PHP - How To Send One Value To Function And Return Two Values
not sure if this is possible but i do not know how to return multiple values and read them where the call was made.
Code: [Select] $string = "clause1"; func1($string); // somehow read the response and use the return values seperately. // both $string2 and $string3 Code: [Select] function func1($input) { if ($input == "clause1") { $string2= "value2"; $string3= "value6"; } if ($input == "clause2") { $string2= "value3"; $string3= "value7"; } if ($input == "clause3") { $string2= "value4"; $string3= "value8"; } if ($input == "clause4") { $string2= "value5"; $string3= "value9"; } return ($requesttype, $messagetitle); } 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, normally when i query mysql, it returns my results line by line. im trying to make a print view now that will put 3 results in a row before going to the next row. any help appreciated. SELECT field, (array of values from table 2) FROM $table1 t1 LEFT JOIN $table2 ON t1.id=t2.id ... is something like this possible? (where I return records from t1 in a LEFT JOIN, but also get an array of any of the fields/records from the table 2) Hey all. I'm a noob so bare with me. Just started studying php and have an assignment due. At this phase of the project I have to create a form where a user can register data and I should retrieve that data and store it in my database. Now I have already created the form (student_reg.php) and a separate file with php code (demo.php) My database is name "registration" and I am trying to put this info in the "student" table. It has about 15 fields but for now I am only trying trying to insert the 1 field of data to test if it works. Dont know if that may cause problems? So far my demo.php connects to the database successfully. The problem I am having is the information I submit does not showing in the database field. However, I have an id field in my case a student number field which is set to auto increment so it automatically updates (1,2,3,4) with each new sumbitted info. When I submit the info the id field gets updated each time with a new row ( so i assume some information is somehow going through ) however the info I entered will remain blank in its field. :banghead: Heres my form (student_reg.php) Code: [Select] <div id="apdiv3"> <FORM action = "demo.php" method ="post "> <p>Course name:</p> <INPUT TYPE = "text" name="input"/> <INPUT TYPE = "Submit" VALUE = "Submit"/> </div> </FORM> and heres my php code to retrieve the form info Code: [Select] <?PHP include 'includes/config.php'; //connect to database $link=mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); //error check if (!$link) { die('could not connect: ' . mysql_error()); } $db_selected=mysql_select_db(DB_NAME, $link); // error check if (!$db_selected) { die('can\t use ' . DB_NAME . ': ' . mysql_error()); } // Connected to database // retrieve form data $value=(isset($_POST['input'])); $sql = "INSERT INTO student (sname) VALUES ('$value')"; if (!mysql_query($sql)) { die ('Error: ' . mysql_error()); } mysql_close(); ?> I also recieved a notice "undefined index" or something like that from this part of the code "$value=(isset($_POST['input']));" before I inserted "isset" to the code. I'm not sure if what I did fixed the problem or only hid it, or whether that notice was related to the problem I'm having of not retrieving the info. Haha sorry guys know this is probably childs play for you but I just started out and really need to fix this so I can catch up with this project. Feel free to add tips and comments on other areas. Thanks :mrgreen: So I have an existing picklist that i am trying to tweak - I am making it into a multiselect and trying to return multiple values
I have been able to accomplish the first part by adding select 'multiple' but when i submit it only returns results for the last value selected.
Here is the code for the form (Left out the picklists that do not need to be multiple select)
<td align="center" valign="top"> <? // base64_encode(base64_encode( 'test1'))?> <br /> <br /> <form id='report_form' action="driverreport_sd.php" method="post" name="report_form"> <!--<div class="criteria_div">--> <table align="center" class="table_border" cellpadding="5" cellspacing="0" width="80%"> <tr> <td align="right" class="first_td"> Status: </td> <td align="left"> <select multiple name='Status' id='Status'> <? $statuslist=$ObjReport->ObjStatus->get_Status_List(); while($status=mysql_fetch_object($statuslist)) { if($status->Status=="Please Select") { ?> <option value="<?=$status->Status?>" <?=($status->Status==$_REQUEST['Status']?"selected=selected":"")?>> <?=$status->Status?></option> <? } else { ?> <option value="<?=$status->Status?>" <?=($status->Status==$_REQUEST['Status']?"selected=selected":"")?>> <?=$status->Status?></option> <? } } ?> </select> </td> </tr> <td align="center" colspan="2"> <input type="hidden" id='sort_by' name='sort_by' value="<?=$_REQUEST['sort_by']?>" /> <input type="hidden" name='search_val' id='search_val' value="<?=$_REQUEST['search_val']?>" /> <input type="hidden" name='search_by' id='search_by' value="<?=$_REQUEST['search_by']?>" /> <input type="hidden" name="report_submited" value='report_submited' /> <button type="submit" id="report_submited" name="" value='submit'>Submit</button> </td> and here is the return request <? if($_REQUEST['report_submited']) { $ReportData=$ObjReport->generate_report($_REQUEST['Status'],$_REQUEST['School'],$_REQUEST['Campaign'],$_REQUEST['State'],$_REQUEST['Stdtype'],$_REQUEST['Primbad'],$_REQUEST['Rep'],$_REQUEST['From_Date'],$_REQUEST['To_Date'],$_REQUEST['sort_by'],$_REQUEST['search_by'],$_REQUEST['search_val']); $currentTotalRow=mysql_num_rows($ReportData); if($currentTotalRow>0) { ?> Any suggestions would be appreciated Thanks! I call a web service and it retuens the values in the following format
