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PHP - Code Not Executing Beyond A Point
I am playing around with php but the script doesnt execute to the last point and with no error...
What i am trying to do is checking the input of the user and comparing it what is in db and then applying conditions. So i have 2 files. update.php and home.php. home.php is my html form page and i included 'update.php' in it In my update.php, i have these codes
<?php
error_reporting(E_ALL);
include_once('database.php');
if(isset($_POST['submit']))
{
$q = $_GET['q'];
$next = $q + 1;
if($q == 26)
{
echo '<h2>You have Completed the Exam.Congrats!!</h2>';
}
elseif($q <= 25){
$sql = "SELECT * FROM Students WHERE Email=?";
$stmt = $conn->prepare($sql);
$stmt->bind_param('s', $_SESSION['login']);
$stmt->execute();
$result = $stmt->get_result(); $row = $result->fetch_assoc();
$Mark = $row['Mark'];
$Correct = $Mark + 1;
$Fail = $Mark + 0;
$sqlb = "SELECT * FROM Answers WHERE qid = ?";
$stmtb = $conn->prepare($sqlb);
$stmtb->bind_param('s', $q);
$stmtb->execute();
$resultb = $stmtb->get_result();
$rowb = $resultb->fetch_assoc();
$Answer = $rowb['Answer'];
$Pick = $_POST['ans'];
if($Pick == $Answer)
{
$sqld = "UPDATE Students SET Mark=? WHERE Email = ?";
$stmtd = $conn->prepare($sqld);
$stmtd->bind_param('ss', $Correct,$_SESSION['login']);
$stmtd->execute();
$resultd = $stmtd->get_result();
echo "correct" ;
}
}
else
{
echo 'Ate';
echo "<script type='text/javascript'>document.location.href='home.php?q='.$next.';</script>";
}
}//post submit
?>
The else statement doesnt run. so it keeps reloading the same page. Also i want to ask why i keep getting the value of 2 for my $Correct variable stored on the database into column Mark for a single correct question instead of 1. What could be wrong? Lastly, i dont want the script to run if the browser is edited by the user by changing the value of $q from the browser, because i am using get method, is there a way to do that. Please be nice with your comments. Thanks in advance.!!! Similar TutorialsI am trying to use some php code inside of an html document and it does not work. I have a script.php file that executes perfectly when I try and use it. Here is the code I am currently using: Code: [Select] <?php echo "<title>"; echo "the Homepage!"; echo "</title>"; ?> Any suggestions would be great. Hi, this is my first time posting here. I am just delving into PHP and I am learning about foreach loops. I have written code in Notepad++ EXACTLY the way I saw it in a tutorial video I watched (I wish I could show the tutorial video to you, but it is on Lynda.com and you have to pay to watch) I attached the file with my code. The example 1 code works just fine. The example 2 code is the one that is not working for some reason. However, it worked for the guy that wrote it in the video, so I am not sure where I am going wrong? *The comments in green are mainly for myself, I explain things to myself so that I don't forget what the code does forloops.php 1.74KB 2 downloads I would appreciate some help. Thank you!!! Hi All Basically I have 4 fields in my database image2, image3, image4, image5 I have an image uploader to upload the files but I need what I need is if image2 is empty put in image3, if image3 is empty put in image4 so far I have this //so first is the query to find the right row which gives the variable $row //now here is what im doing if($row['image2']!==""){ //run this code } elseif($row['image3']!==""){ //run this code } elseif($row['image4']!==""){ //run this code } elseif($row['image5']!==""){ //run this code } Now the problem I'm having is the code that is always executing is the first one of $row['image2'] even if image2 has a value in the database it still runs that code and over writes the field in the database. It never moves on to field 3 or 4. Can someone please tell me if I've done something wrong. I'm not storing the image in the database just the name. Thanks Danny In a PHP file [php_one.php], I wrote some PHP code that contained a form. I used ECHO to write the HTML inside of the PHP tags. On another PHP page [php_two.php], I wanted to show the code that was used to create the first page [php_one.php]. My first effort provided an interesting result, since the PHP code was readable. However, all HTML code was visible in its executed form. There was an actual input field and a submit button (not the code used to produce them) in the context of my coding. I have tried using:
