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PHP - How To "remember" Select Tag Value--please Help
Hi guys,i was trying to figure out how to work with <select> tag, remember/set the value that was being selected before. The values are fetched from the database..From other forums I've been, they said that $_SESSION will do the work but I'm having trouble where to put it inside my code.. Please take a look at this
Code: [Select] <form name="form1" action="do_set.php"> <?php $sql="SELECT school_year FROM admin_sy ORDER BY school_year ASC" or die (mysql_error()); $result=mysql_query($sql); echo "Set schoolyear: <select name=SY value=''>"; while($row=mysql_fetch_array($result)) { echo "<option value=$row[school_year]>$row[school_year]</option>"; } echo "</select>"; ?> <br> Sem: <select name="sem"> <option value="1st">1st</option> <option value="2nd">2nd</option> <option value="Summer">Summer</option> </select> <input type="Submit" name="submit" value="Set"> </form> as you can see, 'school_year' are fetched from a database. now my problem is, when i select a specific year, the value should be "remembered" after i click set or refresh the page or go back from the previous page. this will be useful so the user will be able to know what school year has already been set. hope you guys can help.. thanks in advance. Similar TutorialsCan someone please help me with an array problem i can not figure out. I need the array to be numbered from 1 to how ever many fields that are needed in the form and have a mysql field name and the title of the field also in the array. 1, "mysql_field_name", "Title of form field" 2, "", "" and so on then the form will be shown based on the array. I have the following draft code which I am working with. any suggestions on how i may do this array ? Code: [Select] <?php $options = array( '1'=> array('fieldtext'=>'option1', 'mysqlfield'=>'option1'), '2'=> array('fieldtext'=>'option2', 'mysqlfield'=>'option2'), '3'=> array('fieldtext'=>'option3', 'mysqlfield'=>'option3'), '4'=> array('fieldtext'=>'option4', 'mysqlfield'=>'option4'), ); // $options = array(1 => "option1", "option2", "option3", "option4"); // the line above works but i want to include the name of the mysql field as well. $userid = 1; ?> <div style="align: center; margin: 12px; font-family:Tahoma;"> <br><br><?php if ($_POST['Update'] != "Update") { // check if form submitted yet, if not get data from mysql. $res = db_query("SELECT * FROM `users` WHERE `userid` = '" . $userid . "'"); foreach($options as $key => $value) { $_POST[$key] = mysql_result($res, 0, $value); } $ok_to_update = "no"; } elseif ($_POST['Update'] == "Update") { // check if form submitted yet, if so get POST data. // error checking // foreach($options as $key => $value) { // $_POST[$i] = ""; // } $ok_to_update = "yes"; } if ($_POST['Update'] == "Update" && $ok_to_update == "yes") { // $res = db_query("INSERT INTO `users` () VALUES ()"); // add user details to database. ?><p><br><br><br>Thank you for updating</p><?php } else { ?><form name="form1" method="post" action=""> <?php foreach($options as $key => $value) { ?><p><?php echo($value); ?>: <input type="text" name="<?php echo($key); ?>" value="<?php echo($_POST[$key]);?>"></p> <?php } ?> <input name="Update" type="submit" value="Update"> </form> <?php } ?> </div> What I'm trying to accomplish is your average session login w/ a cookie-based "remember me" feature. I can log in fine, I can set cookies, I can access those cookies, and I can use that data to log in just as if the user had logged in manually. The problem is logging out. But the bigger problem is figuring out why. You see, it works perfect. It logs in and it log out, so long as I am outputting data to the page. I was outputting a session variable and the cookies I set to make sure they were all working right; and they were. But then as soon as I disable those echoes, all of a sudden it won't log out anymore. So then I turn them on to see what the data says and BAM, I'm logged out. I log back in fine, I log back out fine, so I turn em off again. I log in fine. I can't log out. I try multiple times. I close my browser and open a new one. Still logged in. I try a few more times, still logged in. I turn the output back on, load the page again and I'm logged out. So... WTF? (my code to follow) Hello all, I'm using sessions to store a username/password For example, I have the following at the top of all my source files: Code: [Select] <?php session_start(); if(!isset($_SESSION['myusername'])){ header("Location:login.php"); exit; } else { $username = $_SESSION['myusername']; $permissionLevel = $_SESSION['mypermissionlevel']; } ?> Now, I'd like to add a "Remember Me" checkbox so they can store their login info (securely!) so they don't have to log in all the time.... How do I do this? Thanks Hi, bit stuck on how to find and replace "<" and ">" with "<" and ">". I basically have a database record that outputs to screen and I need the code in the <code> tags to be rendered to the screen. I therefore need it to go through the whole array variable from the db and change the symbols just inside the code tags. Please be aware that the code tags might happen more than once here's an example below Code: [Select] <p>blah blah blah</p> <p>blah blah blah</p> <p>blah blah blah</p> <code> <h1>hello</h1> </code> <p>blah blah blah</p> <p>blah blah blah</p> <p>blah blah blah</p> <code> <h1>hello</h1> </code> the desired output would be: <p>blah blah blah</p> <p>blah blah blah</p> <p>blah blah blah</p> <code> <h1>hello</h1 </code> <p>blah blah blah</p> <p>blah blah blah</p> <p>blah blah blah</p> <code> <h1>hello</h1 </code> help on this would be great Cheers Rob Hi guys I am a newbie with PHP and I have been trying to solve this problem for 3 days.. Ive set up a booking form on my website to be sent directly to my email