PHP - Execute Form
How can I get this form to execute? Do I need to use a hidden input or something?
Code: [Select] <form action="partnerRequest.php" method="post"> <a href="" title="Become Associates" id="becomeassoc" class="mll"> <span class="f_right">Become Associates</span><img src="../assets/img/becomeassoc.jpg" class="mrs vmiddle" alt="Become Associates" /> </a> </form> Similar TutorialsHi,
In reference to my first attached image, I have a form which displays two SELECT/drop-down fields (labeled "Store Name" and "Item Description".....and both of which pull-in values from two separate lookup/master tables, in addition to providing an additional option each for "NEW STORE" and "NEW ITEM").
Now, when first-run, and/or if "NEW STORE" and "NEW ITEM" are not selected from the drop-down's then the two fields in green ("New Store Name" and "New Item Name" are hidden, by means of the following code:
<div class="new-store-container" id="new-store-container" name="new-store-container" style="display:none;"> <div class="control-group"> <div class="other-store" id="new_store_name"> <?php echo standardInputField('New Store Name', 'new_store_name', '', $errors); ?> </div> </div> </div>Conversely, if "NEW STORE" and/or "NEW ITEM" are selected from the two drop-down's then one (or both) of the "New Name" fields are unhidden by means of the following two pieces of code, one PHP and the second JS: <select class="store-name" name="store_id" id="store_id" onclick="toggle_visibility('store_id','new-store-container')"> <?php echo $store_options; ?> <?php if($values['store_id'] == "OTH") { echo "<option value='OTH' selected> <<<--- NEW STORE --->>> </option>"; } else { echo '<OPTION VALUE="OTH"> <<<--- NEW STORE --->>> </OPTION>'; } ?> </select> function toggle_visibility(fieldName, containerName) { var e = document.getElementById(fieldName); var g = document.getElementById(containerName); if (e.value == 'OTH') { if(g.style.display == 'none') g.style.display = 'block'; else g.style.display = 'none'; } }All of that is working just fine. The problem I'm having is that when I click the "Create" button, after having left any one of the form fields blank, the two "New Name" fields are hidden again, which I don't want to happen i.e. I want them to remain visible (since the values of "store_id" and/or "item_id" are "OTH"), so that the user can enter values into one or both of them, without havng to click on the drop-down a second time in order to execute the "on-click" code. The second attached image shows how the fields are hidden, after clicking "Create". How can I achieve that? It would be greate if someone could cobble-up the required code and provide it to me, since I'm relatively new to this. Thanks much. Snap1.png 26.14KB 0 downloads Snap2.png 149.47KB 0 downloads Hi,
I have a cron job that executes this script every 2 minutes:
<?php // LOAD WP-LOAD.PHP require('/opt/bitnami/apps/wordpress/htdocs/wp-load.php'); // INCLUDE AND EXECUTE SCHEDULER.PHP include('/opt/bitnami/apps/wordpress/htdocs/wp-content/themes/yeelloe/scheduler.php'); ?>When I try to include; /opt/bitnami/apps/wordpress/htdocs/wp-content/themes/yeelloe/scheduler.php: <?php // EXPLODE AND PARSE WP-CONTENT; FUNCTIONS.PHP $parse_uri = explode( '/opt/bitnami/apps/wordpress/htdocs/wp-content', $_SERVER['SCRIPT_FILENAME'] ); // LOAD WP-LOAD.PHP require_once( $parse_uri[0] . '/opt/bitnami/apps/wordpress/htdocs/wp-load.php' ); // LOAD TEMPLATE FUNCTION CheckFunction(); ?> Not sure if this is possible I am trying to do a multi delete: Code: [Select] <form name="form1" method="post" action=""> <table> while($rows=mysql_fetch_array($result)){ ?> <tr> <td align="center" bgcolor="#FFFFFF"><input name="checkbox[]" type="checkbox" id="checkbox[]" value="<? echo $rows['id']; ?>"></td> <td bgcolor="#FFFFFF"><? echo $rows['id']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['name']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['lastname']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['email']; ?></td> </tr> <?php } ?> // Check if delete button active, start this if($delete){ for($i=0;$i<$count;$i++){ $del_id = $checkbox[$i]; $sql = "DELETE FROM $tbl_name WHERE id='$del_id'"; $result = mysql_query($sql); } <tr><td> <select name="dropdown"> <option value="option1">Choose an action...