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PHP - Need Help Writing A Function(probably Really Easy)
So I know next to nothing about PHP so I need some help with something thats probably incredibly easy.
I just need a function(hope I'm using that right) that checks if a div with a certain class exists in the same page. then based on whether that's true or not changes the class of an image. This is how I imagine it looks except I don't know what to put in the if brackets to make it find the div and evaluate if its there or not. <img if (){ echo(class=\"visible\"); } else{ echo(class=\"hidden\"); } } src=\"icon.png\" /> Similar TutorialsHi, I found this very good tutorial at IBM's site: http://www.ibm.com/developerworks/xml/library/x-validxphp/ BUT my question is, what if any arbitrary XML file is passed passed to this xml validator function as parameter (that I will soon write), so meaning there may not be an xml schema created, so how would I validate it then b/c the link above assumes there a schema is made to use to validate. Any help much appreicated! I am going through the tutorial found at http://goo.gl/69MeY and in my ultimate wisdom have decided to see if I can create a form, which you input some text, and then in the php file that handles the form in puts that variable into a text document using the fwrite function, basically creating a file from the website containing whatever text you put in. There is no specific purpose to this idea, I am just trying to implement everything I have learnt so far. Form: http://pastebin.com/J62UHBeT PHP Handler: http://pastebin.com/FHuaKkxD I someone could point me in the right direction of what I am doing wrong it would be much appreciated Thanks in advance Michael hello, I'm a beginner when it comes to PHP: I'm working on a log in system but i keep on getting the same errors which makes my system unreliable. this is the function I'm trying to use: if(pg_numrows($q) == 1){ echo "<p>Someone took that username</p>"; include "signup.php"; exit; } --> I'm using a postgresql database this is the warning I keep on getting: Code: [Select] Warning: pg_numrows(): supplied argument is not a valid PostgreSQL result resource in /var/ftpdirs/512544/PHP/login/adduser.php on line 13 I have the same problem when I try to look if an username is taken or not... when people try to register this is my adduser.php file: <?php session_start(); include "connectie_db.php"; $User_name = $_POST['user_name'] ; $User_pass = $_POST['user_pass'] ; $User_pass2 = $_POST['user_pass2'] ; $checkUsername= "SELECT user_name from tovanu.users where user_name = '$User_name';"; $q = $db->exec($checkUsername); if(MDB2::isError($q)){ echo "code: ".$q->getUserInfo(); exit(); } if(pg_numrows($q) == 1){ echo "<p>Someone took that username</p>"; include "signup.php"; exit; } If(strlen($User_name > 32)){ echo "<p>The username is too long</p>"; include "signup.php"; exit; } if($User_pass != $User_pass2){ echo "<p>Both passwords must be the same</p>"; include "signup.php"; exit; } $password = md5($User_pass); $add = "INSERT INTO tovanu.users (user_name,user_pass,user_email,user_date,user_level,naam,adres,plaats,postcode) VALUES ('$User_name','$password','email',current_date,1,'jef','jonhstraat','maaseik','3687')"; $execute = $db->exec($add) ; if(MDB2::isError($execute)){ echo "code: ".$q->getUserInfo(); exit(); } $_SESSION['user'] = $username; include "index.php"; ?> any help would be appreciated!!!!!!!!!!! Ok, I know how to write to a file, but what I'm looking for is to check if a line of code exists, and if it is, don't recreate it. It would also be nice to not recreate the file either. Code: [Select] $file = fopen("index.html", "w"); fwrite($file,"This is a line of text"); fclose($file); im trying to make a program that changes some files from 0 to 1,but im having some trouble... heres my code, Code: [Select] $id = $_GET["id"]; $file1 = "http://mysite.co.cc/users/".$id."/file1.txt"; $fh = fopen($file1, 'w'); fwrite($fh, "1"); fclose($fh); I followed this tut: http://www.tizag.com/phpT/filewrite.php I cant find anything wrong with the code,but it will not make the file have 1 in it. I don't think the start of my code is right?! Code: [Select] echo '<button onclick=\"gohere(viewpub.php?PubID='.$row['PubID'].')" id="button" type="button" class="ui-button ui-widget ui-state-default ui-corner-all ui-button-text-only" role="button" aria-disabled="false"><span class="ui-button-text">View Pub</span></button>'; Please help?! Greetings,
I am new to these forums, I am working on this assignment, and these are the current issues I am running into.