$response = $client->submitRequest($requestParams);Which returns stdClass Object ( [return] => stdClass Object ( [result_code] => 0 [result_data] => City[0]=Chicago [message_text] => ) )what I want is just the value for [result_data] => City[0]=Chicagoassigned to a variable Psedo Code $city = [result_data] => City[0]=Chicago;so that $city = Chicago; Hello, My question is how to send values over and over without using forms or even hidden forums. I have tried to use sessions but it didn't work out for me. Any suggestions! let say i have a function function checkme($x) { if(strlen($x)>10) { return $x; } else { return false; } } $variable=checkme("5"); so two questions i wish to ask... 1. If i wish to get the empty string for the $variable, then is it correct to have return false there? 2. If it is true that return shall not be used twice in a function as it will somehow reduce bytes space or something like that?? thanks in advance Hi, i got this function to get the wordCount that i store in a tabel: Code: [Select] function getWordCount($word) { $query = "SELECT count FROM uniqueWords WHERE word='$word'"; $data = execQuery($query); return $data; } Only it returns an array, what i want is to have it return a integer, how can i do that? Hi all, can you give me suggestion how to return value from function like this Code: [Select] function returnArray(){ for($i=1; $i<=3; $i++){ return $i; //how to return this as array?? } } when I code "echo" inside the function then call the function it give me output => 123 but when I code "return" inside the function, it will give output => 1 Can you suggest me how to output it as array? Here's the code that deals with the client side:
<?php session_start(); if(!isset($_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. Hi,Dear Frnds...........................I Hope U r all fine. I am using php.Here is the problem in the Return array from function. This is the code. Code: [Select] <?php $store = array(); $test = get_my_links(); print_r($store); function get_my_links() { $my_array = array('e','f','g'); foreach($my_array as $key=> $list) { $store[] = $list; } return $store; } ?> Suppose the array is maked dynamically. please help me..THANKS... This is object programming right? Is there a performance issue with this? for example: $notfications = ( blah blah ) ? : ''; then if ($notificatiosn != blabla){ echo 'hey'; }or.... $notifications = function(){ do my stuff here then return 'hey'; }Which way is faster or slower? Reason I Ask this is because I've been doing some object orientated programming in javascript, and didn't know you could do it in php. I'd rather do the objective way so I don't have to use so many freaking variables above the regular way, lol. Edited by Monkuar, 22 January 2015 - 12:22 PM. Hello everyone, Pretty much what I am trying to do is run a query and based of the queries result, return a string - as of now it is turning up blank and cant seem to pass any strings through the function. Current function: function addNewBookmark($hash,$url,$title,$username){ Code: [Select] $query = "INSERT INTO bookmarks (hash,url,title,username) VALUES '".md5($hash)."', '".$url."', '".$title."', '".$username."')"; $result = mysql_query(query, $this->connection); $message = ''; if(mysql_affected_rows($this->connection)!=1) { $message = 'This URL already exists in the database!'; return $message; } else $message = 'The URL was shared!'; return $message; } The above code seems to run fine, just cant seem to pass a string...read that you can possibly use a __toString function? but that keeps throwing errors when I try to work with it. Thanks all! My tree gather function gives me multidimensional array. However i want it to be one dimensional . I was getting one dimensional earlier however i forgot how i got it. Can someone please tell me my mistake Code Code: [Select] <?php include'config.php'; function tree_gather($node) { $sql = "SELECT lchild,rchild FROM tree WHERE parent = '$node'"; $execsql = mysql_query($sql); $array = mysql_fetch_array($execsql); if(!empty($array['lchild']) || !empty($array['rchild'])) { $child[] = $array['lchild']; $child[] = $array['rchild']; $eh[] = tree_gather($array['lchild']); $child = array_merge($child, $eh); $eh1[] = tree_gather($array['rchild']); $child = array_merge($child, $eh1); } return $child; } $pair = tree_gather(1000000); echo "<pre>"; print_r($pair); echo "</pre>"; ?