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PHP - Select Name = $ Symbol?
Code: [Select]
$query = "LOWER(name)="$tabB"; on my forums, speakwhatsreal.com new message page, let's say somone on my site has the name "$tabB" how do i make the above query work so it selects a username with the $ symbol in it? because with that query it's linked to Code: [Select] $DB->query("SELECT name, id, mgroup, language, email FROM ibf_members WHERE ".$query); $to_member = $DB->fetch_row(); if (empty($to_member['id'])) { $this->send_form(0, $ibforums->lang['err_no_such_member']); return; } pretty much says no results.. lol any help? Similar TutorialsWhat is the & symbol used for I have seen it in function setname(&$name){ $this->name =& $name; } Thanks I have looked on google and can't find it as i don't know what it is called. thanks Hello, Im getting a name from the url, for instance, index.php?name=C.-and-G The "and" is then converted to a & symbol and dashes converted to spaces. This then uses the new name "C. & G" to select a record from mysql. However, it does not recognise the & symbol, the name in mysql is "C. & G" but returns 0 records. Any ideas? Thanks! I'm a little confused exactly on how using this } works going to be hard to explain what im having problems with. I'm using JEdit for my editor and I notice if I click on one of these { it will put a box around where it ends there is 4 } at the bottom of the code by them selfs "on their own lines" and 2 in the middle by them selfs "on their own lines how does the code know which ones belong to which { appreicate any help in helping me understand this thanks this code is from a tutor online Code: [Select] <? include_once("connect.php"); ?> <html> <body> <?php if(isset($_POST['Login'])) { if(!preg_match('/^[A-Za-z0-9]{5,20}$/',$_POST['Username'])){ // before we fetch anything from the database we want to see if the user name is in the correct format. echo "Invalid Username."; }else{ $query = "SELECT password,id,login_ip FROM users WHERE name='".mysql_real_escape_string($_POST['Username'])."'"; $result = mysql_query($query) or die(mysql_error()); $row = mysql_fetch_array($result); // Search the database and get the password, id, and login ip that belongs to the name in the username field. if(empty($row['id'])){ // check if the id exist and it isn't blank. echo "Account doesn't exist."; }else{ if(md5($_POST['password']) != $row['password']){ // if the account does exist this is matching the password with the password typed in the password field. notice to read the md5 hash we need to use the md5 function. echo "Your password is incorrect."; }else{ if(empty($row['login_ip'])){ // checks to see if the login ip has an ip already $row['login_ip'] = $_SERVER['REMOTE_ADDR']; }else{ $ip_information = explode("-", $row['login_ip']); // if the ip is different from the ip that is on the database it will store it if (in_array($_SERVER['REMOTE_ADDR'], $ip_information)) { $row['login_ip'] = $row['login_ip']; }else{ $row['login_ip'] = $row['login_ip']."-".$_SERVER['REMOTE_ADDR']; } } $_SESSION['user_id'] = $row['id'];// this line of code is very important. This saves the user id in the php session so we can use it in the game to display information to the user. $result = mysql_query("UPDATE users SET userip='".mysql_real_escape_string($_SERVER['REMOTE_ADDR'])."',login_ip='".mysql_real_escape_string($row['login_ip'])."' WHERE id='".mysql_real_escape_string($_SESSION['user_id'])."'") or die(mysql_error()); // to test that the session saves well we are using the sessions id update the database with the ip information we have received. header("Location: Sample.php"); // this header redirects me to the Sample.php i made earlier } } } } ?> <form id="form1" name="form1" method="post" action=""><center> GAME LOGIN <br /> <br /> Username: <input type="text" name="Username" id="Username" /> <br /> <br /> Password: <input type="password" name="password" id="password" /> <br /> <br /> <input type="submit" name="Login" id="Login" value="Login" /> </center> </form> </body> </html> I am running php 5.3.2 on my local dev and on the server I am running php 5.2.5. In my log files I keep getting deprecated messages like: PHP Deprecated: Assigning the return value of new by reference is deprecated in /Users/xxx/dev/git/xxx/cake/dispatcher.php on line 677 Can I just delete & to get rid of this message or will that hose the script locally and on the server? I believe that php 5+ passes all objects as reference. Thanks This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=322426.0 Hi, I'm looking through some code I have for setting up a rating system. It was a free open source code, which I need to change to work for my own site. I have it working, but I am now just looking through the code to try and understand it, and for the most part, I do. There is one part I am not familiar with, if somebody could help me please? Values involved; $current_rating = total rate value of votes $count = total number of votes cast Problem I have is the @ symbol here $rating=@number_format($current_rating/$count, 1); Does it format it to decimal places or something like that? It's probably something simple I know, but I am a relative newb. Thanks. Easy One - What is the symbol for or? Example: && = and, right? What is 'or'? Trying to echo a string that contains a hash symbol. Instead of getting a hash symbol, I get %29. The code I am using is pretty basic -->$idx="#" . $idxA;<--, but as stated, when I echo $idx, it comes out as %29. (All of this is part of a form, get, attempting to pass a bookmark to the receiving program. Thoughts? Code: [Select] $mystring = "|Signature"; $findme = '|'; $pos = strpos($mystring, $findme); echo $pos; echo's 0 supposed to echo out 1....? why not work thanks How can I enable it to check for Code: [Select] | using strpos? What does it mean Like.. !