file_get_contents("FILE") All results were similar and carried the same problem. How can I view code so that even the HTML is visible as code? I have the following query where $userid is about 100,000 userids separated by commas. Code: [Select] <? $getpostid=mysql_query("SELECT DISTINCT post.username,post.threadid,thread.threadid,thread.forumid from post,thread WHERE post.username='$username' AND post.threadid=thread.threadid AND post.userid IN ($userid) AND thread.forumid IN ($forumids)") or die (mysql_error()); ?>can i just print all the usernames from above result separated by ';' without looping it multiple times? Could someone help me out how to do above in an efficient manner? I'm tired and want to sleep so my mistakes are getting lots now. Anybody awake that can explain my problem with the code below. <? function directoryToArray($directory, $recursive) { $array_items = array(); $i = "0"; if ($handle = opendir($directory)) { while (false !== ($file = readdir($handle)) && $i < "3") { if ($file != "." && $file != "..") { if (is_dir($directory. "/" . $file)) { if($recursive) { $array_items = array_merge($array_items, directoryToArray($directory. "/" . $file, $recursive)); } } else { $file = $directory . "/" . $file; $array_items[] = preg_replace("/\/\//si", "/", $file); $i++; } } } closedir($handle); } natsort($array_items); return array_reverse($array_items); } $data = directoryToArray('images/screenshot/', TRUE); print_r($data); ?> What I want is for the while loop to stop after 4 hits and exit and give me the return results. But now it continue until readdir end. Hey there, I recently started working on a PHP calculator, I tryed testing it out and the script is not executing... *NOTE* I found the error, I forgot to assign values to the options, all is good now Code: [Select] <?php /* Script Created by: Far Cry Started: August 20th, 2010 10:51 PM PST Completed */ $title = "Calculator"; $form = "<form action='calculator.php' method='POST'> <table> <tr> <td><input name='firstnumber' type='text'></td> </tr> <tr> <td><input name='secondnumber' type='text'></td> </tr> <tr> <td><select name='operation'> <option name='multiply'>*</option> </select></td> </tr> <tr> <td><input name='submitbtn' type='submit' value='Calculate!'></td> </tr> </table> </form>"; if ($_POST['submitbtn']){ $first = ($_POST['firstnumber']); $last = ($_POST['secondnumber']); $operation = ($_POST['operation']); } if ($operation == 'multiply'){ echo"The answer is " . $first * $last . $form; } echo $form; ?> Thanks in advance. Hey guys! My current employer contracted someone to create a script for him. (The script itself is to create a serial number for a piece of software) It's just a PHP script that executes a file and retrieves its output. Now I've never worked with exec() in PHP, and I don't have a reliable environment to test it in. I compiled apache2 and php 5.0.4 and set them to run on a Macintosh (10.3.9) in the office, but I'm sure I've missed some crucial setup stage (mainly, I'm missing php.ini. I did a php -i | grep php.ini to find it, and it says it is in /etc/, but when I look in my PHP/etc folder, only pear.conf is there!). PHP and apache both appear to be working fine, though. When I try and run the PHP script that executes the .ELF file, I get a blank page. Is this because I'm running it in OS X? should I be running it in Linux? Also, I don't know how to Enable exec() on my server, or to see if it's enabled. Basically, I just need someone to walk me through the basics of this. The script and executable SHOULD work fine because the contractor got it working fine. Below is the script, and attached is the executable. Thanks! :3 <?php /* inUID = 456890 inSerialNumUID = 123456 inMachineID_s = &6dA-8P@K-Xsq */ $p1 = "456890"; $p2 = "123456"; $p3 = "\&6dA-8P@K-Xsq"; exec("./create_activation_code.elf $p1 $p2 $p3", $output); foreach ($output as $output_str) { echo $output_str."<br>"; } ?> Hello, PowerShell script stored locally can never be executed after I click on button:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Testing PowerShell</title>