Notice: Undefined variable: year in G:\EasyPHP-5.3.2i\www\PHP_Projects\ChineseZodiacs\zodiac_year_switch.php on line 77
ie. $year = validateInput($year,"Birth Year");
Notice: Undefined variable: year_count in G:\EasyPHP-5.3.2i\www\PHP_Projects\ChineseZodiacs\zodiac_year_switch.php on line 138
ie. echo "<p>You are person " . $year_count . "to enter " . $year . "</p>\n";
Honestly, I believe they are linked, because what should be happening, as the user enters the year, and hits submit, it should create a file called counts/$year.txt - $year should equal the entered data in the textbox, any help would be appreciated.
Thank you for your help.
<!DOCTYPE html>
<head>
<title>Write to and From a File</title>
<meta http-equiv="content-type" content="text/html; charset=iso-8859-1" />
<?php
$dir = "counts";
if ( !file_exists($dir)) {
mkdir ($dir, 0777);
}
function validateInput($year, $fieldname) {
global $errorCount;
if (empty($year)) {
echo "\"$fieldname\" is a required field.<br />\n";
++$errorCount;
$retval = "";
}
else
{ // if the field on the form has been filled in
if(is_numeric($year))
{
if($year >=1900 && $year <=2014)
{
$retval = $year;
}
else
{
++$errorCount;
echo "<p>You must enter a year between 1900 and 2014.</p>\n";
}
}
else
{
++$errorCount;
echo "<p>The year must be a number.</p>\n";
}
} //ends the else for empty
return($retval);
} //ends the function
function displayForm() {
?>
<form action = "<?php echo $_SERVER['SCRIPT_NAME']; ?>" method = "post">
<p>Year of Birth: <input type="text" name="year" /></p>
<p><input type="reset" value="Clear Form" /> <input type="submit" name="submit" value="Show Me My Sign" /></p>
</form>
<?php
}
function StatisticsForYear($year) {
global $year_count;
$counter_file = "counts/$year.txt";
if (file_exists($counter_file)) {
$year_count = file_get_contents($counter_file);
file_put_contents($counter_file, ++$year_count);
} else {
$year_count = 1;
file_put_contents($counter_file, $year_count);
}
return ($year_count);
}?>
</head>
<body>
<?php
$showForm = true;
$errorCount = 0;
//$year=$_POST['year'];
$zodiac="";
$start_year =1900;
if (isset($_POST['submit']))
$year = $_POST['year'];
$year = validateInput($year,"Birth Year");
if ($errorCount==0)
$showForm = false;
else
$showForm = true;
if ($showForm == true)
{
//call the displayForm() function
displayForm();
}
else
{ //begins the else statement
//determine the zodiac
$zodiacArray = array("rat", "ox", "tiger", "rabbit", "dragon", "snake", "horse", "goat", "monkey", "rooster", "dog", "pig");
switch (($_POST['year'] - $start_year) % 6) {
case 0:
$zodiac = $zodiacArray[0];
break;
case 1:
$zodiac = $zodiacArray[1];
break;
case 2:
$zodiac = $zodiacArray[2];
break;
case 3:
$zodiac = $zodiacArray[3];
break;
case 4:
$zodiac = $zodiacArray[4];
break;
case 5:
$zodiac = $zodiacArray[5];
break;
case 6:
$zodiac = $zodiacArray[6];
break;
case 7:
$zodiac = $zodiacArray[7];
break;
case 8:
$zodiac = $zodiacArray[8];
break;
case 9:
$zodiac = $zodiacArray[9];
break;
case 10:
$zodiac = $zodiacArray[10];
break;
case 11:
$zodiac = $zodiacArray[11];
break;
default:
echo "<p>The Zodiac for this year has not been determined.</p>\n";
break;
} //ends the switch statement
echo "<p>You were born under the sign of the " . $zodiac . ".</p>\n";
echo "<p>You are person " . $year_count . "to enter " . $year . "</p>\n";
} //ends the else statement
?>
</body>
</html>
