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PHP - Displaying Literal Code
in my datebase I have this
<iframe src="https://www.xxxxxx.com/agent.asp?id=xxxxx style="width: 545px; height: 750px;" frameborder="1"></iframe> obviously when I echo it it makes the iframe how can I get it to echo out literally what it is with the < > and all? thanks figured it out I changed the < to < in my db Similar TutorialsI have a bunch of PDF's, some of them having Uppercase first letters and some having lowercase first letters. Right now my code is alphabetizing all of the Uppercase first letters first then all of the lowercase. I want them to be mixed. $dir = "../../forms/pdf/"; foreach(glob($dir.'*.pdf') as $pdf){ $files[] = str_replace($dir, '', $pdf); } then: <select name="current_pdf" id="current_pdf"> <option value="">Please Select a PDF to replace</option> <?php foreach($files as $file) { echo "<option value=\"$file\">" . $file . "</option>"; } ?> </select> This topic has been moved to PHP Regex. http://www.phpfreaks.com/forums/index.php?topic=358198.0 This topic has been moved to PHP Regex. http://www.phpfreaks.com/forums/index.php?topic=345658.0 SQLITE has syntax like ' WHERE MATCH ( 'colname : "one two" )' //My pdo sql query $sql .= MATCH 'colname : "?" )'; $pdo->bindValue(1,$text);` But Pdo placeholders can't have quotes around them. So this does not work. I tried a million variations of the placeholder syntax "?" "" ? "" """ ? """ \" ? \" . But nothing works. Errors I get : General error: 1 near "?" | 25 column index out of range . Also for this query : MATCH ( names: ? AND categoryids: ? ) , the error is: `General error: 1 unrecognized token: ":"` Would really love some help here.. Thanks Edited March 29, 2019 by requinix removing bad styling Hi I have a text area, that I want to display info pulled from a database. I can get the data to show, But can get each entry of the table to display on it's own line. Example: ob1ob2ob3ob4ob5 Should be: ob1 ob2 ob3 ob4 ob5 CODE: Code: [Select] <textarea id="interest" onfocus="clearInterest()" class="textareacss" style="height:212px;overflow:auto;"><?PHP $newInterestSub = Admin_interests_sub::find_by_cat_id($id); foreach($newInterestSub as $newInterestSubs){ echo $newInterestSubs->interest_sub.'\n'; } ?></textarea> Any help would be great. Currently testing a site thats almost built, am going to be including php on a sidebar on all pages so thought it'd be easier to just make all pages .php, however when i upload and try and view, in Firefox it just displays the source code on screen! It does it on other computers, but works fine in IE! ARRGGGHH Anyone have a clue? Kind Regards, Chris Hi guys, I have a problem.. I have a table which shows the results, but the table is showing right at the bottom below the button "delete members" although the code is right above it so I don't really understand.. and hope someone can shed the light? Appreciated Code: [Select] <?php include "connect.php"; session_start(); if(!isset($_SESSION['email'])) { die('You have no access to this page.'); } else{ //Check if records were submitted for deletion if(isset($_POST['id']) && count($_POST['id'])) { //Ensure values are ints $deleteIDs = implode(', ', array_map('intval', $_POST['id'])); $query = "DELETE FROM member WHERE id IN ({$deleteIDs})"; //Debug line echo "Successfuly deleted member(s)"; mysql_query($query) or die(mysql_error()); } ?> <link rel="stylesheet" type="text/css" href="view.css" media="all"> <script type="text/javascript" src="view.js"></script> <body id="main_body" > <img id="top" src="top.png" alt=""> <div id="form_container"> <h1>Consult, Modify or Delete a Member</h1> <form id="form_362567" class="appnitro" method="post" action="searchmember.php"> <div class="form_description"> <h2> Consult, Search, Modify or Delete a Member</h2> <p></p> </div> <ul > <li class="section_break"> <br>Search members with any keyword or character within the options id/email/firstname/lastname/phone/address1/address2/city/county/postcode - relevant matches will show.