooc9uPiedee3oos9xoh4th"); mysql_select_db ("game_over"); $query = "SELECT * FROM game_over WHERE first_name LIKE '$search'"; $result = mysql_query ($query); if ($result) { while ($row = mysql_fetch_array ($result)) { echo "Name: {$row['name']} " . "{$row['lname']} <br>" . "Email: {$row['email]} <br>" . } } ?> Any insight would be most appreciated. Thank you. Hey guys, New to the forum and a newer user of PHP / MySQL. I am having trouble with some code I've written up. I don't seem to get any errors when running it, but it's not updating my database the way that it should. hopefully a simple fix. I am thinking that it must be on the MySQL side of things. Couple of things to start. My html form is comprised completely of drop down list inputs. I'm the only user so I thought this would be the easiest approach. Because of that I've made my PHP as follows: Code: [Select] <?php $season = $_POST['season']; $month = $_POST['month']; $day = $_POST['day']; $year = $_POST['year']; $time = $_POST['time']; $event = $_POST['event']; $game = $_POST['game']; $buyin = $_POST['buyin']; $connect = mysql_connect('localhost','root','') or die('can not connect'); if ($connect) { echo "connected to database"; } $db = mysql_select_db('dpl') or die('can not find database'); if ($db) { echo "DPL Selected"; } $query = sprintf("INSERT INTO events (season , month , day , year , time , event , game , buyin) VALUES ('%s' , '%s' , '%s' , '%s' , '%s' , '%s' , '%s' , '%s')", $season , $month , $day , $year , $time , $event , $game , $buyin ); if ($query) { echo "Your event has been added"; } ?> My connection is working, my database is selected and I'm even now getting confirmation that my query is working, but when i go to check my database there are no entries in it? any thoughts? I've tried the drop down variables as both VARCHAR and TEXT inputs in MySQL, but I can't seem to get it to work. Any help is greatly appreciated. Hi, I have a MySQL database called "2011_database" that has a table called "2011_list." In that table I have fields, amongst others, called "name" and "district." I need to find way to get the data from the table and put them into a drop down list on other PHP page. But they need to be listed as "name - district" on one line. I am PHP beginner and if I understand it correctly there need to be two references to get all the data in all records, a third reference to merge them together with " - " in between; and what eludes me the most, putting them in a drop down menu. Any help is greatly appreciated Thanks How to send multipart/form-data ? via curl or so ... here i am going to send it - http://dezend.me/dezend/ Hello everyone, I am trying to submit a comment in a comment box and send it to the DB but is not happening. The connection is good as I am logging in and all but no data is sent to the DB when I post the comment. It doesn't show in my comment section either.
Form <!--comment section--> <?php if(isset($_SESSION['id'])) { echo "<form method='POST' action='" . setComments($conn) . "'> <input type='hidden' name='uidUsers' value='".$_SESSION['id']."'> <input type='hidden' name='posted' value='" . date('Y-m-d H:i:s') . "'> Comments: <textarea rows = '5' cols = '15' name='body'></textarea><br><br> <button name='commentSubmit' type='submit'>Comment</button> </form>"; }else { echo "Log in to comment!"; } getComments($conn);
Function to set and get comments function setComments($conn) { if (isset($_POST['commentSubmit'])){ $user_id = $_POST['uidUsers']; $body = $_POST['body']; $posted = $_POST['posted']; $sql = "INSERT INTO comments (uidUsers, posted, body) VALUES ('$user_id', '$posted', '$body')"; $result = mysqli_query($conn, $sql); } } function getComments($conn) { $sql = "SELECT * FROM comments"; $result = mysqli_query($conn, $sql); while ($row = $result->fetch_assoc()){ $id = $row['uidUsers']; $sql2 ="SELECT * FROM users WHERE uidUsers='$id'"; $result2 = mysqli_query($conn, $sql2); if($row2 = $result2->fetch_assoc()){ echo "<div class='comment-box'><p>"; echo $row2['uidUsers'] . "<br>"; echo $row['posted'] . "<br>"; echo nl2br($row['body']); echo "</p></div>"; } } }
