PHP - Problem Escaping A Percent Sign
Hi. I am brand new here...
I have a script where I'm trying to parse an email and everything is working fine but it looks like the percent sign (%) is getting stripped out of my message. The percent sign is in the subject message and it gets removed and I also can't use it in any other strings in the script. I've tried just adding a slash in front like \% but that didn't fix it. Is there another way to escape a percent sign when using it in a script? Any help is appreciated. Thanks. -Jeff Similar TutorialsHi, Example: echo percent(5, 10, 2)."%"; gives you 50% Instead of this representation, I want to use a bar like this: So what ever percentage it is, it will show on the bar. (out of 100%) anyone help me achieve this please? Thanks Code: [Select] function percent($sub, $total, $dec) { if ($sub) { $perc = $sub / $total * 100; $perc = round($perc, $dec); return $perc; } else return 0; } Hey, I have a question about doing a % in PHP.. So say a user in my database has a cash amount of $20.00. Their referrer is going to get 20% of that. How would I calculate that in PHP? Stupid Question, and it might even be as simple as adding a zero, but I cannot figure it out.
$prob = number_format(mt_rand(1,100)/100, 2) ; $q = $db->query("SELECT * FROM rpg_monsters WHERE monster_zone = 1 AND chance >= $prob ORDER BY chance,rand() LIMIT 1");As you can see this works out perfectly. But only one issue. I want that 0.05 to be in the thousandths of percent: 0.005%. So it's a higher probability. But if I change it to 0.005, it never shows, even with iterations over 5000 in a for loop? I feel like I need to do something with my mt_rand? I need to simply move a decimal over someplace, but not sure where. Edited by Monkuar, 12 October 2014 - 08:30 PM. Quote from: Mike Solstice on April 14, 2011, 12:24:45 AM Ok, that works if I run it CLI or though phpMyAdmin, but won't if I try it in php Code: [Select] $remcomma = "UPDATE top_train SET deaths = replace( replace(field_name, ',', ''), '"', '' )"; mysql_query($remcomma); I assume because of the quotes used in the regex. I tried escaping the " Code: [Select] $remcomma = "UPDATE top_train SET deaths = replace( replace(field_name, ',', ''), '\"', '' )"; mysql_query($remcomma); Which broke it altogether. Anyone have any ideas? Thanks! Moved here since that thread was marked as solved & it's more of a PHP issue at this point anyway. Any & all help is greatly appreciated! Im getting a syntax error caused by the first line. How do I properly escape the code? echo "<tr class="'.$colors[$i++ % 2].'"> <td align=center nowrap>($phone1) XXX-$phone2</td> <td align=center nowrap>$overall</td> <td align=center nowrap>$hygiene</td> <td align=center nowrap>$attitude</td> <td align=center nowrap>$ethnicity</td> <td align=center nowrap>$body</td> <td align=center nowrap>$city</td> <td align=center nowrap>$user</td> <td align=center nowrap>$date</td> <td align=center nowrap>$comment</td> </tr>"; Hi, I have set up a simple function so that when a user enters a title in a php form, php gives the first letter of each word a capital letter. As follows: function caps($text){ $search_text=$text; $search_text=ucwords(strtolower($search_text)); $look_for = "(a"; $change_to = "(A"; $changed_text = str_replace($look_for, $change_to, $search_text); $search_text=$changed_text; $look_for = "(b"; $change_to = "(B"; $changed_text = str_replace($look_for, $change_to, $search_text); $search_text=$changed_text; $look_for = "(c"; $change_to = "(C"; //...etc...etc.. up to $look_for = "(z"; $change_to = "(Z"; $changed_text = str_replace($look_for, $change_to, $search_text); $search_text=$changed_text; return $search_text; } The trouble is, if I were to enter the following "The secret of DNA", my function would return "The Secret Of Dna" (removes the caps). Any ideas how to get around this would be most useful. Thanks in advance. Russ Hi, newbie here. Could someone show me how to properly escape the quotes in this code so