Edited by mstevens, 16 October 2014 - 06:36 PM. I need help! I cant get this to write the correct way! What i need is based on the value of what is posted to the script it has to write it in the config file and also make a folder! Please help Code: [Select] <?php $start = '$uploadpath=\''; $structure = '../banner/images/'.$_POST['FOLDER'].'\';\n'; $myFile = "PHP/confup.php"; $fh = fopen($myFile, 'w') or die("can't open file"); $stringData = "<?\n"; fwrite($fh, $stringData); $stringData = "$start $structure"; fwrite($fh, $stringData); $stringData = "?>\n"; fwrite($fh, $stringData); fclose($fh); // Desired folder structure // To create the nested structure, the $recursive parameter // to mkdir() must be specified. if (!mkdir($structure, 0777, true)) { die('Failed to create folders...'); } // ... ?> I am using a cache script, well it writes a array to a file like this: Code: [Select] fwrite($fh, '<?php'."\n\n".'define(\'PUN_LOTTERY_LOADED\', 1);'."\n\n".'$lottery = '.var_export($output2, true).';'."\n\n".'?>'); output2 is Code: [Select] $result2 = $db->query('MY QUERY '); $output2 = array(); while ($cur_donors = $db->fetch_assoc($result2)) $output2[] = $cur_donors; Now, I want to ditch the mysql and I want to use this script with 7 variables that I already have loaded, so I dont need to use the mysql, how do I add my 7 variables to my var_export function instead of using mysql to loop them? Hey Guys. I am trying to write to the file depending on which condition is met. The code works fine on my local machiene but not on my remote server. I have also tried to output any error messages to see if it would output anything, and I don't get anyting on my browser. Can anyone help me with this issue? Thanks
<?php
if($_SERVER['REQUEST_METHOD'] == "POST") {
isset($_POST['interfax']) ? $option= "interfax" : $option= "";
isset($_POST['metrofax']) ? $option= "metrofax" : $option= "";
switch ($option) {
case 'interfax':
$file = "fax.php";
$fax_client = "interfax";
if(file_put_contents($file, "<?php ".'$fax_client = "' . $fax_client . '"'." ?>"))
{ echo "Successful"; } else {
die("Can't write file");
}
break;
// By defualt all the orders go to metrofax so by selecting the variable it resets it self
case 'metrofax':
$file = "fax.php";
$fax_client = "metrofax";
file_put_contents($file, "<?php ".'$fax_client = "' . NULL . '"'." ?>");
break;
}
}
?>
<form action="#" method="POST">
<input type="radio" name="interfax" value="interfax">Switch To Interfax<br>
<input type="radio" name="metrofax" value="metrofax">Switch To Metrofax<br>
<input type='submit' name="submit" >
I'm writing a script that takes user input from a html form and updates the database with the new data. In this example the user is updating the data about the university they attend. The problem with this is, the script to update the database with the new information appears to be not working - the new data is not being added to the database. The SQL queries used in the script work in phpMyadmin, and the variables - $uni and $username - contain the data the correct data. This has me stumped, could someone look over this for me please and tell me what is going wrong here. Code: [Select] <?php include 'connect.php'; session_start(); $_SESSION['username']; if(!(isset($_SESSION['login']) && $_SESSION['login']!