<br><br> <b>Keyword/Character:</b><br> <input id="search" name="search" class="element text medium" type="text" maxlength="50" value=""/> <br><br> <b>Search on</b>:<br> <select name="searchvalue" id="searchvalue"> <option name="id" id="id">id</option> <option name="email" id="email">email</option> <option name="firstname" id="firstname">firstname</option> <option name="lastname" id="lastname">lastname</option> <option name="phone" id="phone">phone</option> <option name="address1" id="address1">address1</option> <option name="address2" id="address2">address2</option> <option name="city" id="city">city</option> <option name="county" id="county">county</option> <option name="postcode" id="postcode">postcode</option> </select><br><input type='submit' value='Search' name='go' /><INPUT TYPE="button" VALUE="Cancel" onClick="history.go(-1);return true;"><br> <br> To consult or modify a profile click on the member's E-mail address. <li class="section_break"> </form> <?php $result = mysql_query("SELECT * FROM member"); ?> <table border='1'> <tr> <th>First Name</th> <th>Last Name</th> <th>Phone Number</th> <th>Postcode</th> <th>E-mail</th> <th>Delete</th> </tr> <?php while($row = mysql_fetch_array($result)) { echo "<form action='' method='POST'>"; echo "<tr>"; echo "<td>" . $row['firstname'] . "</td>"; echo "<td>" . $row['lastname'] . "</td>"; echo "<td>" . $row['phone'] . "</td>"; echo "<td>" . $row['postcode'] . "</td>"; echo "<td>"; echo '<a href="consultmember.php?email='; echo $row['email']; echo '">'; echo $row['email']; echo "</a></td>"; echo "<td><input type='checkbox' name='id[]' value='{$row['id']}' /></td>"; echo "</tr>"; } ?> <br><input type='submit' value='Delete Members' name='delete' /><INPUT TYPE="button" VALUE="Cancel" onClick="history.go(-1);return true;"><br> </form> <div id="footer"> </div> </div> <img id="bottom" src="bottom.png" alt=""> </body> <?php } ?> Hi everyone I just getting back to PHP after a break but have forgot how to build a HTML table row. Here is the part of the code that builds up rows for a html table... while($rows = mysql_fetch_array($qry)){ $table .= " <tr> <td><input type=\"checkbox\" name=\"C1\" value=\"ON\"></td> <td width=\"87\">$rows['ref']</td> <td width=\"178\">$rows['transactionReferenceNumber']</td> <td>$rows['totalAmountReceived']</td> <td>$rows['myItemQuantity']</td> <td>$rows['flag_1']</td> <td>$rows['flag2']</td> </tr> "; } This causes an error as you might expect. Please tell me the best way to place the html code into $table Hey guys, Having a slight problem with part of the code in my index.php file Code: [Select] mysql_select_db('db_name', $con); $result = mysql_query("SELECT * FROM spy ORDER BY id desc limit 25"); $resulto = mysql_query("SELECT * FROM spy ORDER BY id desc"); $count = mysql_num_rows($resulto); while($row = mysql_fetch_array($result)) { ?> <div class="contentDiv">Someone is looking at <?=$row[title];?> Stats for "<a href="/<?=$row[type];?>/<?=$row[code];?>/<?=$row[city];?>"><?=$row[code];?> <?=$row[city];?></a>"</div> <?}?> </div> <div id="login"></div> <? include("footer.php"); ?> </div> </body> </html> I'm getting the following error when viewing the file Code: [Select] Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in ~path to file/index.php on line 70 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in ~path to file/index.php on line 71 where lines 70 and 71 are $count = mysql_num_rows($resulto); while($row = mysql_fetch_array($result)) Any ideas on how to fix this? I hav a php page which takes data from the db.But the problem i am facing is for a particular fild in database ,i have long data,which i need to b displayed in a wordwrap way eg my data is " Description need a site for school gave order on 12-10-2010 work started on 12-10-2010 " this shuld be displayed as "Description need a site for school gave order on 12-10-2010 work started on 12-10-2010" Description is the field in the database and rest of them are the datas in it am total newbie to programming, apart from knowing SQL, the thing is i have been given a MYSQL database containing various information about kids diseases and a web interface written in php to create reports from the database that can be accessed via the interface. there are almost 25 different variables that need to be computed in the report, i have written sql queries to compute all these values, but i don't know anything about PHP, if i have all these queries isn't there a way for me to combine all these sql queries to be display results on a webpage and come up with this report without writing PHP code? Thanks for your help very sorry if this is too