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PHP - Variable Displaying Over And Over
<?php include'db.php'; $session= $_SESSION['username']; $query="SELECT * FROM login_ip WHERE login='$session'"; $result= mysql_query($query); $count= mysql_num_rows($result); //$close = die($query); $min= $count - 1; $find= mysql_query("SELECT * FROM login_ip WHERE login= '$session' & times='$min'")or die(mysql_error()); while($row = mysql_fetch_assoc($find)) { $lip=$row['ipaddress']; echo" <br>$min<br>$count"; } ?> i worte the code above to pull the last ip address that a user has login form it is finding the ip adress but it is displaying like this Code: [Select] 72.145.25.457 72.145.25.457 72.145.25.457 72.145.25.457 72.145.25.457 so i remove the $lip form the echo and place $min and $count in there and i am getting Code: [Select] 1 2 1 2 1 2 1 2 Can some one please enplane why i am getting this problem thank you Similar Tutorialsapparently I am doing this wrong. I want my string results from a form that searches DB content to appear within HTML table, tr, td tags. I get the results fine, but the HTML part isn't appearing. How should I be doing this? <?php $string = ''; $result = mysql_query($sql); /// This is the execution if (mysql_num_rows($result) > 0){ while($row = mysql_fetch_object($result)){ echo "<table>"; echo "<tr>"; $string .= "<td>".$row->last_name."</td> "; $string .= "<td>".$row->first_name."</td>"; $string .= "<td>".$row->employee_id."</td>"; $string .= "<td>".$row->title."</b>"; $string .= "<td>".$row->territory."</td>"; $string .= "<td>".$row->district."</td>"; $string .= "<td>".$row->Phase1A_Score."</td>"; $string .= "<td>".$row->Phase1B_Score."</td>"; $string .= "<td>".$row->Phase1_Average."</td>"; $string .= "<td>".$row->Phase1A_HS_Exam."</td>"; $string .= "<td>".$row->Phase1A_HS_Exam_RT."</td>"; $string .= "<td>".$row->Phase1B_HS_Exam."</td>"; $string .= "<td>".$row->Phase1B_HS_Exam_RT."</td>"; $string .= "<td>".$row->Class_Date."</td>"; $string .= "<td>".$row->Awards."</td>"; $string .= "<br/>\n"; echo "</tr>"; echo "</table>"; } }else{ $string = "No matches found!"; } echo $string; ?>
Hi! The file login.php and the config file containing the db connection are in the model folder and the login_view is in view folder. I am trying to make a login form am display some error messages to the user, but I don't get the error messages. Can you help me figure it out? PS: the connection to the db works fine! Thx!
LOGIN.PHP
<?php
//start the session
session_start();
foreach (glob("model/*.php") as $filename) {
include $filename;
}
//form defaults
$error["alert"]= "";
$error["user"]= "";
$error["pass"]= "";
$input["user"]= "";
$input["pass"]= "";
if(isset($_POST["submit"]))
{
//process form
if($_POST["username"] == "" || $_POST["password"] == ""){
//display errors
if($_POST["username"] == ""){$error["user"] = "required";}
if($_POST["password"] == ""){$error["pass"] = "required";}
$error["alert"] = "Please fill in the required fields";
//get the values for the username and password
$input["user"]= $_POST["username"];
$input["pass"]= $_POST["password"];
include("../views/login_view.php");
}
}else{
include("../views/login_view.php");
}
LOGIN_VIEW.PHP
<?php
session_start();
foreach (glob("model/*.php") as $filename) {
include $filename;
}
?>
<link href="//maxcdn.bootstrapcdn.com/bootstrap/4.0.0/css/bootstrap.min.css" rel="stylesheet" id="bootstrap-css">
<script src="//maxcdn.bootstrapcdn.com/bootstrap/4.0.0/js/bootstrap.min.js"></script>
<script src="//cdnjs.cloudflare.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<!------ Include the above in your HEAD tag ---------->
<!--author:starttemplate-->
<!--reference site : starttemplate.com-->
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no">
<meta name="keywords"
content="unique login form,leamug login form,boostrap login form,responsive login form,free css html login form,download login form">
<meta name="author" content="leamug">
<title>Unique Login Form | Bootstrap Templates</title>
<link href="/styles/style.css" rel="stylesheet" id="style" type="text/css">
<!-- Bootstrap core Library -->
<link href="//netdna.bootstrapcdn.com/bootstrap/3.2.0/css/bootstrap.min.css" rel="stylesheet" id="bootstrap-css">
<script src="//netdna.bootstrapcdn.com/bootstrap/3.2.0/js/bootstrap.min.js"></script>
<script src="//code.jquery.com/jquery-1.11.1.min.js"></script>
<!-- Google font -->
<link href="https://fonts.googleapis.com/css?family=Dancing+Script" rel="stylesheet">