it works properly? I'm having major problems with it, thanks. echo "<td style="background-color:#fff" onMouseover="this.style.backgroundColor='#ff9900';" onMouseout="this.style.backgroundColor='#fff';">" Hi guys I am encoding JSON in PHP but want to know how to escape ' and " thats in the content? Thanks Hi Php Freaks I am trying to escape variable but I am lost. I am sure that it is easy. can anyone help ta Code: [Select] $y="\\$GLOBALARRAY = array( \n"; echo $y; Hello! I'm very new to PHP, and I'm sure this is a noob question - still, its got me stuck! How would I best escape this properly? $resizeObj = new resize('C:\xampp\htdocs\images_test\$file'); so that $file is parsed? the whole script is as follows: Code: [Select] <?php // *** Include Nathan's class include("resize-class.php"); // Define the full path to your folder from root $path = 'C:\xampp\htdocs\images_test'; // Open the folder $dir_handle = @opendir($path) or die("Unable to open $path"); // Loop through the files while ($file = readdir($dir_handle)) { if($file == "." || $file == ".." || $file == "index.php" ) continue; // *** 1) Initialise / load image $resizeObj = new resize('C:\xampp\htdocs\images_test\$file'); // *** 2) Resize image $resizeObj -> resizeImage(100, 100, 'auto'); // *** 3) Save image $resizeObj -> saveImage('C:\xampp\htdocs\images_test\$file', 90); echo "<a href=\"$file\">$file</a><br />"; echo $path; } // Close closedir($dir_handle); ?> I have a table with two columns A and B. I'm submitting a form where the user selects different checkboxes. Column A contains the values of all the checkboxes. I want to, after the form is submitted, using the array of values of randomly checked checkboxes, search the table for the corresponding values in column B. Then display the values of selected checkboxes and their cosesponding values from column B. First I did an implode() on the array of submitted checkbox values, Code: [Select] $list = "'".implode("','", $_POST['Ref'])."'"; then did a mysql_real_escape_string on that. Code: [Select] for($i=0;$i<$count;$i++) { $list = mysql_real_escape_string($list[$i]); } I got a Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'SYSTEM'@'localhost' (using password: NO) Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established for each row of $list. What am I doing wrong? Do I need to escape variables passed via link? Is it possible that someone does an SQL Injection via unescaped $_GET variable? (Yes I am using it in an SQL query, not inserting it into a DB tho) I've got this code and it works fine. All I want to do is, when people click on the x to delete something they submitted erroneously, I want a confirmation...really want to delete this? (sometimes that x is too tempting you know) But I *think* I'm having trouble escaping the line (near bottom): onsubmit='".return confirm('Really Delete');."' I'm not sure if it will work anyhow...but right now it doesn't. Thanks for any suggestions. <code> <?php include '../php/config_conn.php'; $querysum = "SELECT SUM(total_time) FROM `coop_hours` where user = '".$_SESSION['user_name']."'"; $resultsum = mysql_query($querysum); $arr = mysql_fetch_row($resultsum); $resulthours = $arr[0]; $querytime = "SELECT * FROM `coop_hours` WHERE user = '".$_SESSION['user_name']."' ORDER BY `date_completed`"; $result = mysql_query($querytime); $num = mysql_num_rows($result); mysql_close(); echo "<table width='1000' cellpadding='0' cellspacing='0' border='0'><tr> <td width='200'><strong>Coop Job</strong></td> <td align=center width='200'><strong>Date Completed</strong></td> <td align=center width='120'><strong>Total Time</strong></td> <td width='380'><strong>Comments</strong></td> <td width='100'>Delete Entry</td></tr>"; $i=0; while ($i < $num) { $hours_id = mysql_result($result, $i, "hours_id"); $user = mysql_result($result, $i, "user"); $coop_job = mysql_result($result, $i, "coop_job"); $date_completed = mysql_result($result, $i, "date_completed"); $start_time = mysql_result($result, $i, "start_time"); $end_time = mysql_result($result, $i, "end_time"); $total_time = mysql_result($result, $i, "total_time"); $comments = mysql_result($result, $i, "comments"); echo "<tr><td>$coop_job</td> <td align=center>$date_completed</td> <td align=center>$total_time</td> <td>$comments</td> <td align=center valign='middle'> <form action='".php/del.php."' method='".post."' onsubmit='".return confirm('Really Delete');."'