basic Hello everyone, I begin in everything web related but I have been programming for years. I tried to code something simple : small Mysql DB (works fine) and to begin a search bar to browse data. I adapted a code that I understood provided here : https://www.cloudways.com/blog/live-search-php-mysql-ajax/. Base principle is simple : as you type in your query, it will pass the text to script.js that will forward this request to ajax.php file. In the ajax.php, a javascript function named “fill()” will pass the fetched results. This function will also display the result(s) into “display” div in the “search.php” file. The problem is that when I type anything it displays, below the search bar, at the moment I type a character: Quote
'; //Fetching result from database. while ($Result = MySQLi_fetch_array($ExecQuery)) { ?> ")'>
instead of the actual answer from my database (no error in the browser console). I tested the SQL query + the user I provide and everything seems fine. Any clue what could be the root cause ? I strongly suspect a mistake in the code as I already corrected one (script.js instead of scripts.js) but I really cannot figure out where. Thanks in advance,
problematic code (ajax.php):
<?php
//Including Database configuration file.
include "db.php";
//Getting value of "search" variable from "script.js".
if (isset($_POST['search'])) {
//Search box value assigning to $Name variable.
$Name = $_POST['search'];
//Search query.
$Query = "SELECT Name FROM search WHERE Name LIKE '%$Name%' LIMIT 5";
//Query execution
$ExecQuery = MySQLi_query($con, $Query);
//Creating unordered list to display result.
echo '
<ul>
';
//Fetching result from database.
while ($Result = MySQLi_fetch_array($ExecQuery)) {
?>
<!-- Creating unordered list items.
Calling javascript function named as "fill" found in "script.js" file.
By passing fetched result as parameter. -->
<li onclick='fill("<?php echo $Result['Name']; ?>")'>
<a>
<!-- Assigning searched result in "Search box" in "search.php" file. -->
<?php echo $Result['Name']; ?>
</li></a>
<!-- Below php code is just for closing parenthesis. Don't be confused. -->
<?php
}}
?>
</ul>
Hello - I'm opening my website up to visitors for free, and trying to bypass a login screen to go straight into the data content that was appearing after a user logged in. I have an index.php file that included the following code at the beginning: <?php session_start(); include("database.php"); include("login.php"); include("/vservers/skyranks/db_connect.php"); ?> <?header("Cache-control: private"); ?> <html> I deleted the "include("login.php"); line, and was successful at bypassing the username and login screen. However, the page that is supposed to display the data content is incomplete. In fact, it only displays my company's logo. Any ideas as to why the data content is not showing up? Thank you for any help with this, as my php is quite novice at this point. Regards - Joe Hello everyone, When I submit my login form to go to my login page it goes to a white page and does not display an error. Code: [Select] <?php include("../secure/database.php"); if(!empty($_POST['loginsubmit'])){ if(!empty($_POST['email'])){ $email = securevar($_POST['email']); if(!empty($_POST['passconf'])){ $pass = securevar($_POST['passconf']); $q = "SELECT * FROM `accountinfo_db` WHERE `email` = '$email' AND `password` = '$pass'"; $res = mysql_query($q) or die(mysql_error()); $login = mysql_fetch_array($res); $id = $login['id']; $active = $login['active']; if($id>=1){ if($active==1){ $_SESSION['logged'] = $login['id']; $q = "UPDATE `accountinfo_db` SET `loggedtimes` = `loggedtimes`+'1' WHERE `id` = '$id'"; $res = mysql_query($q) or die(mysql_error()); $user = $login['username']; header("Location: ../mainframe.php?strmsg=Welcome Back $user"); }else{ header("Location: activate.php"); } }else{ header("Location: ../index.php?strmsg=Login Information Incorrect!"); } }else{ header("Location: ../index.php?strmsg=You did not enter a valid password!"); } }else{ header("Location: ../index.php?strmsg=You did not enter a valid email!"); } }else{ header("Location: ../index.php?strmsg=We are sorry but you are not allowed viewance of that directory!"); } ?> any help would be great. Brian All I am trying to do is add a record on a page without the page refreshing. For that ajax is used. Here is the code.