= " ")){ header("Location: login.php"); } $username = $_SESSION['username']; $tablename = 'usr_test'; $uni = $_POST['uni']; $uni = stripslashes($uni); $uni = mysql_real_escape_string($uni); $uni = trim($uni); if(isset($uni)) { $username = $_SESSION['username']; $loc = mysql_query("SELECT * FROM usr_test WHERE usr = '$username'"); if (mysql_num_rows($loc) == 0) header("Location:notlogged.php"); else { extract(mysql_fetch_array($loc)); mysql_query("UPDATE usr_test SET uni = ('$uni') WHERE usr = '$username'") or die (mysql_error()); header("Location:profile.php"); } } ?> Greetings, I am trying to write XML using SimpleXML for a web service call. I was having success until the XML got a little more complicated. Here is the XML format where I am running into problems: Code: [Select] <Agent> <Person Last='Smith' First='John'> <Addresses /> <PhoneNumbers> <Phone Type='1' Number='888-555-1212'> </PhoneNumbers> </Person> </Agent> Here is my php: Code: [Select] $xmlOutput = new SimpleXMLElement('<?xml version="1.0"?><ReportRequest> </ReportRequest>'); $xmlOutput->addAttribute('CID','9200'); $xmlOutput->addAttribute('Diagram','1'); $xmlOutput->addAttribute('DueDate','2011-11-15'); $xmlOutput->addAttribute('NumPhotos','6'); $xmlOutput->addAttribute('InspectAfter',''); $xmlOutput->addAttribute('PolicyNumber','JTC0004425'); $reportType = $xmlOutput->addChild('ReportType'); $reportType->addAttribute('CPType','Commercial'); $reportType->addAttribute('SectionIDs',''); $reportType->addAttribute('Description',''); $reportType->addAttribute('ReportTypeID','123'); $locations = $xmlOutput->addChild('Locations')->addChild('Addresses')->addChild('Address'); $locations->addAttribute('Zip','91216'); $locations->addAttribute('City','Chatsworth'); $locations->addAttribute('Line1','123 Main St'); $locations->addAttribute('Line2','Suite 201'); $locations->addAttribute('State','CA'); $locations->addAttribute('Latitude',''); $locations->addAttribute('Longitude',''); $agent = $xmlOutput->addChild('Agent')->addChild('Person')->addChild('Addresses')->addChild('PhoneNumbers')->addChild('Phone'); $agent->addAttribute('Last','Smith'); $agent->addAttribute('Email',''); $agent->addAttribute('First','John'); $agent->addAttribute('Title',''); $agent->addAttribute('Type','1'); $agent->addAttribute('Number','888-555-1212'); $agent->addAttribute('TypeName','Office'); $agent->addAttribute('Extension',''); Header('Content-type: text/xml'); echo $xmlOutput->asXML(); Everything outputs as it should until the line that starts with $agent. I want the output to match the section above. I have tried several variations but I cannot seem to figure it out. I know the current PHP sample doesn't work. Help?? Thanks in advance, John I have a web app hosted on Just Host that I have nearly finished writing but still needs to have coded the ability to write a text file to user's computer. the user should be able to specify where on their computer they would like the file to be stored. Is this possible ? And can you tell me how it's done, or point me to a souce that can tell me I rarely ever ask for help regarding programming, but this has flumoxed me. If it is not possible to do this, then would I have to generate this file on the Host's server (Just Host in my case), then download it ? If this can be done then can you please tell me how, as any info I have found related to file downloads seems a bit obscure thanks in advance I'm trying to write to a file with the following code. $fh = fopen("../inc/config.php", "a") or die("\r\nCan't open file."); $write = "\$config['database_host'] = {$mysqlh}; \$config['database_user'] = {$mysqlu}; \$config['datbase_pass'] = {$mysqlp}; \$config['datbase_name'] = {$dbn}; \$config['table_prefix'] = {$tp};"; fwrite($fh, $write); fclose($fh); It is printing out "Can't open file", which I guess means it can't find the file or I've given the wrong path. The file that is executing this code is /install and the file I'm trying to write to is /inc/config.php. I thought .. put you up a directory so if I do .. I will be at the root and then do /inc I will be in inc. Can someone please guide me on what I'm doing wrong? Thanks. I have written some code to past information from one page to a diffrent page however all the variables are sent over but they are not writting to the database could anyone help me out please .... $ud_P_Id = $_POST['ud_P_Id']; $ud_LastName = $_POST['ud_LastName']; $ud_FirstName = $_POST['ud_FirstName']; $ud_GroupCode=$_POST['ud_GroupCode']; $ud_P_Unit1=$_POST['ud_P_Unit1']; $ud_P_Unit2=$_POST['ud_P_Unit2']; $ud_P_Unit3=$_POST['ud_P_Unit3']; $ud_P_Unit6=$_POST['ud_P_Unit6']; $ud_P_Unit14=$_POST['ud_P_Unit14']; $ud_P_Unit20=$_POST['ud_P_Unit20']; $ud_P_Unit27=$_POST['ud_P_Unit27']; $ud_P_Unit28=$_POST['ud_P_Unit28']; $ud_P_Unit42=$_POST['ud_P_Unit42']; $ud_P_Unit10=$_POST['ud_P_Unit10']; $ud_P_Unit11=$_POST['ud_P_Unit11']; $ud_P_Unit12=$_POST['ud_P_Unit12']; $ud_P_Unit13=$_POST['ud_P_Unit13']; $ud_P_Unit1454=$_POST['ud_P_Unit1454']; $ud_P_Unit15=$_POST['ud_P_Unit15']; $ud_P_Unit16=$_POST['ud_P_Unit16']; $ud_P_Unit17=$_POST['ud_P_Unit17']; $ud_P_Unit18=$_POST['ud_P_Unit18']; $con = mysql_connect('localhost', 'lccstude_progre', '********'); $db= "lccstude_pro"; if (! $con) die("Couldn't connect to MySQL"); mysql_select_db($db , $con) or die("Couldn't open $db: ".mysql_error()); mysql_query("UPDATE BTECL31113 SET FirstName='$ud_FirstName' , LastName='$ud_LastName' , GroupCode='$ud_GroupCode' , Unit1='$ud_P_Unit1' , Unit2='$ud_P_Unit2' , Unit3='$ud_P_Unit3' , Unit6='$ud_P_Unit6' , Unit14='$ud_P_Unit14' , Unit20='$ud_P_Unit20' , Unit27='$ud_P_Unit27' , Unit28='$ud_P_Unit28' , Unit42='$ud_P_Unit42' , Unit10='$ud_P_Unit10' , Unit11='$ud_P_Unit11' , Unit12='$ud_P_Unit12' , Unit13='$ud_P_Unit13' , Unit1454='$ud_P_Unit1454' , Unit15='$ud_P_Unit15' , Unit16= $ud_P_Unit16' , Unit17='$ud_P_Unit17' , Unit18='$ud_P_Unit18' WHERE P_Id='$ud_P_Id'"); echo "Record Updated"; mysql_close($con); Please any help would be great I have a code for writing a log file... The code is working fine but it inserts the new details at the bottom of the file but i want to insert in the top... So what function should i use... Here is the file my_log.php Code: [Select] <?php $usname = $_SESSION['usname']; date_default_timezone_set('Asia/Calcutta'); $date = date("l dS \of F Y h:i:s A"); $file = "log.php"; $open = fopen($file, "a+"); fseek($open,289); fwrite($open "<b><br/>USER NAME:</b> ".$usname . "<br/>"); fwrite($open, "<b>Date & Time:</b> ".$date. "<br/>"); fwrite($open, "<b>What have they done :</b> ".$reason . "<br/><br/>"); fclose($open); ?> and here is my log.php file : Code: [Select] <?php session_start(); if($_SESSION['stage']!=1 || $_SESSION['stage2']!=2) {header('location:index.php'); die(" "); } ?> <?php if($_GET['valu']=="view_log") { $reason=" ".