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PHP - How Do I Find The Entry Page?
Whenever I see a PHP website (and given ftp access) I often struggle to find the entry page.
Most I see don't have index.php in the public_html directory - or that isn't the main entry page. Htaccess and adding Wordpress, shopping carts etc. also messes around with which file is the entry page. Is there a quick checklist of what a simpleton should do to find the entry page? E.g. (1) Check .htaccess................ etc. Similar TutorialsThis topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=312600.0 Hi All, I was wondering if there is any way that a PHP page can update itself when a row in a DB is added or updated? I am trying to get a feed up and running and want it to update when a row is updated/inserted. Thanks Matt Below is my code... it writes to the SQL database but does not show on the HTML page. Any advice? Code: [Select] <?php $query = mysql_query("SELECT * FROM `entries`"); $query = mysql_query("SELECT *, UNIX_TIMESTAMP(`date`) as date FROM `entries` ORDER BY `date` DESC LIMIT"); ?> <table width="100%" cellpadding="0" cellspacing="0" border="0"> <tr> <td valign="top"> <table width="100%" cellpadding="0" cellspacing="0" border="0"> <tr> <td valign="top" class="entrysmall" align="right">Posted on: <?php echo date("d/m/y g:i a", $row['date']); ?></td> </tr> <tr> <td valign="top"> <table width="100%" cellpadding="0" cellspacing="0" border="0"> <tr> <td valign="top" style="padding-right: 10px;"><span class="entrytitle">Name:</span></td> <td valign="bottom" width="100%"><?php echo htmlspecialchars(stripslashes($row['name'])); ?></td> </tr> <tr> <td valign="top" style="padding-right: 10px;" nowrap="nowrap"><span class="entrytitle">E-Mail:</span></td> <td valign="bottom" width="100%"><?php $email = explode('@',stripslashes($row['email'])); echo $email[0].' *at* '.$email[1]; ?></td> </tr> <td valign="top" style="padding-right: 10px;"><span class="entrytitle">Hometown:</span></td> <td valign="bottom" width="100%"><?php echo htmlspecialchars(stripslashes($row['hometown'])); ?></td> <tr> <td valign="top" style="padding: 5px 10px 0px 0px;" colspan="2"><span class="entrytitle">Message:</span></td> </tr> <tr> <td valign="top" style="padding-right: 10px;" colspan="2"><?php echo smilies(htmlspecialchars(stripslashes($row['message']))); ?></td> </tr> </table> </td> </tr> </table> </td> </tr> <tr> <td valign="top" height="10"></td> </tr> </table> Hello dear friends, I've very annoying problem my website is for child drawing (draw.php) after child do drawing will click on submit (form) by sending it to another page (thanks.php) | | | | data will be submitted to database and gives message saying ( thank you for ...blah blah blah) here is the problem if he refresh the page , it will also add entry to the database so imagine if someone did many many refresh, i will get many many empty entry into database how to stop this ? here is simple code based on this problem Code: [Select] <form name="frm" method="post" action="thanks.php"> <input type="text" name="name" id="name" value=""> <input type="text" name="email" id="email" value=""> <button type="submit">Submit</button> </form> and the (thanks.php) file code *assume we have connection to db Code: [Select] $sql = "INSERT INTO $table (name, email) VALUES ('$name', '$email')"; mysql_query($sql, $conn) or die(mysql_error()); echo "Thank you kid..nice drawing"; now my problem if (thanks.php) got refreshed it will also will add empty entry to database can anyone please help me how to stop it. Alright, wasn't quite sure how to summarize this in the title, but I want to: Check if a user status is "active" or not based on the UserName input. I have a table witch holds: Code: [Select] VarChar Username Var CharPassWord int Active Ted TedsPW 1 something like the above(assuming it formatted correctly. In my php script I will want to input a variable for Username to check for: inputUN in this example would be "Ted". $UserNameToCheck = $_GET['inputUN']; Then I want to check for that UserName in the database, if it exists, I want pull the value for the "Active" field for just that UserName and echo it. Thanks for any help. When there is a duplicate entry in my database, I would like it to go to the another page that indicates the error, rather than the "Error: Duplicate entry 'username' for key 1". It happens if someone is going to join and there is a duplicate entry. I would rather it show one of my customized php pages rather than a black page stating the error. I've tried to catch the entry with some php code but it bypasses it and continues with the duplicate error. I would appreciate any direction with this. Thanks! Hello,
I've tried to get a dynamic table from an external page, and searching for entries in it, so i used a dynamic XLS file using php excel reader. I only exported the file, but i couldn't search for data.