@file_get_contents('url'); or.. !file_get_contents or @file_get_contents Hi i want to match $ symbol from the string. I am using the following script Code: [Select] $str="ckj"; if(preg_match("/^$/",$str)) { echo "Dollar is found"; } else { echo "Not Found"; } Its not working fine. Can anyone help me out please. So I have a simple query that adds a text record into a MySQL database table. It works great with exception of one thing. I noticed that if I have "&" symbol in my paragraph, the text after that symbol won't be inserted into the table. The text before that symbol will insert fine. It seems to be doing that only with & symbol; other symbols insert and show up fine. can anyone tell me why this is happening? hello i can't figure out how to modify a value from assoc array so with each recursion to add additional string to his: for example i have attribute=""; then call recursive function and i try to modify its value by concatenating "0"; or "1"; on the next loop to add again "0" or "1"; hirealimo.com.au/code1.php this works as i want it: Quote SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1 but apparelty selecting * is not a good thing???? but if I do this: Quote SELECT priceID, price FROM price INNER JOIN vehicle....etc it works but i lose the info from the vehicle table. but how do i make this work: Quote SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc so that i am specifiying which colums from which tables to query?? thanks So i'm having some sort of an issue. what i'm trying to do is to get the name from an URL using GET that includes the "&" symbol. I have a link and that link looks like this: name_of_company_&_member. What im trying to do is to get that whole line using Get and echoing on a page. The problem is that the line cuts off after "company_" because it reads the & symbol as something different. Does anyone know how i can be able to retrieve the whole line including everything after the & symbol? i tried using the special character "%26" but i still get the same result. I have 2 queries that I want to join together to make one row
Dear All, I wish to have 2 drop down boxes, Country Select Box and Locality Select Box. The locality select box will be affected by the value chosen in the country select box. All is working fine except that the locality select box is not being populated. I know that the problem is in the sql statement WHERE country_id='$co' because i am having an error that $co is an undefined variable. All the rest works fine because i have replaced the $co variable directly with a number (say 98) for a particular country id and it worked fine. In what way can i define this variable $co so that it is accepted by my sql statement? Thank you for your help in advance. MySQL Tables indicated below: CREATE TABLE countries( country_id INT(3) UNSIGNED NOT NULL AUTO_INCREMENT, country_name VARCHAR(30) NOT NULL, PRIMARY KEY(country_id), UNIQUE KEY(country_name), INDEX(country_id), INDEX(country_name)) ENGINE=MyISAM; CREATE TABLE localities( locality_id INT(10) UNSIGNED NOT NULL AUTO_INCREMENT, country_id INT(3) UNSIGNED NOT NULL, locality_name VARCHAR(50), PRIMARY KEY (locality_id), INDEX (country_id), INDEX (locality_name)) ENGINE=MyISAM; Extract PHP script included below: // connect to database require_once(MYSQL); if(isset($_POST['submitted'])) { // trim the incoming data /* this line runs every element in $_POST through the trim() function, and assigns the returned result to the new $trimmed array */ $trimmed=array_map('trim',$_POST); // clean the data $co=mysqli_real_escape_string($dbc,$trimmed['country']); $lc=mysqli_real_escape_string($dbc,$trimmed['locality']); } ?> <form action="form.php" method="post"> <p>Country <select name="country"> <option>Select Country</option> <?php $q="SELECT country_id, country_name FROM countries"; $r=mysqli_query($dbc,$q) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc)); while($row=mysqli_fetch_array($r)) { $country_id=$row[0]; $country_name=$row[1]; echo '<option value="' . $country_id . '"'; if(isset($trimmed['country']) && ($trimmed['country']==$country_id)) echo 'selected="selected"'; echo '>' . $country_name . '</option>\n'; } ?> </select> </p> <p>Locality <select name="locality"> <option>Select Locality</option> <?php $ql="SELECT locality_id, country_id, locality_name FROM localities WHERE country_id='$co' ORDER BY locality_name"; $rl=mysqli_query($dbc,$ql) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc)); while($row=mysqli_fetch_array($rl)) { $locality_id=$row[0]; $country_id=$row[1]; $locality_name=$row[2]; echo '<option value="' . $locality_id . '"'; if(isset($trimmed['locality']) && ($trimmed['locality']==$locality_id)) echo 'selected="selected"'; echo '>' . $locality_name . '</option>\n'; } // close database connection mysqli_close($dbc); ?> </select> </p> <p><input type="submit" name="submit" value="Submit" /></p> <input type="hidden" name="submitted" value="TRUE" /> </form> Hi all, Im trying to get the requested genre and compare it with the <SELECT> list to add the word SELECTED on the option. So if the requested genre is the same as the option name them make that the SELECTED option. Cheers. Here's a bit of code I made that does not work Code: [Select] <? function selected(){ if ($_REQUEST['genre'] = $name){ echo"SELECTED"; }} ?> Gen <SELECT class="sort" align="right" onChange="window.location.href=this.options[this.selectedIndex].value;"> <option value="<?=$_SERVER['PHP_SELF'];?>" <?$name = ''; selected(); ?>>Any</option> <option value="<?=$_SERVER['PHP_SELF'];?>?genre=Action" <?$name = 'Action'; selected(); ?>>Action</option> <option value="<?=$_SERVER['PHP_SELF'];?>?genre=Adventure" <?$name = 'Adventure'; selected(); ?>>Adventure</option> <option value="<?=$_SERVER['PHP_SELF'];?>?genre=Animation" <?$name = 'Animation'; selected(); ?>>Animation</option> </SELECT> |
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