|
PHP - Variable On Radio Buton
I am wondering if this is posible. I have a group of radio butons. I would like to make them as variables so that I can use an "if then" statment in another part of the page to show an image depending on which button is selected.
<form id="form1" name="form1" method="post" action=""> <p> <label> <input type="radio" name="color" value="red" id="color_0" /> red</label> <br /> <label> <input type="radio" name="color" value="blue" id="color_1" /> blue</label> Similar TutorialsI 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]' ");}?> Hey all, I'm almost too embarrased to post this as I'm sure it is very simple - but for the life of me I can't get this to work. I am a complete newbie to PHP so please bear with me.. What I'm trying to do is to write a program that will go on to calculate a user's Body Mass Index (BMI) based on user inputted data of height and weight. However, I want to be able to accept heights and weights in a number of different units for maximum ease of use. I am trying to write some PHP code that will handle this , the main goal is to convert everything into 'cm' and 'kg' before going on and doing the simple BMI calculation later on. However I am stuck at this point: I have a text field for users to input their weight into. This is immediately followed by two radio buttons for the user to select which units they are inputting their weight in (kilograms or pounds). A third option to input their weight is given after this for those wishing to input their weight in 'stones and pounds'. My problem is as follows: I can't get my code to recognise which radio button (either 'kg' or 'lbs') has been pressed. What should happen is that my code can tell something has been inputted in the text box AND which of the radio buttons has been selected. From this it does one of the two; for the kilogram option it leaves the value as it is, but if the weight has been inputted in pounds (and the 'lbs' radio button selected) then I want the code to convert this into kilos by multiplying by 0.4535. Here is what I have so far (sorry its a bit messy - like i say I am a newbie): <?php //convert.php $ft = $cm = $inches = $weight = $stones =$pounds = $kilos = $units = ""; if(isset($_POST['cm'])) $cm = sanitizeString($_POST['cm']); if(isset($_POST['ft'])) $ft = sanitizeString($_POST['ft']); if(isset($_POST['inches'])) $inches = sanitizeString($_POST['inches']); if(isset($_POST['weight'])) $weight = sanitizeString($_POST['weight']); if(isset($_POST['stones'])) $stones = sanitizeString($_POST['stones']); if(isset($_POST['pounds'])) $pounds = sanitizeString($_POST['pounds']); if(isset($_POST['kilos'])) $kilos = sanitizeString($_POST['kilos']); if(isset($_POST['units'])) $units = sanitizeString($_POST['units']); if ($ft != '') { $height = intval(($ft * 30.48) + ($inches * 2.54)); $out = "you are $height cm tall"; } elseif($cm != '') { $height = intval($cm); $out = "you are $height cm tall"; } else $out = ""; if ($stones != '') { $kilos = intval((($stones * 14) + $pounds) * 0.45359237); $out2 = "you weigh $kilos Kg"; } elseif($weight != '' AND $units = "kg") { $kilos = $weight; $out2 = "you weigh $kilos Kg"; } elseif($weight != '' AND $units = "lbs") { $kilos = ($weight * 0.45359237); $out2 = "you weigh $kilos Kg"; } else $out2 = ""; echo <<<_END <html><head><title>Height & Weight Converter</title> </head><body><pre> Please enter your details below <b>$out$out2</b> <form method="post" action="convert.php"> Height: <input type="text" name="cm" size="3"> cm OR <input type="text" name="ft" size="1">ft <input type="text" name="inches" size="2">inches Weight: <input type="text" name="weight" size="4" /> Kg<input type="radio" name="units" value="kg" /> lbs<input type="radio" name="units" value="lbs" /> OR <input type="text" name="stones" size="2">stone <input type="text" name="pounds" size="2">pounds <input type="submit" value="Submit" /> </form></pre></body></html> _END; function sanitizeString($var) { $var = stripslashes($var); $var = htmlentities($var); $var = strip_tags($var); return $var; } ?> what would be the $_POST variable name for a radio button option? 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! 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); 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 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. 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 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? 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? 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>"; } } 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 + $var2 = 2012-02-24 08:38:22 Thanks for the help! I just moved my code from Appserv to EasyPHP and it gave me this error, it was working fine on Appserv...what's with easyPHP ?? I have a function that get's a quick single item from a query:
function gimme($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($name); isn't working outside the function. Is there a way to return a variable variable? In other words, is there a way to make a function that creates a variable variable that will available outside of the function?