> Output Code: [Select] Array ( [0] => 3632873 [1] => 5951538 [2] => Array ( [0] => 8930480 [1] => 7563232 [2] => Array ( [0] => 1144744 [1] => 4386810 [2] => [3] => ) [3] => ) [3] => ) This is my error: Code: [Select] Fatal error: Can't use function return value in write context in /home/a8152576/public_html/SecretSanta.html on line 41 confused about why. This is the line 41: if(isset($_POST('submit')){ this is the php of my page (without the mysql data): <?php class functions{ public static function chooseGiftee($SS){ $done = false; while($done=false){ $sqllink= mysql_connect($mysql_host,$mysql_user,$mysql_password); if(!$sqllink){ echo "Could not connect! Please Try Again"; mysql_close($sqllink); } else { $selectGiftee = "SELECT * FROM SecretSantas ORDER BY RAND() LIMIT 0,10;"; while($row=mysql_fetch_array($result)){ $results = mysql_query($selectGiftee); if(!$row['Gifter']=''){ $done=false; } else { $gifteeResults = $row['Name']; $addSS = "INSERT INTO SecretSantas (Gifter) VALUES ('" .$SS. "')"; $done=true; return $gifteeResults; } } } mysql_close($sqllink); } } } if(isset($_POST('submit')){ $echoName = functions::chooseGiftee($_POST['gifterName']); echo $echoName; } ?> This is html: <form id="form1" name="form1" method="post" action=""> <p> <label>Your Name: <input type="text" name="gifterName" id="gifterName" /> </label> <input type="submit" name="submit" id="submit" value="Submit" /> </p> <p> </p> </form> Confused about the issue, first time using functions. Thanks in advance, Brandon I hate array... 😞 So I had a block of code inside my photo-gallery.php script that took the path to my photos directory, and went to that directory, and then read all of the photo filenames into an array. Then in my HTML, I iterate through this array to display all of the photos for my gallery. Now I would like to move that code to an included file so multiple scripts can access it and always be working with the same array. It seems to me that I need to encapsulate my code inside a function? Then I could call my getPhotoFilesArray back to my callings cript, and use that array for whatever. I haven't coded PHP in like 4 years and I am struggling to return the entire array back to my caling script. This is what I have so far... function getPhotoFilesArray($photoPath){ $photoFiles = array(); <code to find corresponding files> $photoFiles gets populated in a loop return $photoFiles; }
Then in my calling script, I have... <?php require_once('../../../secure_outside_webroot/config.php'); require_once(WEB_ROOT . 'utilities/functions.php'); getPhotoFilesArray($photoPath); var_dump($photoFiles);
I get some error...
Notice: Undefined variable: phtoFiles in photo-gallery.php line 133 (which is my var_dump).
<br> Would appreciate help getting this to work!
Edited December 6, 2019 by SaranacLake Hi, I have some strange issues with my code: Code: [Select] <?PHP session_start(); $loginid = $_SESSION["valid_id"]; // First check if we are guest or user. if (!$_SESSION["valid_email"]) { $visitor = "yes"; } else { $visitor = "no"; $email = $_SESSION['valid_email']; $userid = $_SESSION['valid_id']; } //Load Header (blue menu) require("./inc/header.php"); //Load Sub-acc (silver account menu) require("./inc/sub-group.php"); //Load nav-group (Tabs) require("./inc/nav-group.php"); //Load Config file require("./inc/config.php"); //Set & get profile ID $getid = $_GET["id"]; //check ID $result = mysql_query("SELECT * FROM profiles WHERE id=('$getid') LIMIT 1"); $row = mysql_fetch_array($result); IF (mysql_num_rows($result) != 1) { exit("Invalid ID"); } //If we are guest, do we allow anon access to the profile IF ($row["privacy"] <= 10 && $visitor = "yes") { exit("You may not view this profile as a visitor, due to the users privacy settings"); } //Let's check if we are friends ELSEIF ($visitor = "no") { $result2 = mysql_query("SELECT * FROM profiles_friends WHERE user=('$getid') AND target=('$loginid') LIMIT 1"); $row2 = mysql_fetch_array($result2); $friends = $row2["status"]; if (mysql_num_rows($result2) = 0) { $friends = "no"; } } //If we are friend, do we allow access to the profile IF ($row["privacy"] >= 9 && $friends != 1) { exit("You may not view this profile because of the privacy settings."); } $row = mysql_fetch_array($result); $memgroup = $row["group"]; IF ($row["activated"] != 1) { exit("This account is suspended and cannot be viewed."); } //Check what group member is in. $result2 = mysql_query("SELECT * FROM profiles_groups WHERE id=('$memgroup') LIMIT 1"); $row2 = mysql_fetch_array($result); ?> Alright, so the error: Fatal error: Can't use function return value in write context in C:\xampp\htdocs\prog\profile.php on line 45 Code: [Select] 42. $result2 = mysql_query("SELECT * FROM profiles_friends WHERE user=('$getid') AND target=('$loginid') LIMIT 1"); 43. $row2 = mysql_fetch_array($result2); 44. $friends = $row2["status"]; 45. if (mysql_num_rows($result2) = 0) Alright, this is one thing that bothers me, the other is: Code: [Select] //check ID $result = mysql_query("SELECT * FROM profiles WHERE id=('$getid') LIMIT 1"); $row = mysql_fetch_array($result); IF (mysql_num_rows($result) != 1) { exit("Invalid ID"); I tried to put an invalid ID, and already here the script should have died/exited before executing the parts of the code that doesn't work. I tested my code on another page and it works flawlessly, perhaps this error is just generated before it actually exists i dunno.. Any way, any help is much appreciated } |