once the client clicked on the send button.. The problem I am having while checking these pages on wamp is that the booking form is ok but when I click on the send button the following message error appears "Warning: mail() [function.mail]: Failed to connect to mailserver at "localhost" port 25, verify your "SMTP" and "smtp_port" setting in php.ini or use ini_set() in C:\wamp\www\fluffy_paws\booking_form2.php on line 27" When I check online I have this message "500 - Internal server error. There is a problem with the resource you are looking for, and it cannot be displayed." Are my codes wrong??? My webhosting is Blacknight.com , Im on Windows Vista, my internet is a dongle with O2 ireland Thanks for your help! Hi guys I'm struggling a bit, I need to replace a word that occurs multiple times in text with an array("up","down","forward","backwards") of words. $find = "left"; $replace = array("up","down","forward","backwards"); $text = "left left left left"; echo str_replace($find,$replace,$text); The Output is: array array array array Did try this with a foreach statement as well, but no luck. Is there a better way of doing this? Thanks cant work out this mysql syntax error "operation":"medupdate","medid":"","name":"ibo","medyear":"5","medmonth":"21","medday":"1","recuser":1,"SuccFail":"fail","SuccFailMessage":"error occured You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '''WHERE med_id=' at line 2","ResultData":""} Code: [Select] $sql="update metodology set med_description='".$req['name']."', med_year='".$req['medyear']."', med_month='".$req['medmonth']."', med_day='".$req['medday']."', med_recorddate= now(), med_recorduserid= 1'"; $sql.= " WHERE med_id=".$req['medid']; I am currently creating a form and I want to populate a drop down selection menu with data from two fields in a form. For example, I want it to pull the first and last name fields from a database to populate names in a drop down menu in a form. I want the form to submit to the email address of the person selected in the drop down. Is this possible to do? The email is already a field in the record of the person in the database. Can anyone give me some pointers or advice on how I should go about setting up the "Select" box drop down? I am not sure how to code it to do what I am wanting. Any links to relevant help would be appreciated too. Thanks in advance! Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/aci/docs/admin/hours.php on line 20 Code: [Select] <?php $user = $_SESSION['myusername']; $result = mysql_query("SELECT * hours WHERE member='$user'"); while ($row = mysql_fetch_array($result)){ echo $row['date'] . " " . $row['time'] . "<br />"; } ?> Code: [Select] CREATE TABLE `hours` ( `hoursID` int(5) NOT NULL auto_increment, `member` varchar(20) NOT NULL default '', `date` date NOT NULL default '0000-00-00', `time` time NOT NULL default '00:00:00', PRIMARY KEY (`hoursID`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ; -- -- Dumping data for table `hours` -- INSERT INTO `hours` VALUES (4, 'aci', '2010-11-30', '14:31:39'); INSERT INTO `hours` VALUES (2, 'aci', '2010-11-30', '14:31:08'); INSERT INTO `hours` VALUES (3, 'aci', '2010-11-30', '14:31:23'); INSERT INTO `hours` VALUES (5, 'aci', '2010-11-30', '14:31:40'); I am trying to print the list of a table which I requested with "SELECT DISTINCT" as below Code: [Select] $db_connect = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); $sql_get = "SELECT DISTINCT category FROM con"; $sql_run = mysqli_query($db_connect, $sql_get) or mysqli_error($db_connect); $sql_assoc = mysqli_fetch_assoc($sql_run); What is now needed to print the list of the table data by this conditions? I tried the while loop, but I seem to approach wrong, and get endless loops or errors. Hi, I need to call to this text to add a all with value "blank/empty" or "0" to the dropdownlist code below.. it should be inserted int the first position in the dropdownlist and have a value blank and call to this text from the language files: $searchAll = gettext('_LANGUAGE_SEARCH_ALL',_LANGUAGE_SEARCH_ALL,false,false); example is: ratestitle=>0 gettext('_LANGUAGE_SEARCH_ALL',_LANGUAGE_SEARCH_ALL,false,false); the words it calls is "ALL" when the user selects or leaves the "ALL" in the list the value remains empty this allows the search to search everything.. basically im stuck how to get it in the dropdown list as the first default position: can anyone help? here is my code://this is a search function which calls to the field rate_title the text from the select box is matched to the database table and the search pages shows the search option is the search is successful.. I have 5 of these types of searches but cant give a broader search option to my users as the values must be selected to search--- i need the option "ALL" to have a "EMPTY" value or no value.... if (in_array("ratestitle",$searchOptions)&& $showSearchOptions ) { if (empty($sch->filter['ratestitle']) ) $selectOption=$output['LANGUAGE_SEARCH_RATESTITLE']; else $selectOption=$sch->filter['ratestitle']; $showButton=true; $query = "SELECT DISTINCT rate_title FROM #__rates_table ORDER BY rate_title ASC"; $dropDownList ="<select class=\"inputbox\" name=\"ratestitle\">"; $ratesitles =doSelectSql($query); foreach ($ratesitles as $ratetitle) { $selected=""; $rate_title=$ratetitle->rate_title; if ($ratetitle==".$sch->filter['ratestitle'].") $selected="selected"; $dropDownList .= "<option ".$selected." value=\"".$rate_title."\">".$rate_title."</option>"; } $dropDownList.="</select>"; $output['RATESTITLE']=$dropDownList; } I am getting the following error when using composer: "continue" targeting switch is equivalent to "break". Did you mean to use "continue 2"? Just started. I am pretty sure composer.json wasn't changed. journalctl doesn't show anything. Maybe Doctrine related, however, nothing seems to help.