$_SESSION['usname']." Viewed the LOG"; include('my_log.php'); header('location:log.php'); die(""); } // bytes till here are 289 //LINE 1 : Log details have to insert here I want to insert every detail from the top of the page but just below the php code.. What should i do... ANy helo would be appreciated Pranshu Agrawal pranshu.a.11@gmail.com I am using a script I adapted from a tutorial to print the contents of a text box to a txt file. Basically, it's a really simple way of seeing who has logged in. I only have a handful of users. The problem is, although the text file is being created in the proper folder, it isn't being written to and just remains blank. <div align="center"> <table width="300" border="2" bordercolor="#FFFFFF" style="-moz-border-radius: 18px; -webkit-border-radius: 18px;" height="120" cellpadding="0" cellspacing="0"> <tr> <form name="form1" method="post" action="checklogin.php"> <td> <table width="100%" border="0" cellpadding="3" cellspacing="1" background="images/loginbg.jpg" style="-moz-border-radius: 15px; -webkit-border-radius: 15px;"> <tr align="center"> <td colspan="3"><font color="#FFFFFF"><strong>Family Login </strong></font></td> </tr> <tr> <td width="78"><font color="#000000">Username</font></td> <td width="6">:</td> <td width="294"><input name="myusername" type="text" id="myusername"> <?php $myusername = $_POST['myusername']; $data = "$myusername\n"; //open the file and choose the mode $fh = fopen("logs/login.txt", "a"); fwrite($fh, $data); fclose($fh); ?></td> </tr> <tr> <td><font color="#000000">Password</font></td> <td>:</td> <td><input name="mypassword" type="password" id="mypassword"></td> </tr> <tr> <td> </td> <td> </td> <td><input type="submit" name="Submit" value="Login"> </td> </tr> </table> </td> </form> </tr> </table> </div> I'm not sure what's going wrong but I'm guessing it's the placing of the php, or at least some of it. I'd quite like to add the time they logged in as well. Any idea's anyone? Hi everyone. Still new to PHP and to Object-Oriented Programming. My goal for today is to write a few dummy Classes in PHP that do enough so that I can see something on my webpage. I am wondering if there is someone here who would be willing to help in either the forums or one-on-one. It may sound like a silly request, but this is all new to me as a Procedural Programmer who has actively programmed in about 10 years!! Thanks, TomTees ob_start(); include('file.php'); $content = ob_get_clean(); ok so this is the code i found for adding a file to a variable.. now how would i use it in term of this: <?php $id = NULL; if(isset($_GET['id']) && $_GET['id'] == "1"){ $id = 'Marc'; } else if(isset($_GET['id']) && $_GET['id'] == "2"){ $id = 'Glenn'; } else if(isset($_GET['id']) && $_GET['id'] == "3"){ $id = 'Nathan'; } else if(isset($_GET['id']) && $_GET['id'] == "4"){ $id = 'Scott'; } ?> ok so the get variable checks to see whose id youve requested. then when for example id = 4 i want to display an include file somewhere else on the page so i would use.. <?php echo $id; ?> now insted of it being "nathan" i want it to be a file i didnt really understand the first code i found I found this bug when creating my topic he
http://forums.phpfre...ted-rectangles/
I spent 10 minutes writing up my issue after I copy and pasted my code inside the [code=auto:0] brackets. Then, what do you know? I submitted my post with all my writing below my [code=auto:0] brackets and BOOM it was all gone. Only my code was showing. That's what issued Requinex to reply like that, it's a nasty forum bug.
Here is a video I made to help explain and re-create the issue. Hopefully you guys fix it, thanks!
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