</head>
<body>
<?php
// If there was no submit variable passed to the script (i.e. user has visited the page without clicking submit), display the form:
if(!isset($_POST["submit"]))
{
?>
<form name="testForm" id="testForm" action="get-process.php" method="post" />
Your name: <input type="text" name="username" id="username" maxlength="20" /><br />
<input type="submit" name="submit" id="submit" value="Do stuff NowNew" />
</form>
<?php
}
// Else if submit was pressed, check if all of the required variables have a value:
elseif((isset($_POST["submit"])) && (!empty($_POST["username"])))
{
// Display the alert box
echo '<script>alert("Welcome to Geeks for Geeks")</script>';
// Get the variables submitted by POST in order to pass them to the PowerShell script:
$username = $_POST["username"];
// Best practice tip: We run out POST data through a custom regex function to clean any unwanted characters, e.g.:
// $username = cleanData($_POST["username"]);
$psPath = "C:\\Windows\\SysWOW64\WindowsPowerShell\\v1.0\\powershell.exe";
$psDIR = "C:\\TestNew\\";
$psScript = "pscripta.ps1";
$runScript = $psDIR. $psScript;
$runCMD = $psPath." ".$runScript." 2>&1";
echo "\$psPath $psPath <br>";
echo "\$psDIR $psDIR <br>";
echo "\$psScript $psScript <br>";
echo "\$runScript $runScript <br>";
echo "\$runCMD $runCMD <br>";
exec( $runCMD,$out,$ret);
echo "<pre>";
print_r($out);
print_r($ret);
echo "</pre>";
}
// Else the user hit submit without all required fields being filled out:
else
{
echo "Sorry, you did not complete all required fields. Please go back and try again.";
}
?>
</body>
</html>
Thank you for your help! OK, so I have the following file that has database functions: <?php function dbDelete($param){ $conDelete = mysql_connect($url,'username','password',true); mysql_select_db('my_db',$conDelete); mysql_query($param,$conDelete); mysql_close(); if(isset($conDelete)){ mysql_close($conDelete); } } ?> I include the above file in a session handling file. But for some reason the new file can't call the above function as follows: <?php include_once 'the above stated file'; function timeOut(){ dbDelete("DELETE FROM acctussessi WHERE acctussessi_usid = '$sessius[0]'"); $_SESSION = array(); setcookie(session_name(),'',time()-4200); session_destroy(); } //CHECK FOR TIMEOUT REQUEST switch($_GET['xyz']){ case 'timeout': timeOut(); header("Location: /?xyz=timedout"); break; case 'logout': timeOut(); header("Location: /?xyz=loggedout"); break; case 'refresh': dbUpdate("UPDATE acctussessi SET acctussessi_time = ".time()." WHERE acctussessi_unid ='".$token."'"); break; } ?> dbDelete isn't being called inside of the TimeOut() function. The only way it works is by doing the following: <?php function timeOut(){ $_SESSION = array(); setcookie(session_name(),'',time()-4200); session_destroy(); } //CHECK FOR TIMEOUT REQUEST switch($_GET['xyz']){ case 'timeout': dbDelete("DELETE FROM acctussessi WHERE acctussessi_usid = '$sessius[0]'"); timeOut(); header("Location: /?xyz=timedout"); break; case 'logout': dbDelete("DELETE FROM acctussessi WHERE acctussessi_usid = '$sessius[0]'"); timeOut(); header("Location: /?xyz=loggedout"); break; case 'refresh': dbUpdate("UPDATE acctussessi SET acctussessi_time = ".time()." WHERE acctussessi_unid ='".$token."'"); break; } ?> But this is annoying and repetitive. What is going on? Thanks in advance. Looking to INSERT a row in a data table and keeping the User on the current page. Will it work like what I have below, or do I need to send it to its own page then return the User back to where he started? What I have below isn't inserting a row. No errors showing.