It does not add the record to mysql table. Can anyone tell me what I am doing wrong?
record.php
<!DOCTYPE HTML>
<html lang="en">
<head>
<script type="text/javascript" src="js/jquery-1.11.0.min.js"></script>
<script type="text/javascript" >
$(function() {
$(".submit_button").click(function() {
var textcontent = $("#content").val();
var name = $("#name").val();
var dataString = 'content='+ textcontent + '&name='+name;
if(textcontent=='')
{
alert("Enter some text..");
$("#content").focus();
}
else
{
$("#flash").show();
$("#flash").fadeIn(400).html('<span class="load">Loading..</span>');
$.ajax({
type: "POST",
url: "action.php",
data: dataString,
cache: true,
success: function(html){
$("#show").after(html);
document.getElementById('content').value='';
$("#flash").hide();
$("#content").focus();
}
});
}
return false;
});
});
</script>
</head>
<body>
<?php
$record_id = $_GET['id']; // getting ID of current page record
?>
<form action="" method="post" enctype="multipart/form-data">
<div class="field">
<label for="title">Name *</label>
<input type="text" name="name" id="name" value="" maxlength="20" placeholder="Your name">
</div>
<div class="field">
<label for="content">content *</label>
<textarea id="content" name="content" maxlength="500" placeholder="Details..."></textarea>
</div>
<input type="submit" name="submit" value="submit" class="submit_button">
</form>
<div id="flash"></div>
<div id="show"></div>
</body>
</html>
action.php
if(isset($_POST['submit'])) {
if(empty($_POST['name']) || empty($_POST['content'])) {
$error = 'Please fill in the required fields!';
} else {
try {
$name = trim($_POST['name']);
$content = trim($_POST['content']);
$stmt = $db->prepare("INSERT INTO records(record_id, name, content)
VALUES(:recordid, :name, :content");
$stmt->execute(array(
'recordid' => $record_id,
'name' => $name,
'content' => $content
));
if(!$stmt){
$error = 'Please fill in the required fields.';
} else {
$success = 'Your post has been submitted.';
}
} catch(Exception $e) {
die($e->getMessage());
}
}
}
I have the following code in html: <html> <head> <script type="text/javascript"> <!-- function delayer(){ window.location = "http://VARIABLEVALUE.mysite.com" } //--> </script> <title>Redirecting ...</title> </head> <body onLoad="setTimeout('delayer()', 1000)"> <script type="text/javascript"> var sc_project=71304545; var sc_invisible=1; var sc_security="9c433fretre"; </script> <script type="text/javascript" src="http://www.statcounter.com/counter/counter.js"></script><noscript> <div class="statcounter"><a title="vBulletin statistics" href="http://statcounter.com/vbulletin/" target="_blank"><img class="statcounter" src="http://c.statcounter.com/71304545/0/9c433fretre/1/" alt="vBulletin statistics" ></a></div></noscript> </body> </html> Is a basic html webpage with a timer redirect script and a stascounter code. I know a bit about html and javascript, but almost nothing about php. My question is: How a can convert this html code into a php file, in order to send a variable value using GET Method and display this variable value inside the javascript code where says VARIABLEVALUE. Thanks in adavance for your help. Hi, I have some code which displays my blog post in a foreach loop, and I want to add some social sharing code(FB like button, share on Twitter etc.), but the problem is the way I have my code now, creates 3 instances of the sharing buttons, but if you like one post, all three are liked and any thing you do affects all of the blog post. How can I fix this? <?php include ("includes/includes.php"); $blogPosts = GetBlogPosts(); foreach ($blogPosts as $post) { echo "<div class='post'>"; echo "<h2>" . $post->title . "</h2>"; echo "<p