</option> <option value="edit">Edit</option> <option value="delete">Delete</option> </select> <a class="button" href="#">Apply to selected</a> if($result){ echo "<meta http-equiv=\"refresh\" content=\"0;URL=nameslist.php\">"; } } mysql_close(); I have written a Validation class that checks to see if a file being uploaded to the server meets certain conditions. That works a treat. The next step is to actually upload it to the server and I have an Upload class that can do that. Again, that works perfectly fine. Once the file uploads, I am passing the $location of that of that file to my DB class. The DB class is full of methods that prepare and then execute strings that are SQL queries that are required in other areas of my application. Nothing I have at the moment is suitable for just running an SQL file so I don't know what to do now.. My procedural code, that works, looks like this; $dbh2 = new PDO("mysql:host=localhost;dbname=DB360transfer", $login_user, $login_password); $sql = file_get_contents($path.$new_file_name); $qr = $dbh2->exec($sql);I'm not sure how to replicate this up in a PDO/OOP application. My DB Class script is attached. The run_from_file code starts on line 98. I've left in the other stuff as I suspect the answer has something to do with using the $this or the self:: - but really I have no idea. So my question is, what is the correct syntax for executing a file in OOP? Attached Files DB.php 2.92KB 6 downloads Hi, I am trying to create a PHP script ("repair-correct.php") in order to run some CLI commands without using PuTTy - The CLI commands are needed to repair/correct the execution of a web application named Mautic. Shared web host account with PHP 7.0 URL: https://www.myserver.com Mautic directory: https://www.myserver.com/mautic
What I want to do is:
Step1: Change ownership of files and folders To find out which user Apache is running as, I want to execute the following command and take note of the first entry in the line which is returned: ps aux | grep apache2 I want to use this information to find the groups with the following command groups apache_user (where apache_user is the user I identified from the first step above) To reset the ownership of files and folders, I want to use the following command (ensuring that I replace apache_user and apache_group with the values identified in the steps above): sudo chown -R apache_user:apache_group With this command I want to change ownership, using the -R flag which means recursively - including all files/folders within that location.
Step 2: Reset the file and folder permissions
find . -type f -not -perm 644 -exec chmod 644 {} +
Thanks so much for your help in solving this problem!
Best, Tony
Hi, Below i have some sample code and am confused over execute(), i have the code below in a try and catch block, in the catch block i call a function i created to log any error that is caught in catch block to a .txt file. I then looked online and it seems that i should do an if statement check on execute to ensure it executed the query, the part that confuses me if the execute failed i thought it would be caught in the catch block but it seems that is not the case. To explain better i have commented the code in depth on the area that i am confused about. Any help in making me understand would be great. I have not included all code above try and catch to keep things simple Code: [Select] <?php try { // connect to database $dbh = sql_con(); // checke if username exists on users table or users banlist table $stmt = $dbh->prepare(" SELECT users.user_login FROM users WHERE users.user_login = ? UNION ALL SELECT users_banlist.user_banlist FROM users_banlist WHERE users_banlist.user_banlist = ?"); // this is the part i am confused with, why is it i would use an if statement on execute() ? // i thought using a try and catch block any errors would be caught in the catch block. // using an if statement to check if execute() worked, i thought if execute failed it would // be handled by the catch block, i mean in my exmaple code here, what could cause the execute to fail ? // and why if execute failed it would not be caught by catch block ? // i am looing at exmaple online and i am reading different things and its all confusing me // execute query if(!$stmt->execute(array($username, $username))){ echo 'something went wrong .. '; } else { // execute worked } } // if any errors found log them in my ExceptionErrorHandler() function and display friendly message catch (PDOException $e) { // this function catches an error and logs them to file ExceptionErrorHandler($e); require_once($footer_inc); exit; } } ?