function bookmark_add () {
if (isset($_POST['submit'])) {
include("/home2/csi/public_html/resources/con.php");
$query = "INSERT INTO a_player_bookmark (bookmark)
VALUES ('1')";
$results = mysqli_query($con,$query);
echo mysqli_error($con);
}
$output = '<form name="bookmark_add" method="post" action="" class="bookmark-plus">';
$output .= '<span><input type="submit" name="Bookmark"></span>';
$output .= '</form>';
return $output;
}
hey guys im sure this is possible im wondering how a user can execute a mysql query by a click of a button but without the page reloading...thanks guys In my php web program I'm trying to execute a bash shell script on the web server passing to it some parameters gathered within the program. Ideally I wouldl ike to run this script as another user. However, I get only a return code of "1" and no other error messages that I can see. I tried using system and passthru as well and I'm sure the script is executable by anyone. I've also tried the sticky bit on the target shell script. Can someone/anyone please provide some gotcha pointers about doing this? As can be expected, after playing with this for a week now I'm extremely frustrated and nearly exhausted my resources (except for this forum). Ideas/suggestions please? Thanks Hello all, I have been working with PHP for almost a year now, so somewhat still new to it all. OK, this may be an Apache issue, but this involves a php script I wrote that tries to execute on pages that do NOT call it (via the include command). So, here is a little info on how I have set things up. I use WAMP Server to test php and MySQL code for myself (for fun and for practice) and for clients and their web sites. I run WAMP Server off of my laptop, so I can code on the go and because it is on my laptop, I decided to create a PHP & MySQL user authentication setup to keep out unwanted visitors, but allow trusted friends and fellow developers to view my work on client web sites. Let's say the authentication script is called "auth.php" and for all of the pages I want private, I include that script at the top of those pages, if the visitor is not logged in, redirect them to the login page, otherwise, let them view the page (provided their permission level is adequate). Now, for the client web sites that I am working on, I do not call auth.php anywhere on those pages. Now, the problem I notice is that when I visited a client's web site that I am working on (on my WAMP Server), I reviewed my Apache error log and it gave me a PHP error stating the following: Code: [Select] Maximum execution time of 60 seconds exceeded in E:\\wamp\\www\\assets\\auth.php on line 3, referer: http://192.168.1.4/Web_Projects/NCCCLP/home.php Of course I am able to view the client web site pages without having to login, however, the auth.php script tries to run nonetheless. So my question is, does anyone know why this is happening and if and how I can stop it? This is not a serious (urgent) issue, but it does clog up my error logs and so forth and I prefer to have as few errors as possible. Plus, I would just like to know why and how it is running without being called. Any and all help is greatly appreciated and if I need to provide anymore details or be more specific, please let me know. Also, if an issue like this has already been solved and I missed that thread, please let me know and have my apologies. Thanks in advance. Hi, I am using cpanel, where i am want to run a php file, to delete the old records, i used cronjob in cpanel to do this. In time columns i have made : * * * * * to execute each and every minute the script.. In the command textbox i have made: /home/server25/public_html/auto_cancel.php in the email box: admin_email@gmail.com I am getting Permission denied to my email address: The error report am getting is : /bin/sh: /home/server25/public_html/auto_cancel.php: Permission denied Please help me to solve this. Thanks, Hi guys, im having a problem, in my phpfile were is generated the action code form the contacts form is giving me a lot of trouble, im receiving a lot of blank emails, first i thiked was bots,im usign a captcha, but then i notice another thing, when i put the name of the file that generates the action in the browser this loads it and send it the black email. How can i prevent for my php action file to dont be loadit in the browser? Here is the code to you guys have a idea <?php $datahora = "DATA: <B>" . date("d/m/y - H:i:s") . "</B><BR><BR>"; foreach ($_POST as $campo => $valor) { if (($campo == 'imageField2_x') or ($campo == 'imageField2') or ($campo == 'imageField2_y') or ($campo == 'distrito') or ($campo == 'subimit_y') or ($campo == 'codigo') or ($campo == 'seguranca')) {}else { if ($valor <> '') { $campo = str_replace("_", " ",$campo); $campos .