> <input type='".hidden."' name='".hours_id."' value='".$hours_id."'> <input TYPE='image' SRC='images/del.png' width='11' height='11' border='0' alt='Delete'> </form> </td> </tr>"; $i++; } echo "<tr><td colspan=5><hr></td></tr>"; echo "<tr><td></td><td align=right>Total hours:</td><td align=center>$resulthours</td><td></td></tr>"; echo "<table>"; ?> </code> I am passing this value: Quote \\"''"""\""\\\\""""''''\"""\\\""''""""'/'/'/\"'\'\\'\\''\\''''' the the following code is not handling it correctly: $data = mysql_real_escape_string(strip_tags(htmlspecialchars($data))); here is what it looks like when the server gets it: Quote \\"''"""\""\\\\""""''''\"""\\\""''""""'/'/'/\"'\'\\'\\''\\''''' and here is after the code executes: Quote \\\\"\'\'"""\\""\\\\\\\\""""\'\'\'\'\\"""\\\\\\""\'\'""""\'/\'/\'/\\"\'\\\'\\\\\'\\\\\'\'\\\\\'\'\'\'\ Notice if I have Quote \' it bugs out. I have a html form with a wusiwug editor when i selete an image it generates the code for that img like <img src="http://www.abc4blinds.co.uk/editor/elfinder/files/icons%20social.png" width="33" height="35"> but when i submint it it gives me <img src="%5C" http:="" www.abc4blinds.co.uk="" editor="" elfinder="" files="" icons%20social.png\="" 33\="" 35\="" width="\" height="\"> why i am not running any vailidation all i am doing is accessing it in $_POST['article'] <td><textarea name="article" cols="85" rows="5" id="article"><?php echo $_POST['article']; ?></textarea></td> I've got a file with some strings that have both types of quotes in them. And I seem to have managed to get the data, display it in my html, store it in a js array (using a json_encode in php and then simply inserting it into my js) but I cannot seem to pass the string as a parameter form an onclick function call to js.
For most strings the addslashes makes it work in the function call. But for those with both sets of quotes it won't work. My console tells me there are "unterminated string constants..". I've experimented with many silly changes but none make it work.
Ex. of the strings:
What do you mean "It's crooked"?
Of course I could remove the contraction and that would probably work, but that would be a hack, would not it?
This should be really simple but I can't get my head round it. Please see the PHP/HTML below. <?php $a = array( 'test' => "'data'" ); $json = json_encode($a); ?> <!DOCTYPE html> <html lang="en"> <head> <meta charset="utf-8"> <title>Unknown Page</title> <script type="text/javascript" src="http://ajax.microsoft.com/ajax/jquery/jquery-1.5.min.js"></script> <script type="text/javascript"> var j = jQuery.parseJSON('<?php echo $json; ?>'); </script> </head> <body> </body> </html> You will notice that data is stored by " then ' I have tried how it is and also with add slashes $a = array( 'test' => addslashes("'data'") ); It either outputs without slashes (which is clearly wrong) Code: [Select] var j = jQuery.parseJSON('{"test":"'data'"}');Or which I cant work out why dosent work Code: [Select] var j = jQuery.parseJSON('{"test":"\\'data\\'"}'); Now firebog throws the following error Code: [Select] missing ) after argument list How should I be escaping this variable? This topic has been moved to PHP Regex. http://www.phpfreaks.com/forums/index.php?topic=353729.0 What is the correct way to echo this line??? Quote <php? echo "proceed to <a href= <?php echo urldecode($_GET['url']); ?>> <?php echo urldecode($_GET['title']); ?> </a>"; ?> My Parse error says: Quote Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' in C:\Program Files\EasyPHP5.3.0\www\URL_tester\page2.php on line 16 |