> Thanks or any help! Hello, I've been assigned the task of creating an internal site for our office that has links to all the websites we use as well as the applications we use. I'm very comfortable using php with mysql for handling forms and displaying database information but I'm not sure how to go about this. So my question is, is there anyway I can use php to open an exe file? Hi Guys, I have the following script which works fine for GoogleMaps Code: [Select] <?php require'EasyGoogleMap.class.php'; $gm = & new EasyGoogleMap("rYvyXhRFPYW5tkV_IE5hWxecUidHAYjJhBSp59xzQuNZcoxNYCiaum4_Xb66Fw"); $gm->SetMarkerIconStyle('STAR'); $gm->SetMapZoom(10); $gm->SetAddress("10 market st, san francisco"); $gm->SetInfoWindowText("This is the address # 1."); $gm->SetAddress("Manila, Philippines"); $gm->SetInfoWindowText("This is Philippine Country."); $gm->SetSideClick('Philippines'); ?> <html> <head> <title>EasyGoogleMap</title> <?php echo $gm->GmapsKey(); ?> </head> <body> <?php echo $gm->MapHolder(); ?> <?php echo $gm->InitJs(); ?> <?php echo $gm->GetSideClick(); ?> <?php echo $gm->UnloadMap(); ?> </body> </html> I've now coded that static page to take the following, and within the GoogleMaps database field entered in exactly the following $gm->SetAddress("10 market st, san francisco"); $gm->SetInfoWindowText("This is the address # 1."); $gm->SetAddress("Manila, Philippines"); $gm->SetInfoWindowText("This is Philippine Country."); $gm->SetSideClick('Philippines'); Code: [Select] <?php $maps_find = $OO_Open->newFindCommand('Maps'); $maps_find->AddFindCriterion('AccountName','xx'); $maps_result = $maps_find->execute(); $maps_row = current($maps_result->getRecords()); $Gdata = $maps_row->getField('GoogleMaps'); //echo $Gdata; //exit; ?> <?php require'EasyGoogleMap.class.php'; $gm = & new EasyGoogleMap("my key"); $gm->SetMarkerIconStyle('GT_FLAT'); $gm->SetMapZoom(10); $gm->SetMapWidth(1080); # default = 300 $gm->SetMapHeight(550); # default = 300 echo $maps_row->getField('GoogleMaps'); //eval("\$Gdata = \"$Gdata\";"); ?> <html> <head> <title>EasyGoogleMap</title> <?php echo $gm->GmapsKey(); ?> </head> <body> <?php echo $gm->MapHolder(); ?> <?php echo $gm->InitJs(); ?> <?php echo $gm->GetSideClick(); ?> <?php echo $gm->UnloadMap(); ?> </body> </html> Is there way of getting the $maps_row->getField('GoogleMaps'); command to behave as if it was php (when I echo it, it does display the code as text). I'm not using the FileMaker db to store the code, I thought eval function was the answer but not had much luck. Many Thanks for the advice Jalz I need to execute a program in the command prompt in windows using php, which php functions would be best to do this? and would i need to set any type of permissions to be able to use them? hi, I am having problems setting a session from a SQL query, please help... here is my code: Code: [Select] $query1 = "SELECT * FROM members_copy WHERE rsUser = '$username' AND rsPass = '$password'"; $result1 = mysql_query($query1); echo "<br>".$result1['USERID']; $_SESSION['s_logged_n'] = 'true'; $_SESSION['s_username'] = $username; $_SESSION['USERID']=$result1['USERID']; $_SESSION['RSTOWN']=$result1['RSTOWN']; $_SESSION['RSEMAIL']=$result1['RSEMAIL']; $_SESSION['RSUSER']=$result1['RSUSER']; no sessions are currently being set?! How can i run this javascript command show_seasons(some_value), in php code, i mean without any action ? I have this php code: if ($sid == $s_id) { javascript:show_seasons($sid); } How to do that? Thanks.. Hello, I am mounting google drive to my raspberry pi with this command from command line; sudo gdfs -o allow_other /var/www/html/gdfs.creds /media/pi/gdrives İt is working from command line, but it is not work when i execute it from web browser. Here php content; shell_exec("sudo gdfs -o allow_other /var/www/html/gdfs.creds /media/pi/gdrives"); and i changed my sudoers file giving permission www-data. here is my sudoers file content # This file MUST be edited with the 'visudo' command as root. # # Please consider adding local content in /etc/sudoers.d/ instead of # directly modifying this file. # # See the man page for details on how to write a sudoers file. # Defaults env_reset Defaults mail_badpass Defaults secure_path="/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin" # Host alias specification # User alias specification # Cmnd alias specification # User privilege specification root ALL=(ALL:ALL) ALL # Allow members of group sudo to execute any command %sudo ALL=(ALL:ALL) ALL www-data ALL=NOPASSWD: ALL # See sudoers(5) for more information on "#include" directives: #includedir /etc/sudoers.d Can anyone tell me what i am doing wrong ?
PHP is owner of image directory with "permissions 660." My script is: $imagePath = '../imageDirectory/'.gif'; $image = imagecreatefromstring(base64_decode($raw_image_data)); $rotate = imagerotate($image,-90,0); imagegif($rotate,'../imageDirectory/'.gif'); Is it because the imagerotate and imagegif functions need the execute requirements to be able to monkey with the image? Thank you.