= strtoupper($campo) . ": <b>" . $valor . "</b><Br>"; } } } $www = "WWW.USA.COM"; $assunto = "CONTACT - USA - " . $www; $conteudo = "CONTACT - USA<br><br>" . $datahora . ($campos) . "<br>" . $www; $para = "madness@hotmail.com"; $headers = "MIME-Version: 1.0\n"; $headers .= "Content-type: text/html; charset=iso-8859-1\n"; $headers .= "From: USA <info@usa.com>\r\n"; $headers .= "Reply-To: ".$_POST['email']."\r\n"; if ($_POST['codigo'] == $_POST['seguranca']) { if (mail($para,$assunto,$conteudo,$headers) == true){ ?> <script> alert('Sent sucess!'); window.location = 'contact.php'; </script> <? }} ?> Hope for some help Hi, In some cases I need to execute a default query , so I am storing the SELECT statement in a variable like this and executing it: Code: [Select] $default_query="SELECT * from user where userid='$userid'"; $user_query=mysql_query($default_query); The above code returns an empty results. But if I execute it without the variable it works fine. Code: [Select] $user_query=mysql_query("SELECT * from user where userid='$userid'"); Am I syntatically wrong somewhere? Is there a way or function that stops the execution of a required script but continues to execute the parent script, for example: if this is a.php <?php echo "<br>Good morning"; // function to stop execution of only this script to appear here echo "<br>Good Afternoon"; ?> and then this is b.php <?php require("a.php'); echo "<br>Good Evening"; ?> So if the file b.php is called it is supposed to display: Good morning Good evening So it has to skip "Good afternoon" because of the anonymous function which I need Hi all,
This is just an experiment to make sure that I can get data from a Stored Proc 'IF' and when I need to, so it's basically a THOUGHT EXERCISE.
MSSQL 2014
I'm a noobie and need a hand, I've managed to get my Instance connected to the internet and I can query it using PHP and SQL, I can also look at views with no problem.
I have it working as an "ADODB.Connection" and like I said it connects and I can query data and display results.
Now I have coded a Stored Proc "GetMonthDays" in Sql Server:
Which returns days 1 through xxx in a given month and also returns the Day name eg... Sat for each date
2014-01-01 Thurs
2014-01-02 Fri
etc...
It works perfectly and very fast so All cool with that side BUT...
I want to be able to query the Database through a Stored Proc, I've spent all day trying to find a way to get this to work and I've hit a wall
Can anyone please tell me IF it's A. possible and B. maybe if they are really really nice and share with me a code example that I can change to fit my needs ?
This T-SQL returns what it needs to
Begin EXEC dbo.qselGetMonthDays '2015-01-01' End Cheers JD Edited by juddadredd, 02 January 2015 - 04:55 PM. [About the code] Basically what I have is a web page that de-codes a complex XML document, and displays it as a table. This is working successfully. One of the columns in the tables displays a "rank" of an object as a number (0-9). This is working successfully. What I did was write a code that would translate that rank number into a specified text, depending what number it was. [The Code] (This is only the partial code with the problem in it, the entire code of the page is very extensive.) Code: [Select] $RankNum = "{$Member->Rank}"; // {$Member->Rank} successfully displays the rank number on it's own. Below 'should' translate that number into the name of the rank. if('$RankNum' == 0) { echo('test1');}; if('$RankNum' == 1) { echo('test2');}; if('$RankNum' == 2) { echo('test3');}; if('$RankNum' == 3) { echo('test4');}; if('$RankNum' == 4) { echo('test5');}; [The Problem] All this code keeps displaying "test1" for every value in the entire column, regardless if it equaled 1 or 2 or higher... I know this has to be some simple coding mistake, but I've been sitting here for an hour trying to figure out what I had done wrong. Thanks in advance :) |
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