Can i get some help please ?
I am using php to upload a file to my server, and at the same time inserting the files name and url into my mysql database.
$sql = "UPDATE uploads SET name = '$name', url='$target_path'";
$statement = $dbh->prepare($sql);
$statement->execute();
This is working, however, when I upload a new file, rather than making a new entry in my database, it just overwrites the first one.I'm quite new at mysql so was wondering how I would make it add new entrys instead of overwriting the current one? i have the following code which should redirect the user to the index.php page, but instead it shows error 404 page not found....Please if anyone see the problem help me login.php <?php session_start(); ?> <html> <head> <title>Basic CMS | Admin area Login</title> </head> <body> <?php if (isset($_SESSION['user'])) { header("Location: index.php"); } else { ?> <form action="dologin.php method="post"> <table> <tr> <td><span>Username:</span></td> <td><input type="text" name="username" /></td> </tr> <tr> <td><span>Password:</span></td> <td><input type="password" name="password" /></td> </tr> <tr> <td colspan="2" align="right"><input type="submit" name="login" value="Login"/></td> </tr> </table> </form> <?php } ?> </body> </html> dologin.php <?php session_start(); ?> <html> <head> <title>Basic CMS | Admin area Login</title> </head> <body> <?php if (isset($_SESSION['user'])) { header("Location: index.php"); } else { ?> <form action="dologin.php method="post"> <table> <tr> <td><span>Username:</span></td> <td><input type="text" name="username" /></td> </tr> <tr> <td><span>Password:</span></td> <td><input type="password" name="password" /></td> </tr> <tr> <td colspan="2" align="right"><input type="submit" name="login" value="Login"/></td> </tr> </table> </form> <?php } ?> </body> </html> index.php <?php session_start(); if(isset($_SESSION['user'])) { ?> <html> <head> <title>Basic CMS | Admin area</title> </head> <body> </body> </html> <?php } else { header("Location: login.php"); } ?> Hello! I am executing an external command from a PHP script, using the exec function. Since this program can take more than 1-2 minutes to run, I thought I should use a "Loading page...please wait". What I need is to be able to get the process id from the external program that is being run and, when this finishes, I'll start outputting the results. Is there a way to do this? Im dealing with html links in a database where I need to execute a small piece of php code. Well mysql doesnt execute php code thats held in a table. The easiest way for me to do this is just add the links without the code, & set up a function that will replace anything I set to for example in my case a link lookin like this: http://mysite.com/index.php?page=offers&blah=blah&sid= needs to look like this; http://mysite.com/index.php?page=offers&blah=blah&sid=<?php echo $_SESSION['login']; ?> so I would need a function like <?php $replace('sid=','sid=<?php echo $_SESSION['login'];?> ?> something like that. And I know its possible & I could create a function like this but honestly right now I am so tired after working 12 hours im lazy dont want to think to hard while I know there are a lot of people out there that have a function like this already. If you dont have one & dont feel like writing one for me could you atleast point me in the right direction so I dont have to think to hard?! Thanks!! Hi.. So im currently working on a script.. My script generates a "oid" based on timestamp. Ive made the "oid" field unique in my db, so if i do a quick refresh i get the message: Duplicate entry '1283195988' for key 'oid' Is there some way i can check if its a dublicate, and if it is + it with 1 or something? Hey folks, Sorry for being a pain in the ass. I am trying to submit data to my database via a form and when I click Submit, I get: Duplicate entry '' for key 1 I understand that it means I have a duplicate entry with the ID of 1 or something like that. I can't find where the issue is. Here is the form: <form actin="" id="settings" name="settings"> <table class="listing form" cellpadding="0" cellspacing="0"> <tr> <th class="full" colspan="2"><?php echo $lang_settings; ?