class='postnote'>" . $post->post . "</p"; echo "<span class='footer'>Posted By: " . $post->author . "</span>"; echo "<span class='footer'>Posted On: " . $post->datePosted . "</span>"; echo "<span class='footer'>Tags: " . $post->tags . "</span>"; echo ' <div class="addthis_toolbox addthis_default_style "> <a class="addthis_button_facebook_like" fb:like:layout="button_count"></a> <a class="addthis_button_tweet"></a> <a class="addthis_counter addthis_pill_style"></a> </div> <script type="text/javascript">var addthis_config = {"data_track_clickback":true};</script> <script type="text/javascript" src="http://s7.addthis.com/js/250/addthis_widget.js#username=webguync"></script>'; echo "</div>"; } ?> Advance thank you. Can you help please. The error..... Warning: mysql_fetch_assoc() expects parameter 1 to be resource, string given in C:\wamp\www\test_dabase.php on line 24 code. Code: [Select] <?php //database connection. $DB = mysql_connect("localhost","root") or die(mysql_error()); if($DB){ //database name. $DB_NAME="mysql"; //select database and name. $CON=mysql_select_db($DB_NAME,$DB)or die(mysql_error()."\nPlease change database name"); // if connection. }if($CON){ //show tables. $mysql_show="SHOW TABLES"; //select show and show. $mysql_select2="mysql_query(".$mysql_show.") or die(mysql_error())"; } //if allowed to show. if($mysql_select2){ //while it and while($data=mysql_fetch_assoc($mysql_select2)){ //show it. echo $data; } } ?> hey gurus, i am a newbie php coder.. i am learning by example. what i am trying to do is write a piece of code which will alter 3 tables (user, bonus_credit, bonus_credit_usage) ---------------------------------------------------------------- the table structure that will be used is as follows: user.bonus_credit user.ID bonus_credit.bonusCode bonus_credit.qty bonus_credit.value bonus_credit_usage.bonusCode bonus_credit_usage.usedBy ---------------------------------------------------------------- so lets say, in bonus_credit i have the following bonusCode = 'facebook' (this is the code they have to type to redeem the bonus qty = '10' ( number of times the bonusCode can be redeemed, but same person can't redeem it more than once) value = '5' (this is the amount of bonus_credit for each qty) Now, I need to write a code that check to see if the code has been redeemed in the bonus_credit_usage table and if the user.ID exists in this table as bonus_code_usage.usedBy, then give an error that its already been used and if it hasn't been used, then subtract 1 from qty, add ID to usedBy and then add the value to the bonus_credit ----------------------- i have started the steps just to create a simple textbox and entering a numeric value to bonus_credit, and that works.. but now i have to use JOIN and IF and ELSE.. which is a little too advanced for me.. so i'd appreciate a guide as i write the code. if(isset($_REQUEST['btnBonus'])) { $bonus_credit = addslashes($_REQUEST['bonusCode']); $query = "update user set bonus_credit=bonus_credit+'".$bonus_credit."' where id='".$_SESSION['SESS_USERID']."'"; echo "<script>window.location='myreferrals.php?msgs=2';</script>"; mysql_query($query) or die(mysql_error()); } Hi, I need to insert some code into my current form code which will check to see if a username exist and if so will display an echo message. If it does not exist will post the form (assuming everything else is filled in correctly). I have tried some code in a few places but it doesn't work correctly as I get the username message exist no matter what. I think I am inserting the code into the wrong area, so need assistance as to how to incorporate the username check code. $sql="select * from Profile where username = '$username'; $result = mysql_query( $sql, $conn ) or die( "ERR: SQL 1" ); if(mysql_num_rows($result)!