<!-- Font Awesome-->
<link href="//maxcdn.bootstrapcdn.com/font-awesome/4.2.0/css/font-awesome.min.css" rel="stylesheet">
</head>
<body>
<!-- Page Content -->
<div class="container">
<div class="row">
<div class="col-md-offset-5 col-md-4 text-center">
<h1 class='text-white'>Login Form</h1>
<form action ="" method="POST">
<div class="form-login"></br>
</br>
<?php
if($error["alert"] != ''){
echo "<div class='alert'>".$error["alert"]."</div>";
}
?>
<div>
<input type="text" name ="username" id="userName" class="form-control input-sm chat-input" placeholder="Username" value= "<?php echo $input['user'] ?>" required/>
<?php echo $error['user'] ?>
</div>
</br></br>
<input type="text" name ="password" id="userPassword" class="form-control input-sm chat-input" placeholder="Password" value= "<?php echo $input['password'] ?>" required/>
<div class="error"> <?php echo $error['pass'] ?>
</br></br>
<div class="wrapper">
<input type="submit" class="btn btn-primary" value="submit">
</div>
</div>
</div>
</form>
</div>
</div>
</body>
</html>
Edited October 28, 2020 by Alexa On my web page, there is a variable called $submission. I would like to display exactly 11 rows from the query below: the row where $submission equals $row["title"], 5 rows above it, and 5 rows below it. All ranked by points descending. How can I do this? Code: [Select] $sqlStr = "SELECT title, points, submissionid FROM submission ORDER BY points DESC"; $result = mysql_query($sqlStr); $arr = array(); $count=1; echo "<table class=\"samplesrec\">"; while ($row = mysql_fetch_array($result)) { echo '<tr >'; echo '<td>'.$count++.'.</td>'; echo '<td class="sitename1">'.$row["title"].'</td>'; echo '<td class="sitename2"><div class="pointlink2">'.number_format($row["points"]).'</div></td>'; echo '</tr>'; } echo "</table>"; I am trying to allow the user to update a variable he chooses by radio buttons, which they will then input text into a box, and submit, to change some attributes. I really need some help here. It works just fine until I add the second layer of variables on top of it, and I can't find the answer to this question anywhere. <?PHP require('connect.php'); ?> <form action ='' method='post'> <select name="id"> <?php $extract = mysql_query("SELECT * FROM cars"); while($row=mysql_fetch_assoc($extract)){ $id = $row['id']; $make= $row['make']; $model= $row['model']; $year= $row['year']; $color= $row['color']; echo "<option value=$id>$color $year $make $model</option> ";}?> </select> Which attribute would you like to change?<br /> <input type="radio" name="getchanged" value="make"/>Make<br /> <input type="radio" name="getchanged" value="model"/>Model<br /> <input type="radio" name="getchanged" value="year" />Year<br /> <input type="radio" name="getchanged" value="color" />Color<br /><br /> <br /><input type='text' value='' name='tochange'> <input type='submit' value='Change' name='submit'> </form> //This is where I need help... <?PHP if(isset($_POST['submit'])&&($_POST['tochange'])){ mysql_query(" UPDATE cars SET '$_POST[getchanged]'='$_POST[tochange]' where id = '$_POST[id]' ");}?> Hello all. I am very new to PHP, and I am not sure where to look or what I'm looking for in my current assignment. My task is to take in two numbers between 0-100. Once I take in that number, it should state beside it "The __ was accepted." The program should not accept any numbers greater than 100 or any characters. Once I do this, I must take a second number and do a similar thing. Finally, I must have a statement show up at the bottom stating which number is greater. Essentially, I need help in determining what I should use to place parameters, and how I can keep the program from echo ing any statement until input has been taken and tested for parameters. Any help you can provide will be greatly appreciated! My login script stores the user's login name as $_SESSION[ 'name'] on login. For some unapparent reason, i'm getting errors stating that $user and $priv are undefined variables, though I've attempted to define $user as being equal to $_SESSION['name'], using $user to look up the the user's privilege level (stored as the su column ) in the SQL table, and then where the result of the sql query is $priv which is then evaluated in an if statement. I can't seem to figure out why this might not be working. The code I'm using:
<?php
session_start();
function verify() {
//verify that the user is logged in via the login page. Session_start has already been called.