></th> </tr> <tr> <th colspan="2"><?php echo $lang_settings_description; ?></th> </tr> <tr> <td><?php echo $lang_sitename; ?>: </td> <td><input type="text" name="sitename" value="<?php echo $site_name; ?>" width="172" /> <em>Site name for logo</em></td> </tr> <tr> <td><?php echo $lang_email; ?>: </td> <td><input type="text" name="email" value="<?php echo $site_email; ?>" width="172" /> <em>Your email address</em></td> </tr> <tr> <td><?php echo $lang_yourname; ?>: </td> <td><input type="text" name="name" value="<?php echo $your_name; ?>" width="172" /> <em>Your own name</em></td> </tr> <tr> <td><?php echo $lang_meta_description; ?>: </td> <td><input type="text" name="meta-description" value="<?php echo $description; ?>" width="172" /> <em>SEO</em></td> </tr> <tr> <td><?php echo $lang_keywords; ?>: </td> <td><input type="text" name="meta-keywords" value="<?php echo $keywords; ?>" width="172" /> <em>Separate with Commas</em></td> </tr> <tr> <td><input type="submit" class="button" name="submit" value="<?php echo $lang_button_savesettings; ?>"></td> </tr> </table> </form> Here is the Insert code: $insert = "INSERT INTO settings (site_name, description, keywords, email, name) VALUES ('$sitename', '$meta_description', '$meta_keywords', '$site_email', '$your_name')"; mysql_query($insert) or die(mysql_error()); Can anyone please tell me where I am going wrong here? Much appreciated. Hello there, I have some code here which sends a number of variables from flash to SQL... I would simply like to add the functionality to overwrite records which have the same 'name' or 'pseudo'... can anyone help me please ? Thanks in advance Martin <?php $pseudo=$_POST['var1']; $score=$_POST['var2']; $table = $_POST['tab']; $dategame = $_POST['tempjoueur']; //$micro = microtime(); //$dategame = time()."".substr($micro, 2, 6); $_COOKIE['User'] = $_SERVER['REMOTE_ADDR']; $envoie = InsertDatas($table, "name, score, dategame", "'".$pseudo."','".$score."','".$dategame."'"); if ($envoie) { print_r("OK, $pseudo, $score, $dategame,$ipclient"); } else { echo "BAD, $pseudo, $score, $dategame,$ipclient"; } ?> $insertCount=0; foreach($results[1] as $curName) { if($insert){$insertCount++;} echo <<< END $curName<BR> END; } Right now the results would show up as... Bill Fred Jessica James John How do you make them show up like... 1 Bill 2 Fred 3 Jessica 4 James 5 John Hey guys! I know, I know this problem is EVERYWHERE but i just dont understand! I have a solid knowlage of php but my SQL skills are low, so i dont know too much about Keys and stuff. But my error is: Duplicate entry '' for key 2. The thing that im working on at this section is logging in with facebook. The code that presents my error is: $sql = "SELECT * FROM users WHERE uid=".$uid; $fbid = mysql_query($sql); $num_rows = mysql_num_rows($fbid); if(mysql_num_rows($fbid) < 1) { echo "You are not logged in. "; mysql_query("INSERT INTO `users` (`uid`) VALUES ('".$uid."')") or die(mysql_error()); } else { mysql_query("UPDATE users SET logged = '1' WHERE uid=".$uid); //mysql_query("UPDATE users SET full_name = $me WHERE uid=".$uid); echo "Your Logged in "; echo $me['name']; ?> Continue to <a href="removed :)"> My Settings </a>. <? } Any help is welcome Well this sounds weird, but it happens. I have an itemshop script, and I wanna code a system that automatically delete the database entry once the variable item amount falls to 0. However, this doesnt happen when I check phpmyadmin, and the item remains in the database even after the amount becomes 0. In fact, the amount can become negative at times, which annoys me. The code of deleting sql entry looks like this: if($continue == "yes"){ $query = "SELECT * FROM ".