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PHP - Entering Ips Into A Database With Inet_aton.
So after surfing the net I found that the 'best' way to enter ip addresses into a database was by using the INET_ATON function because you can then put it into a unsigned interger column and that will save space. But after looking around I can't find a way of fitting it into php.
This is my attempt: $ip = getenv("REMOTE_ADDR"); $query = sprintf("UPDATE INTO user ( name, password) VALUES ('%s','%s','%s')", mysql_real_escape_string($_SESSION['name']), mysql_real_escape_string(md5($_SESSION['values']['password'])) INET_ATON('$ip')); //line 37 $result2 = mysql_query($query, $db) or die(mysql_error($db)); I get a parse error: Parse error: syntax error, unexpected T_STRING in C:\x\xampp\htdocs\pages\login.php on line 37. This worked until I wanted to add the ip into the database. Any help appreciated. Similar Tutorials I have written code in php to connect and insert rows into a MSSQL database. i used odbc to connect database.user can enter his details through the form. after submitting the form the details are getting stored into a database. while inserting rows into a database am not trying to insert duplicate values . for this i have given if conditions.these conditions are able to notice the user cname and name exist in the database if the same name exist. but the else part after these conditions are not working i.e rows are not getting inserted. i put everything inside the while loop. how can i correct it? This is my php code. $connect = odbc_connect('ServerDB','sa', 'pwd'); //connects database $query2="select count(*) from company";//this is needer for loop through $result2=odbc_exec($connect,$query2); while(odbc_fetch_row($result2)); { $count=odbc_result($result2,1); echo "</br>","$count"; } $query1="select * from company"; $result1 = odbc_exec($connect, $query1); # fetch the data from the database while(odbc_fetch_row($result1)) { $compar[$count] = odbc_result($result1, 1); $namearray[$count] = odbc_result($result1, 2); if($compar[$count]==$_POST['cname']) { echo "<script> alert(\"cname Exists\") </script>"; } else if($namearray[$count]==$_POST['name']) { echo "<script> alert(\"Name Exists\") </script>"; } else { $query=("INSERT INTO company(cname,name) VALUES ('$_POST[cname]','$_POST[name]') "); $result = odbc_exec($connect, $query); echo "<script> alert(\"Row Inserted\") </script>"; } } What's the best way for putting actual quotations into a database? I was using... $quote = htmlspecialchars(mysqli_real_escape_string($dbc, $_POST['quote'])); Should I be? $name value is coming through from form submission.
<?php } If statement is not working properly. Whether $name is empty or not, it adds the where clause. ?>
<?php I have a text firld to enter amount af a book,But i need to check whether the data entering is numeric only eg:70.98 etc. how can i achieve this. my code is echo"<table><tr><td></td><td>Amount in USD</td><td></td><td><input type='text' name='amnt' id='amnt' /> </td><td></td></tr></table>"; How would I incorporate a function to simply check the "name" and "message" for a certain amount of chars, like 15 & 150? Code: [Select] <form method="post" action="chat.php"> <p><input name="name" type="text" id="name" value="your name" size="10" maxlength="15"> <input name="message" type="text" id="message" value="your message" size="20" maxlength="150"> <input name="submit" type="submit" id="submit"></p> </form> </body> </html> <?php // when the submit button is clicked if(isset($_POST['submit'])) { // strip any html tags before continuing $name=strip_tags($_POST['name']); $message=strip_tags($_POST['message']); // stop if nothing was entered if($name!='') if($message!='') { // trim any extra whitespace $data=trim($name)."\n"; $data.=trim($message)."\n"; //open the text file and enter the data $file_ar=file("db.txt"); $fp=fopen("db.txt","w"); fputs($fp,$data); if($file_ar!=NULL) { $loop=0; foreach($file_ar as $line) { // do not store more than 20 messages if($loop>=19*3) break; fputs($fp,$line); $loop++; } } fclose($fp); } } // display the messages $fp=fopen("db.txt","r"); while(!feof($fp)) { $name=trim(fgets($fp,999)); $message=trim(fgets($fp,999)); if($name!='') { echo "<p><b>$name: </b>$message</p>"; } } fclose($fp); ?