Hello, i want to send and receive data from https url. how it can be done in php? Thanks I am successfully doing this, however, I believe I am doing it wrong. I am currently doing the following. How should I be doing this? My thought is getObject.php should return JSON, not JavaScript, but I don't know how to assign the received JSON to the myObj variable. Thanks I needed a server side PHP validation script for my form to use on top of client side Javascript as a backup. I found a nice package where I just include the validator and set a few options then it does the rest. However it doesn't seem to work with the action="whateveraction.php" for the form. It will skip all the validation and immediately go to the action script. So I left that blank as you can see in the form below and figured I would send the post data somehow to the script using PHP when the validation is successful. However I am stumped on how to do that can I get some coding help? This is the small validation script. I will attach the formvalidator.php in this post. Code: [Select] <?php require_once "formvalidator.php"; if(isset($_POST['submit'])) { $validator = new FormValidator(); $validator->addValidation("age","req","Please fill in your Age"); $validator->addValidation("mcid","req","Please fill in your Minecraft Username"); $validator->addValidation("description","req","Please fill in a Description"); if($validator->ValidateForm()) { //need something here to have it send the post data to /mcbuildapp/form_submit.php } else { echo "<div class=blockrow><b><font size=5>Form Errors:</font><b></div>"; $error_hash = $validator->GetErrors(); foreach($error_hash as $inpname => $inp_err) { echo "<div class=blockrow><p><font color=red>$inp_err</font></p>\n</div>"; } } } Then the form if it matters.... Code: [Select] <div class="blockrow"> <font size=4><b>Minecraft Building Rights Application</b></font> <br /> <br /> <b>Failing to fill in any of these fields sufficiently will result in an automatic denial of your application!</b><br /> <b>Think of it this way, being lazy and applying wrong is only wasting your time. Take the time to do it correctly the first time.</b> <br /> <br /> <b><font color=red>To ensure proper user authentication and security you must first join the server at least once before applying for your building rights to work!</font></b> <br /> <br /> <form action="" id="mcbuilderapp" name="mcbuilderapp" method="post"> <label><b>Your Minecraft Username (required)</b><br /> Use exact capitalization and punctuation that matches your Minecraft.net account!</label> <br /> <br /> <input type="text" name="mcid" id="mcid" maxlength="25"/> <br /> <br /> <label><b>Your Age (required)</b></label> <br /> <br /> <input type="text" name="age" id="age" maxlength="3"/> <br /> <br /> <label><b>Describe Yourself (required)</b><br /> Describe yourself in <b>no less</b> than 3 sentences using <b>correct grammar</b>.</label> <br /> <br /> <textarea name="description" id="description" minlength="10" maxlength="1000" cols=60 rows=5 wrap="physical"> </textarea> <br /> <br /> <input type="submit" name="submit" value="Submit" /> </form> </div> MOD EDIT: PHP manual tags changed to code tags . . . [attachment deleted by admin] I am retrieving Google Books info in JSON format and displaying it inside a div. I would like to send the contents of this div (name, title, description) to my database using Ajax.
Currently only the ISBN field sends because I have it declared as a variable. However my question is, how do I send the other fields (name, title, author). How do I declare these also, I'm not sure what format they need to be in etc.
My JS
$(document).ready(function() { $('#submit').click(function(ev) { ev.preventDefault(); var isbn = $('#isbn_search').val(); //get isbn direct from input var url='https://www.googleapis.com/books/v1/volumes?q='+isbn; $.getJSON(url,function(data){ $.each(data.items, function(entryIndex, entry){ var html = '<div class="results well">'; html += '<h3>' + entry.volumeInfo.title + '</h3>'; html += '<div class="author">' + entry.volumeInfo.authors + '</div>'; html += '<div class="description">' + entry.volumeInfo.description + '</div>'; }); }); }); });My Ajax; $.ajax({ type: 'POST', url: 'addIsbnScript.php', data: { 'isbn' : isbn, 'title' : title 'subtitle' : subtitle, 'authors' : authors, 'description' : description }, success: function () { $.growl({ message: " Record added" }); } });Note, if i manually set the vars like below they all do successfully send to my database, so I know my query is working ok var title = "some text", var author = "some text", Var description = "some text"Thanks in advance for any help, newbie here (incase it wasn't obvious!). J Quick question for you guys. I am collecting payment data for a product. Is it possible to send data from my form via http directly into another payment form via the post http headers? So say if i had all the details payment info and address info and i matched my form exactly to theirs. Is it possible to write a php script to send the data from my form directly to another websites form to make an order, say amazon or a big site like this? ?first_name=samuel&last_name=east&action=Submit Any Help? Thanks |