Sub question (maybe more important than the first question): Thank you. Hey everyone, I have an SQL file called "Cype.sql" and I was wanting to have it run when the install feature is going on. However I can't seem to figure out why exactly the code is not working. is it possible to have it run without getting too in depth the PHP coding? I'm not OOP Literate yet. //Connection is opened //While in installation file $sqlFile = "Cype.sql"; if(!file_exists($sqlFile)){ echo "File not found"; } else{ $openFile = fopen($sqlFile, "r"); $tryQuery = mysql_query($openFile); } Now obviously, I'm not the best coder. I've been out of the works of PHP for quite some time now as well. However, I have researched it and found no answers to my issue. Any help would be greatly appreciate. Thank you. Hi, I have a strange problem using the exec command. I have the following php code : <?php error_reporting(E_ALL); ini_set('display_errors','On'); $accountBase = "HORAIREMOBILE"; $primaryUser = "DUM"; $secondaryUser = "DUM"; $result = exec(escapeshellcmd("/home/evidian/utils/getAccount ".escapeshellarg($accountBase)." ".escapeshellarg($primaryUser)." ".escapeshellarg($secondaryUser)),$output,$return_val); echo $result; ?> When I execute the command from the CLI, with any user, it just works fine, and shows my the result (basically a JSON formated output). However, when I call the code frome the apache server, it simply returns nothing. Could anybody help me with this issue ? Hi, Im starting to write a simple (so I thought! ) project at work, I have no problems with the HTML, CSS or even the php mysql setup, but I am struggling with a simple piece of code! I have a list box, that populates from a table column using the following code: Code: [Select] <?php function select(){ $query="SELECT appname FROM kpe_apps"; $result = mysql_query($query); echo '<select name="item" onchange="this.form.submit()">'; while($nt=mysql_fetch_array($result)){ echo '<option value="' . $nt['id'] . '">' . $nt['appname'] . '</option>'; } echo '</select>'; } ?> I then called the function on the page I needed. which works fine, now all I need is whatever is selected in the listbox other columns in the same table relating to the selected item are seen. I just cant seem to do it, loads of people are using jquery and other code, surely this can be done in php? any help would really be appreciated.. Many Thanks MOD EDIT: code tags fixed, linefeeds and indenting added. I'm looking to get php to write to file when someone visits one of my sites without javascript.
From the research I've done, it seems that an css hidden, iframe within the noscript tag would work, and the php file would then write to a log.
I just want a simple tick log that increments by 1, but perhaps also lists the user-agent.
My one concern is that I want to include this same code on multiple sites, all of which could be trying to access, and write a tick to the same "no JS log". Is that going to cause problems the way php writes to files?
Can someone help me create the code that will write to this log file, that won't cause errors or problems when multiple sites/visitors could be making writes to the log?
I'm a php noob, so while I can research how to get php to write to a file and save, most of the tutorials don't mention if it would be a safe method to use, if multiple domains were trying to write to that same file, at the same time, and trying to add their own "tick" to increment the number within the flat file.
Extended help that I'm looking for is getting ticks per user-agent/search bot, so bots/web users that visit my pages, that don't run JS, could have ticks by them. eg:
Google Bot: 3
Mozilla Firefox 31.0: 15
Any help on this would be appreciated, or any flaws that you see in this type of tracking, thanks!
private function _authenticate() { // if there's already an auth error if ( $this->_checkForMessageType('auth') ) { $this->_addMessage('auth', 3); self::__destruct(); return false; } $stmt = $this->_dbh->prepare("SELECT shopID FROM api_users WHERE shopID = ? AND API_key = ? LIMIT 1"); var_dump($stmt->execute(array($this->_shopID, $this->_key))); echo $stmt->rowCount(); // authenticate key / shop id if ( !$stmt->rowCount() ) { $this->_addMessage('auth', 3); self::__destruct(); return false; } $this->_addMessage('auth', 4); } I am using PDO with MySQL driver and ATTR_EMULATE_PREPARES => true, however, when I run this code I get the output: Code: [Select] bool(false) 0{"auth":{"3":""}} Any ideas why PDOStatement::execute is returning false? I get no connection errors, no PDOExceptions, the db structure is correct, and there is valid data in the database. Any help appreciated, thanks. Hi. I have a query that determines if user is online/offline. Works perfectly when page loads. I want to change to auto run query every X seconds. NOTES: - The query is on a .php page with multiple other queries. -IF it is possible to still run the query with-in this page? or make a separate file and include in an iframe? i would like to avoid the iframe and just $output the result of query every X seconds. Here is my $query code: $query = "SELECT `manager_id` FROM #__profiles_xref WHERE profileid = '$profid'"; $onofflineid=doSelectSql($query,1); foreach ($onofflineid as $isonoffline){ $profmanagerid=$isonoffline->manager_id; } $query = "SELECT `session_id` FROM table_session WHERE userid = '$propmanagerid'"; $onofflinestatus=doSelectSql($query,1); if (count($onofflinestatus)>0) { $output['PROPSTATUS']='Online'; } else { $output['PROPSTATUS']='Offline'; } <?php error_reporting(E_ALL); $psPath = "powershell.exe"; $psDir = "C:\\wamp\\www\\ps\\"; $psScript = "SampleHTML.ps1"; $runScript = $psDir. $psScript; $prem = "-Action enable"; $runCMD = $psPath. " " .$runScript. " " .$prem; //var_dump($runCMD); $output = exec($runCMD); echo $output; ?>Hello, I am working on a small project to get results from powershell script by using PHP. For some reason in PHP logs I get Exec unable to fork. Above is the script I wrote to execute powershell script within php. My webserver is IIS 7, and app pool is using a domain user that has full rights for Powershell to execute and get remote server results. |