> The dreadful apostrophie problem... This search form returns an error whenever searching with an apostrophie (') Here's the code on the form (html) <td align="center" width="135"><form method="post" action="srch_advert.php"><input type=text name='search' size=15 maxlength=255><br><input type=submit></form></td> <td align="center" width="135"><form method="post" action="srch_details.php"><input type=text name='search' size=15 maxlength=255><br><input type=submit></form></td> <td align="center" width="135"><form method="post" action="srch_artist.php"><input type=text name='search' size=15 maxlength=255><br><input type=submit></form></td> <td align="center" width="135"><form method="post" action="srch_track.php"><input type=text name='search' size=15 maxlength=255><br><input type=submit></form></td> and heres the code on srch_advert.php if ($search) // perform search only if a string was entered. { mysql_connect($host, $user, $pass) or die ("Problem connecting to Database"); $srch="%".$search."%"; $query = "select * from tvads WHERE advert LIKE '$srch' ORDER BY advert, year DESC, details ASC LIMIT 0,30"; $result = mysql_db_query("cookuk_pn", $query); if(mysql_num_rows($result)==0) { print "<h2>Your search returned 0 Results</h2>"; } else if ($result) { I know this problem comes up a lot but it can be for various reasons so after reading up I can't decide what might be wrong. I'm still learning! I have a simple php registration form (first name, second name, email address) and every time a user submits an entry a second blank record is created in the MYSQL database after it. Any help would be great. (php code in red) Code: [Select] [color=red]<? $firstname=$_POST['firstname']; $surname=$_POST['surname']; $email=$_POST['email']; mysql_connect(localhost,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $query = "INSERT INTO register VALUES ('','$firstname','$surname','$email')"; mysql_query($query); mysql_close(); ?>[/color] Here is the section on the html HTML page that submits the form Code: [Select] <form action="register.php" method="post" class="BagTitle"> <table width="700" border="0" cellpadding="0" cellspacing="2"> <tr> <td width="100">First Name: </td> <td width="594"><input name="firstname" type="text" size="50" /></td> </tr> <tr> <td>Last Name: </td> <td><input name="surname" type="text" size="50" /></td> </tr> <tr> <td>E-mail:</td> <td><input name="email" type="text" size="50" /></td> </tr> <tr> <td colspan="2"> </td> </tr> <tr> <td colspan="2"><input type="Submit" value="Send" /> <input name="Reset" type="reset" value="Reset Form" /></td> </tr> </table> </form></p> Also I have no idea about securing this information. Are there any basic steps I can take? Hi, i need to run this past you guy's. I need a way to effectivly check if a user has entered correct "real" details in my registeration form and check if email is already in my database to prevent the user from registering twice? I need this as some unknown person is using the form and entering there details like this: Username: 1 Password: 1 email: 1 It's really annoying as they are doing this several times so i have around 13 rows of just "1" in my table now! grr!! so far ive got: Form: <input name="upassword" type="password" class="style7" id="upassword"> <span class="style7">Email</span> <span class="style7">:</span> <input name="email" type="text" class="style7" id="email"> <span class="style7"></span> <span class="style7"></span> <input name="Submit" type="submit" class="style7" value="Register" /> </form> Script: Code: [Select] <?php if ($db_found) { $SQL = "INSERT INTO users (username, upassword, email) VALUES ('" .$username. "', '" .$upassword. "', '" .$email. "')"; $result = mysql_query($SQL); header( 'Location: http://www.removalspace.com/index.php' ); exit(); } else { print "Database NOT Found "; mysql_close($db_handle); } ?> But i need it to perform the checks aswell as drop the "real" data in to MySQL? Any ideas please? after cloasing connection of database i still got the values form database. Code: [Select] <?php session_start(); /* * To change this template, choose Tools | Templates * and open the template in the editor. */ require_once '../database/db_connecting.php'; $dbname="sahansevena";//set database name $con= setConnections();//make connections use implemented methode in db_connectiong.php mysql_select_db($dbname, $con); //update the time and date of the admin table $update_time="update admin set last_logged_date =CURDATE(), last_log_time=CURTIME() where username='$uname'limit 3,4"; //my admin table contain 5 colums they are id, username,password, last_logged_date, last_log_time $link= mysql_query($update_time); // mysql_select_db($dbname, $link); //$con=mysql_connect('localhost', 'root','ijts'); $result="select * from admin where username='a'"; $result=mysql_query($result); mysql_close($con); //here i just check after closing data baseconnection whether i do get reselts but i do, why? echo "after the cnnection was closed"; if(!