=0) { process form } else { echo "That username already exist!"; } the current code of the form <?PHP //session_start(); require_once "formvalidator.php"; $show_form=true; if (!isset($_POST['Submit'])) { $human_number1 = rand(1, 12); $human_number2 = rand(1, 38); $human_answer = $human_number1 + $human_number2; $_SESSION['check_answer'] = $human_answer; } if(isset($_POST['Submit'])) { if (!isset($_SESSION['check_answer'])) { echo "<p>Error: Answer session not set</p>"; } if($_POST['math'] != $_SESSION['check_answer']) { echo "<p>You did not pass the human check.</p>"; exit(); } $validator = new FormValidator(); $validator->addValidation("FirstName","req","Please fill in FirstName"); $validator->addValidation("LastName","req","Please fill in LastName"); $validator->addValidation("UserName","req","Please fill in UserName"); $validator->addValidation("Password","req","Please fill in a Password"); $validator->addValidation("Password2","req","Please re-enter your password"); $validator->addValidation("Password2","eqelmnt=Password","Your passwords do not match!"); $validator->addValidation("email","email","The input for Email should be a valid email value"); $validator->addValidation("email","req","Please fill in Email"); $validator->addValidation("Zip","req","Please fill in your Zip Code"); $validator->addValidation("Security","req","Please fill in your Security Question"); $validator->addValidation("Security2","req","Please fill in your Security Answer"); if($validator->ValidateForm()) { $con = mysql_connect("localhost","uname","pw") or die('Could not connect: ' . mysql_error()); mysql_select_db("beatthis_beatthis") or die(mysql_error()); $FirstName=mysql_real_escape_string($_POST['FirstName']); //This value has to be the same as in the HTML form file $LastName=mysql_real_escape_string($_POST['LastName']); //This value has to be the same as in the HTML form file $UserName=mysql_real_escape_string($_POST['UserName']); //This value has to be the same as in the HTML form file $Password= md5($_POST['Password']); //This value has to be the same as in the HTML form file $Password2= md5($_POST['Password2']); //This value has to be the same as in the HTML form file $email=mysql_real_escape_string($_POST['email']); //This value has to be the same as in the HTML form file $Zip=mysql_real_escape_string($_POST['Zip']); //This value has to be the same as in the HTML form file $Birthday=mysql_real_escape_string($_POST['Birthday']); //This value has to be the same as in the HTML form file $Security=mysql_real_escape_string($_POST['Security']); //This value has to be the same as in the HTML form file $Security2=mysql_real_escape_string($_POST['Security2']); //This value has to be the same as in the HTML form file $sql="INSERT INTO Profile (`FirstName`,`LastName`,`Username`,`Password`,`Password2`,`email`,`Zip`,`Birthday`,`Security`,`Security2`) VALUES ('$FirstName','$LastName','$UserName','$Password','$Password2','$email','$Zip','$Birthday','$Security','$Security2')"; //echo $sql; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } else{ mail('email@gmail.com','A profile has been submitted!',$FirstName.' has submitted their profile',$body); echo "<h3>Your profile information has been submitted successfully.</h3>"; } mysql_close($con); $show_form=false; } else { echo "<h3 class='ErrorTitle'>Validation Errors:</h3>"; $error_hash = $validator->GetErrors(); foreach($error_hash as $inpname => $inp_err) { echo "<p class='errors'>$inpname : $inp_err</p>\n"; } } } if(true == $show_form) { ?> |
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