if (!isset($_SESSION['loggedin'])) {
header('Location: /index.html');
exit;
}
//if user is logged in, we then lookup necessary privleges. $_SESSION['name'] was written with the login name upon login. Privleges // are written in db as a single-digit integer of of 0 for users, 1 for administrators, and 2 for special users.
$user === $_SESSION['name'];
//Connect to Databse
$link = mysqli_connect("127.0.0.1", "database user", "password", "database");
if (!$link) {
echo "Error: Unable to connect to MySQL." . PHP_EOL;
echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
exit;
}
//SQL Statement to lookup privlege information.
if ($result = mysqli_query($link, "SELECT su FROM accounts WHERE username = $user", MYSQLI_STORE_RESULT)) {
//LOOP TO CYCLE THROUGH SQL RESULTS AND STORE Privlege information as vairable $priv.
while ($row = $result->fetch_assoc()) {
$priv === $row["su"];
}
}
// close SQL connection.
mysqli_close($link);
// Verify privleges and take action. Only a privlege of "1" is allowed to view this page. A privlege of "2" indicates special
//accounts used in other scripts that have certain indermediate additional functions, but are not trusted administrators.
if ($priv !== 1) {
echo $_SESSION['name'];
echo "you have privlege level of $priv";
echo "<br>";
echo 'Your account does not have the privleges necessary to view this page';
exit;
}
}
verify();
?>
I have just re-installed Xampp and suddenly my sites are now displaying lots of: Notice: Use of undefined constant name - assumed 'name' in ... Notice: Use of undefined constant price - assumed 'price' in ... this is an example of the line its refering too: $defineProducts[1001] = array(name=>'This is a product', price=>123); I have a form that creates rows of data input textboxes depending on a user input number of things. I have a naming convention for all these textboxes that basically just keeps incrementing a number suffix for each row. All this is working fine. My problem is I need to get the data inserted into this table of textboxes into an array. Here's my code where I attempt to to this (it does not work): Code: [Select] $temp = $_SESSION['Num_Part']; $count = 1; while ($count <= $temp){ $temp2[$count] = "'Participant_P".$count."'"; //echo $temp2[$count]."<br/>"; $temp3[$count]=$_POST[$temp2[$count]]; //here's the problem $temp4[$count] = "'Result_P".$count."'"; $temp5[$count]=$_POST[$temp4[$count]]; //here's the problem //echo $temp4[$count]."<br/>"; $count++; } The problem is that the $_POST does not work with the variable in the argument position - even though the argument is formatted with single quotes. Can a variable be used in a POST argument and if so what is the correct syntax? If not, is there some other simple solution to harvest the data into an array. I understand I can harvest by explicitly accessing each key in the post assoc array. But this could be dozens of rows of input fields. Thanks in advance for your help here. I couldn't find anything online re this topic. I have a script that adds points together based upon the placing. This is the actual script: Code: [Select] <? $points = 0; if($place === '1st') {$points = $points + 50;} elseif($place === '2nd') {$points = $points + 45;} elseif($place === '3rd') {$points = $points + 40;} elseif($place === '4th') {$points = $points + 35;} elseif($place === '5th') {$points = $points + 30;} elseif($place === '6th') {$points = $points + 25;} elseif($place === '7th') {$points = $points + 20;} elseif($place === '8th') {$points = $points + 10;} elseif($place === '9th') {$points = $points + 10;} elseif($place === '10th') {$points = $points + 10;} elseif($place === 'CH') {$points = $points + 50;} elseif($place === 'RCH') {$points = $points + 40;} elseif($place === 'TT') {$points = $points + 30;} elseif($place === 'T5') {$points = $points + 30;} elseif($place === 'Champion') {$points = $points + 50;} elseif($place === 'Reserve Champion') {$points = $points + 40;} echo "Total HF Points: $points"; ?>What it *should* do (my friend's script works the same way and it works) it starts at points = 0, than if there is a first place, it adds 50, and so forth until it reaches the end. It is included into a file, in this area: Code: [Select] <div class="tabbertab"> <h2>Records</h2> <? $query92 = "SELECT * FROM THISTABLE WHERE VARIABLE='$id' OR VARIABLE = '$name' ORDER BY ABS(VARIABLE), VARIABLE"; $result92 = mysql_query($query92) or die (mysql_error()); echo "<table class='record'> <tr><th>Show</th> <th>Class</th> <th>Place</th></tr> "; while($row92 = mysql_fetch_array($result92)) { $class = $row92['class']; $place = $row92['place']; $entries = $row92['entries']; $race = $row92['show']; $purse = number_format($row92['purse'],2); echo "<tr><td>$race</td> <td>$class</td> <td>$place</td></tr>"; } ?> <tr><td colspan='3'><div align='right'><? include('includes/points.php'); ?></div></td></tr> </table> </div> This is the code that is relevant. When ended here, it echoes the last place that appears in the results (such as a 5th place echoing 30 points). When I move it to be included in the while loop, it shows Total Points: 50 Total Points: 25 Total Points: 10 (depending on the results displayed on that page). What am I doing wrong? Hello everyone, I can get Test 2 to successfully operate the if statement using a variable variable. But when I try the same method using a session variable (Test 1) the if statement is not executed. Please could you tell me why the if statement in Test 1 is not being executed? Code: [Select] <?php # TEST 1 $_SESSION[test_variable] = "abcd"; $session_variable_name = "_SESSION[test_variable]"; if ($$session_variable_name == "abcd") { echo "<br>line 373, abcd<br>"; } # TEST 2 $test_variable = "efgh"; $test_variable_name = "test_variable"; if ($$test_variable_name == "efgh") { echo "<br>line 379, efgh<br>"; } ?> Many thanks, Stu I have a function that get's a quick single item from a query:
function gimme($sql) {
global $mysqli;
global $mytable;
global $sid;
$query = "SELECT ".$sql." FROM ".$mytable." WHERE sid = ".$sid; $result = $mysqli->query($query);
$value = $result->fetch_array(MYSQLI_NUM);
$$sql = is_array($value) ? $value[0] : "";
return $$sql; // this is what I've tried so far
$result->close();
}
It works great as:
echo(gimme("name"));
Then I realized that I could use that as a variable ('$name' in this case) elsewhere. However, I can't figure out how get that new, variable variable 'outside' of the function. As such, echo($name); isn't working outside the function. Is there a way to return a variable variable? In other words, is there a way to make a function that creates a variable variable that will available outside of the function?