$result){ echo "cont fetch data"; }else{ $row= mysql_fetch_array($result); echo "id".$row[0]."usrname".$row[1]."passwped".$row[2]."date".$row[3]."time".$row[4]; } // echo "<html>"; //echo "<table border='1' cellspacing='1' cellpadding='2' align='center'>"; // echo "<thead>"; // echo"<tr>"; // echo "<th>"; // echo ID; // echo"</th>"; // echo" <th>";echo Username; echo"</th>"; // echo"<th>";echo Password; echo"</th>"; // echo"<th>";echo Last_logged_date; echo "</th>"; // echo "<th>";echo Last_logged_time; echo "</th>"; // echo" </tr>"; // echo" </thead>"; // echo" <tbody>"; //while($row= mysql_fetch_array($result,MYSQL_BOTH)){ // echo "<tr>"; // echo "<td>"; // echo $row[0]; // echo "</td>"; // echo "<td>"; // echo $row[1]; // echo "</td>"; // echo "<td>"; // echo $row[2]; // echo "</td>"; // echo "<td>"; // echo $row[3]; // echo "</td>"; // echo "<td>"; // echo $row[4]; // echo "</td>"; // echo "</tr>"; // } // echo" </tbody>"; // echo "</table>"; // echo "</html>"; session_destroy(); session_commit(); echo "session and database are closed but i still get values from doatabase session is destroyed".$_SESSION['admin']; ?> session is destroyed but database connection is not closed. thanks At the moment I am creating a search function for my website. The approach I have in mind is a pseudo-PHP database. To give an example: A HTML form will submit the results to a PHP file. HTML FORM - Colour: Black PHP RESULT PAGE - if ($_POST['color'] == 'Black') {readfile("./products/black/*.html");} HTML FORM - Price: <$50 PHP RESULT PAGE - if ($_POST['Price'] == '<$50') {readfile("./products/less50/*.html");} The problem here is if there is an item that is black and costs less than $50, then its going to be listed twice. There is probably some code I can write to ommit the listing of duplicate entries, but it is probably going to be messy, so I am wondering if its better to use a centralized MySQL database, rather than a pseudo-PHP database? I've never used MySQL and don't know much about it and this is my first real attempt at using PHP. hello everyone, I am about to start coding my pages to display results from a database but before i do i want to know information about the following : Is it best to upload images to a database?and display them accordingly? or is it best to use images from a directory? What is most commonly used and or more reliable? Another topic i have trouble finding information on is actually positioning the output from you mysql database, is this practice done with tables?fields and rows? What is this method called? And is there more then one way to go about controlling result layout on your page? Sorry about the 1001 questions , but i am unable to find a clear answer on the topic ..especially question two . Thanks in advance. I have a very tricky php problem here. At least for me. On a page that is editing a job entry. I have a database generated group of checkboxes some of which have been checked when the job was entered. So the script has to do two things generate the complete list of checkboxes and check the ones that have already been selected in the job request. The below script is what I've done trying to figure it out. Description of what I'm trying to do: I'm generating the the boxes with a call to my database for the list of checkboxes. Then I'm building an array with all the possible items that could be checked to compare against. Ok here is where I think my first problem is. I need to make a second query to another table "worklog" in the same database. This is where the list of checked items are stored. What is the best way to do this? A second query seems wrong "or at least not efficient" Code: [Select] <div class="text">Job Type (<a class="editlist" href="javascript:editlist('listedit.php?edit=typelist',700,400);">edit list</a>):</div> <div class="field"> <?PHP $connection=mysql_connect ("localhost", "user", "password") or die ("I cannot connect to the database."); $db=mysql_select_db ("database", $connection) or die (mysql_error()); $query = "SELECT type FROM typelist ORDER BY type ASC"; $sql_result = mysql_query($query, $connection) or die (mysql_error()); $i=1; echo '<table valign="top"><tr><td>'; while ($row = mysql_fetch_array($sql_result)) { $type = $row["type"]; if ($i > 1 && $i % 26 == 0) echo '</td><td>'; else if ($i) echo ''; ++$i; $aTypes = array ("Spec Ad Campaign", "100 x 100 Logo", "100 x 35 (Featured Developer/Broker)", "100 x 40 (Featured Lender)", "120 x 180 (Home Page Auto)", "120 x 45 (Half Tile - Jobs)", "120x60 (Section Sponsor)", "125 x 40 (Profile Page logo)", "135 x 31 (Autos)", "135 x 60 (Autos)", "135 x 60 (Logotile - Jobs)", "150 x 40 (Auto Logo)", "160 x 240 (monster tile)", "160 x 400 (Skyscraper)", "160 x 600 (Tower)", "170x30 (Section Sponsor)", "240 x 180 JPG/GIF Auto Video", "240 x 180 size FLV video ", "300 x 250 (Story Ad)", "300 x 600 (Halfpage)", "468 x 60 (Banner Jpg/Gif)", "728 x 90 (Leaderboard)", "95 x 75 (Sweeps Logo) ", "Agent Profile Page", "Contest", "Creative Change", "Employer Profile", "Holiday Shop", "Home of the Week", "HP Half Banner (234 x 60 Static)", "Mobile Ads (320x53 *300x50*216x36*168x28)", "Newsletter", "Peelback", "Pencil Ad", "Print", "Real Deals", "Resize ad campaign", "Roll-Over Skyscraper", "Site Sponsor logo (170x30)", "Splash Page Production", "Travel Deals Update", "Video Ad", "Web Development", "Web Maintenance", "Web Quote (or Design)"); $dbTypes = explode(',',$row['type']); foreach ($aTypes as $type){ if(in_array($aType,$dbTypes)){ echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type' CHECKED><span style='color:#000;'>$type</span></input><br/>"; }else { echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type'><span style='color:#000;'>$type</span></input><br/>"; } } } echo '</td></tr></table>'; ?> </div> I have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.
my form page test.php
<?PHP
require_once("./include/membersite_config.php");
if(!$fgmembersite->CheckLogin())
{
$fgmembersite->RedirectToURL("login.php");
exit;
}
?>
<form action="editform.php?id_user=<?= $fgmembersite->UserId() ?>" method="POST">
<input type="hidden" name="id_user" value="<?= $fgmembersite->UserId() ?>"><br>
Name:<br> <input type="text" name="name" size="40" value="<?= $fgmembersite->UserFullName() ?>"><br><br>
Email:<br> <input type="text" name="email" size="40" value="<?= $fgmembersite->UserEmail() ?> "><br><br>
Address:<br> <input type="text" name="address" size="40" value="<?= $fgmembersite->UserAddress() ?> "><br><br>
<button>Submit</button>
my editform.php
<?php
$con = mysqli_connect("localhost","root","user","pass");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"UPDATE fgusers3 SET name = '".$_POST['name']."', email= '".$_POST['email']."', address= '".$_POST['address']."' WHERE id_user='".$_POST['id_user']."'");
header("Location: test.php");
?>
Hi guys, I was just wondering if anyone could help me. I've got a My_SQL database containing articles, a summary for the article and a date. I have a basic CMS system set-up, but I want to create a script that when users sign up to a mail list it forwards the summary and dates of the articles database. If that makes sense? But I only want it to forward the most recent 5 rows. I'm pretty new to PHP and I've been mostly following tutorials thus far, but this is quite specific. Thanks in advance! hello I want query from one table and insert in another table on another domain . each database on one domain name. for example http://www.site.com $con1 and http://www.site1.com $con. can anyone help me? my code is : <?php $dbuser1 = "insert in this database"; $dbpass1 = "insert in this database"; $dbhost1 = "localhost"; $dbname1 = "insert in this database"; // Connecting, selecting database $con1 = mysql_connect($dbhost1, $dbuser1, $dbpass1) or die('Could not connect: ' . mysql_error()); mysql_select_db($dbname1) or die('Could not select database'); $dbuser = "query from this database"; $dbpass = "query from this database"; $dbhost = "localhost"; $dbname = "query from this database"; // Connecting, selecting database $con = mysql_connect($dbhost, $dbuser, $dbpass) or die('Could not connect: ' . mysql_error()); mysql_select_db($dbname) or die('Could not select database'); //query from database $query = mysql_query("SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1"); while($row=mysql_fetch_array($query)){ $result=$row[0]; $text=$row[1]."</br>Size:(".$row[4].")"; $alias=$row[2]; $link = '<a target="_blank" href='.$row[3].'>Download</a>'; echo $result; } //insert into database mysql_query("SET NAMES 'utf8'", $con1); $query3= " INSERT INTO `jos_content` (`id`, `title`, `alias`, `) VALUES (NULL, '".$result."', '".$alias."', '')"; if (!mysql_query($query3,$con1)) { die('Error: text add' . mysql_error()); } mysql_close($con); mysql_close($con1); ?> the second database found on the cloud