Thanks
Is it possible we can have radio mp3 in php where we control the mp3 as the admin while user who listen the mp3 can only hear but cannot modify song do you has any example of this radio mp3? Code: [Select] if ($cartype = 1 ) { echo "Compact <br/>"; } elseif ($cartype == 2 ) { echo "Saloon <br/>"; } else if ($cartype == 3 ) { echo "SUV <br/>"; } This code will only display the first radio button atm. How do I fix this? Hello guys. I am learning PHP little by little and am writing a VERY basic quiz. For functionality in the future I want to be able to change questions and answers in one "included" php doc. The problem i am having with my current code is that I would like to place a variable in the "value" definition on the radio input. I am unable to do this I guess. Any suggestions to accomplish the same thing? Sorry for my horrible code, this is the first thing I have written from scratch. The form page: Code: [Select] <?php include ("test.php"); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <title>Untitled Document</title> </head> <body> <h2><? print $q1; ?></h2> <form method="post" action="test.php"> <input type="radio" name="q1answered" value="Basketball"/> <?php print $q1ans[0]; ?> <input type="radio" name="q1answered" value="Football"/> <?php print $q1ans[1]; ?> <br /><br /> <input type="submit" value="Submit Your Answer!" name="submit" /> </form> </body> </html> The php Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <title>Untitled Document</title> </head> <body> <?php //Enter all questions below $q1 = "What is your favorite Sport?"; $q2 = "What is your favorite hobby?"; //End Questions //Enter Asnwer choices for each question $q1ans = array("Basketball", "Football"); //End Answers //All question defined variables here to make grading easier $qbank = array ($q1, $q2,); if (isset($_POST["submit"])) { $choice = $_POST["q1answered"]; if ($choice == $q1ans[0]){ print "Your Favorite Sport is $q1ans[0]."; } else { print "Your Favorite Sport is $q1ans[1]."; } } ?> </body> </html> Thanks! Hello, I had 3 radio buttons with the same name, well, I had created 3 text box that are next to those 3 radio buttons. When I click save to the form, I need to know which radio button had been pressed, so that I can detect & retrieve its value. Any help? Thanks What am I doing wrong with this selected radio button? <form action="Process/privacy.php" method="post" enctype="multipart/form-data" id="privacy"> <div id="privacy_vis">Your profile is visible to:</div> <div id="privacy_select"> <label> <input name="RadioGroup1" type="radio" id="RadioGroup1_0" value="1" <?if($visible=="1")echo 'checked="checked"';?> /> Everyone</label> <label> <input name="RadioGroup1" type="radio" id="RadioGroup1_1" value="0"/> Members Only</label> </div> <br /> <div id="privacy_vis">Your Password to view your private pictures is:</div> <div> <input name="pictures_pw" type="text" id="pictures_pw" value="<? echo "$xxx_pw" ?>" size="50" maxlength="20" /> </div> <br /> <div> <input name="submit" type="submit" class="button" id="submit" value="Save Changes" /> </div> </form> </div> I have a function to display some images and allow a user to select a choice via a radio button.The value is then passed to another function. I have 2 problems My selection is not being passed back and I cant assign the value to a query sring in a link to another page. On the form do i need to add checked? Any help appreciated. Code: [Select] function cover() { echo'<h2>Please upload your book cover or choose one from the selection</h2> <p>All book ideas require a cover. Your choice can be changed at a later date.</p>'; //process selection of cover if (isset($_POST['submit'])) { $coverID=$_POST['cover']; echo $coverID; ?>   <ul><li class="demo-menu-link"><a href="?page=submenu7&coverID=$coverID">Finnished Cover</a></li></ul><?php exit(); } else{ echo "not Submitted"; } global $wpdb; $query = "select * from wp_cover"; $result = mysql_query($query) or trigger_error("Query: $query\n<br />MySQL Error: " . mysql_error()); echo mysql_error(); if (!$result){ return false; } echo'<div class="wrap"><p>choose from one of the covers below</p></div>'; ?><form action="<?php $_SERVER['PHP_SELF'] ?>" method="POST" id="cover" enctype="multipart/form-data"> <table><tr><?php /* display picture and radio button */ while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { $coverID=$row['coverID']; echo"<img src=\"/wordpress_3/wp-content/plugins/Authors2/jackets/{$row['pix']}\" />"; echo"<td>"; echo $coverID; echo "<Input type = 'Radio' Name ='cover' value= '$coverID'/>"; ?> </tr></table> <input type=submit value=submit> </form><?php } } |
|||||||