Thanks
hi all, I have an language pack for example: languages/en.php: Code: [Select] $en['mail']['letter closing'] = "regards,\n your friend!"; and in my config: Code: [Select] $language = "en"; $include_language = @include("languages/".$language.".php"); if(!($include_language)) { $try_default_language = @include("languages/nl.php"); if(!($try_default_language)) { echo "kan de taalpakket niet vinden<br>"; echo "Could not find the language pack.<br>"; echo "example on error: ".$test." shows nothing"; exit; } } In my function I want to include the language pack for example i have $language = 'en' so I want to include $en['general']['letter closing'] I will do this: Code: [Select] global $language,${$language}['general']; But that gives an error unexpected '[' blah blah. How can i call the variable variable array in the valid php way? Probably something simple but I have searched high and low and can't figure this one out. I have a variable that is of the datetime format. I have another variable that is of the time format. I need to add them together. Example: $var1 = 2012-02-24 06:38:22 $var2 = 02:00:00 $var3 = $var1 + $var2 = 2012-02-24 08:38:22 Thanks for the help! I just moved my code from Appserv to EasyPHP and it gave me this error, it was working fine on Appserv...what's with easyPHP ?? Quote i need to store a variable from database like if i have "copies" in one of my column in my database then i have to store a particular value for copies store it to $copies here i want that i can store value of copies into $copies $update_book="update book set copies=copies-1 where bookid='$bookid'"; $result=mysql_query($update_book,$linkID1); if($result) { print "<html><body background=\"header.jpg\"> <p>book successfully subtracted from database</p></body></html>"; } else { print "<html><body background=\"header.jpg\"> <p>problem occured</p></body></html>"; } } I'm trying to loop through a database and assign the results to a variable that is created with the loop with the variables name based on which instance is currently running in the loop. I'll go ahead and include the code I thought might work, but I obviously have no idea what I'm doing in this case $i = 0; while ($i < $num_results) { $ip = ($i++); $result = ("$res_" . "$ip"); $loop_date = ("$res_" . "$ip" . "_date"); $get_result = "SELECT * FROM table WHERE id = '$result'"; $que_result = mysql_query($get_display); while ($sub_res = mysql_fetch_array($que_result, MYSQL_BOTH)) { $loop_date = $sub_res['date']; } $i++; } echo $res_1_date; Can anyone please clue me in on how I can get this to work please. It works with $result but it isn't working with the last $res_1_date... As I'm writing this I see where the problem is but just not quite sure how to make it work... Hi, so I have an easy problem that for some reason I couldn't find an answer to anywhere. I have a bunch of variables like this: $pic1fileName $pic2fileName $pic3fileName $pic4fileName $pic5fileName...you get the idea So I have another variable I'm pulling from a database that specifies which number to show, so I need a variable something like this: $pic($pic_number)fileName I just don't know what the proper syntax is. Anyone? Thank you freaking much for any help; this is a lame problem. Hi i have this scroolbar javascript where you implement the text inside the java variable but because i want the data to come from the database i wanted to use a php example <? print($marker); ?> so i was wondring if it is possible to add a php variable into a javascript like this my javascript at moment the javascript variable is like this Code: [Select] var myMainMessage=" Try this out! The message just keeps repeating. Use it for announcements, news and other items! ";i want to change to this Code: [Select] var myMainMessage="<? print($marker); ?>";is it possible if so can someone show me an example thanks the whole javascript Code: [Select] <SCRIPT LANGUAGE="JavaScript"> <!-- This script and many more are available free online at --> <!-- The JavaScript Source!! http://javascript.internet.com --> <!-- (c) http://www.wyka-warzecha.com --> <!-- Begin // THESE VARIABLES where i want to add the php // var myMainMessage=" Try this out! The message just keeps repeating. Use it for announcements, news and other items! "; var speed=150; var scrollingRegion=50; // END CHANGEABLE VARIABLES // var startPosition=0; function mainTextScroller() { var mainMessage=myMainMessage; var tempLoc=(scrollingRegion*3/mainMessage.length)+1; if (tempLoc<1) {tempLoc=1} var counter; for(counter=0;counter<=tempLoc;counter++) mainMessage+=mainMessage; document.mainForm.mainTextScroller.value=mainMessage.substring(startPosition,startPosition+scrollingRegion); startPosition++; if(startPosition>scrollingRegion) startPosition=0; setTimeout("mainTextScroller()",speed); } // End --> </script> Is if($variable) the same as if(!empty($variable)), just different syntax? Code: [Select] $word = 'numbers'; $numbers= array('1', '2', '3', '4'); echo $$word[0]; I expected the output to be '1'. It ended up being nothing... Why does this not work? Is it not possible to have a variable variable array? And if not, is there a workaround? Cheers, Joe |
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