i try to get JSON data but how to insert and update them to another online database with the same table my php script to return json data <?php include_once('db.php'); $users = array(); $users_data = $db -> prepare('SELECT id, username FROM users'); $users_data -> execute(); while($fetched = $users_data->fetch()) { $users[$fetchedt['id']] = array ( 'id' => $fetched['id'], 'username' => $fetched['name'] ); } echo json_encode($leaders);
i get
{"1":{"id":1,"username":"jeremia"},"2":{"id":2,"username":"Ernest"}} Edited March 24 by mahenda <?php $image_url = 'images/'; //User defined variables for page settings $rows_per_page = 2; $cols_per_page = 4; //Master array of ALL the images in the order to be displayed $images = array( 'image1.jpg', 'image2.jpg', 'image3.jpg', 'image4.jpg', 'image5.jpg', 'image6.jpg', 'image7.jpg', 'image8.jpg', 'image9.jpg', 'image10.jpg', 'image11.jpg', 'image12.jpg', 'image13.jpg', 'image14.jpg', 'image15.jpg', 'image16.jpg', 'image17.jpg', 'image18.jpg', 'image19.jpg' ); //END USER DEFINED VARIABLES //System defined variable $records_per_page = $rows_per_page * $cols_per_page; $total_records = count($images); $total_pages = ceil($total_records / $records_per_page); //Get/define current page $current_page = (int) $_GET['page']; if($current_page<1 || $current_page>$total_pages) { $current_page = 1; } //Get records for the current page $page_images = array_splice($images, ($current_page-1)*$records_per_page, $records_per_page); //Create ouput for the records of the current page $ouput = "<table border=\"1\">\n"; for($row=0; $row<$rows_per_page; $row++) { $ouput .= "<tr>\n"; for($col=0; $col<$cols_per_page; $col++) { $imgIdx = ($row * $rows_per_page) + $col; $img = (isset($page_images[$imgIdx])) ? "<img src=\"{$image_url}{$page_images[$imgIdx]}\" />" : ' '; $ouput .= "<td>$img</td>\n"; } $ouput .= "</tr>\n"; } $ouput .= "</table>"; //Create pagination links $first = "First"; $prev = "Prev"; $next = "Next"; $last = "Last"; if($current_page>1) { $prevPage = $current_page - 1; $first = "<a href=\"test.php?page=1\">First</a>"; $prev = "<a href=\"test.php?page={$prevPage}\">Prev</a>"; } if($current_page<$total_pages) { $nextPage = $current_page + 1; $next = "<a href=\"test.php?page={$nextPage}\">Next</a>"; $last = "<a href=\"test.php?page={$total_pages}\">Last</a>"; } ?> <html> <body> <h2>Here are the records for page <?php echo $current_page; ?></h2> <ul> <?php echo $ouput; ?> </ul> Page <?php echo $current_page; ?> of <?php echo $total_pages; ?> <br /> <?php echo "{$first} | {$prev} | {$next} | {$last}"; ?> </body> </html> This current code provides an easy way of uploading pictures and having the other pictures move automatically without the need of a database. But I must wonder, if doing this on a database would be faster. Would it be? I mean the example adobe shows I'm only using 19 pictures, but if I ever reach say 1000 pictures, would the php file be too big? Should I just make a database now or it doesn't matter? Also, the code above, which I took from a phpfreak member, has the pagination not working. Anyone care to do a diagnosis? Thanks! Not sure if I can post this here, or if I should go to a mySQL forum but... I'm trying to keep track of my user's "score". Each time they do a certain thing they get a preset amount of points, depending on what they did. The way I see it would be having a table that is joined to the user's ID, and have a column for each task. Then when they do a task the number in that column will go up. Then I will multiply that number by amount the task is worth. I can't just adjust there score because they need to be able to see what they did. So I hope that made sense. And if you know a better way to deal with this please let me know! |
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