$prefix."user_inventory WHERE item_owner = '$loggedinname' AND item_name = '$item_name'"; $result = mysql_query($query); $num = mysql_numrows($result); $item_data = mysql_fetch_array($result); $item_amount = $item_data['item_amount']; $newquantity = $item_amount - $quantity; $query = "UPDATE ".$prefix."user_inventory SET item_amount = '$newquantity' WHERE item_owner = '$loggedinname' AND item_name = '$item_name'"; $result = mysql_query($query); if($newquantity == "0"){ $query = "DELETE FROM ".$prefix."user_inventory WHERE item_owner = '$loggedinname' AND item_name = '$item_name'"; $result = mysql_query($query); } What part of the codes should I edit to fix this issue? Thanks. For some reason this only allows one SQL to be added... // SQL Connection $username="monstert_admin"; $password="admin"; $database="monstert_admin"; $connection = mysql_connect("localhost", $username, $password) or die("Connection Failure to Database"); // Select Database mysql_select_db($database, $connection) or die ($database . "No Database" . $username); //Select everything from the the table $MyQuery = "SELECT * FROM photos"; $retrieve = mysql_query($MyQuery) or die(mysql_error()); if(mysql_num_rows($retrieve) != 0): $row = mysql_fetch_assoc($retrieve); else: echo ''; endif; if(isset($_POST['Submit']) && !$errors) { $url = $newname; include('img.php'); $image = new SimpleImage(); $image->load($url); $image->resize(500,315); $image->save($newname); mysql_query("INSERT INTO photos (url) VALUES ('$url')"); echo "File Uploaded Successfully as <i> "; echo $newname; echo "</i>"; } What would the issue be? I only have two columns - ID and url Thanks in advance! My query is not finding the last recieptnum entry, it is finding the number 9 everytime for some odd reason. Im trying to incrementally increase this each time a reciept is created. $getreceiptnum = mysql_query("SELECT receiptnum FROM accounting WHERE agency = '$agency' ORDER BY receiptnum DESC LIMIT 1") or die(mysql_error()); $recieptarray = mysql_fetch_array($getreceiptnum); $recieptnum = $recieptarray['receiptnum']; echo $recieptnum; I have a code where i can edit or delete certain details from the database. Right now, if the user clicks on the edit button it takes him edit page where he can edit the details. But, I am not able to Incorporate a Delete button such that, when the user clicks on a delete button, it should ask for a confirmation box. If the user clicks YES, then do the following: Code: [Select] DELETE from emp WHERE emp_id='$emp_id'; When there are multiple entries and I click on delete it deletes everything from the database. how can i make it to delete only the entry that is besides the delete button? Code: [Select] if(mysql_num_rows($emp_query) > 0){ echo "<table border='1'>"; echo "<th>Employee Id </th>"; echo "<th>Employee Name </th>"; while($get_emp = mysql_fetch_assoc($emp_query)){ $emp_id = $get_emp['emp_id']; $emp_name = $get_emp['first_name']." ".$get_emp['last_name']; echo "<tr>"; echo "<td width='100'>"; echo $emp_id; echo "</td>"; echo "<td width='400'>"; echo $emp_name; $edit_path = 'edit_employee.php?id='.$emp_id; ?> <INPUT TYPE="button" style="display:inline;" value="VIEW / EDIT" onClick="location.href='<?php echo $edit_path; ?>'"> <form style='margin: 0; padding: 0; display:inline;' method="post" action="<?php echo $_SERVER['PHP_SELF'];?>" onSubmit="return confirm('Are you sure this is correct?');"> <input style='display:inline;' name="delbutton" type="submit" value="DELETE"> <?php if(isset($_POST['delbutton'])){ $del_emp = mysql_query("DELETE from employee WHERE emp_id = '$emp_id'") or die(mysql_error()); //header('Location:view_employee.php'); } echo '</form>'; echo "</td>"; echo "</tr>"; } } |
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