Any ideas? [michael@devserver www]$ php -v PHP 7.3.4 (cli) (built: Apr 2 2019 13:48:50) ( NTS ) Copyright (c) 1997-2018 The PHP Group Zend Engine v3.3.4, Copyright (c) 1998-2018 Zend Technologies with DBG v9.1.9, (C) 2000,2018, by Dmitri Dmitrienko [michael@devserver www]$ composer -V Composer version 1.4.2 2017-05-17 08:17:52 [michael@devserver www]$ yum info composer Loaded plugins: fastestmirror Loading mirror speeds from cached hostfile * base: mirror.us.oneandone.net * epel: mirror.rnet.missouri.edu * extras: mirror.us.oneandone.net * remi-php73: mirror.bebout.net * remi-safe: mirror.bebout.net * updates: mirror.us.oneandone.net Installed Packages Name : composer Arch : noarch Version : 1.8.4 Release : 1.el7 Size : 1.8 M Repo : installed From repo : epel Summary : Dependency Manager for PHP URL : https://getcomposer.org/ License : MIT Description : Composer helps you declare, manage and install dependencies of PHP projects, : ensuring you have the right stack everywhere. : : Documentation: https://getcomposer.org/doc/ [michael@devserver www]$ composer update Loading composer repositories with package information Updating dependencies (including require-dev) [ErrorException] "continue" targeting switch is equivalent to "break". Did you mean to use "continue 2"? update [--prefer-source] [--prefer-dist] [--dry-run] [--dev] [--no-dev] [--lock] [--no-custom-installers] [--no-autoloader] [--no-scripts] [--no-progress] [--no-suggest] [--with-dependencies] [-v|vv|vvv|--verbose] [-o|--optimize-autoloader] [-a|--classmap-authoritative] [--apcu-autoloader] [--ignore-platform-reqs] [--prefer-stable] [--prefer-lowest] [-i|--interactive] [--root-reqs] [--] [<packages>]... [michael@devserver www]$
I want to post some form identifyer when I hit submit, so that I can use this in an SQL query. Previously I used the below code and would retrieve either page1.php or page2.php etc from $_POST['subject']); Code: [Select] <td> <select name="subject" id="subject" onchange="gotourl(this.value)"> <option value="" >Please Select</option> <option value="page1.php" >page1</option> <option value="page2.php">page2</option> <option value="page3.php">page3</option> </select> </td> will this work? Code: [Select] <form id="form1" name="form1" value="tableName" method="post" action=""> $_POST['form1']) returns tableName...... Thanks I am attempting to use use two tables from the same database as described below. My db connection is made and is good. As part of a larger application I am creating a comments feature. Each row is a separate comment that was left by a user. I've no problem getting each row from the database and printing each out as I want formatted. The second table I am using contains all of the personal info of each member including their username, their displayed name, location and among other things the URL to an image they want associated with their account. Each row in the comments table contains the following id # of the comment, username of the person that left it, the date and of course the body of the message itself. Now I am trying to get the comment to show the image found at the URL in the second table (personal info). So for each row in the comments table I want it to pull the commentor's img_url from the second table. $result = mysql_query("SELECT * FROM page_comments WHERE page_id = '$id'"); if(mysql_num_rows($result)==0) { echo "There are currently no comments."; }else{ while($row = mysql_fetch_array($result)) { echo "<table>"; echo "<tr>"; echo "<td>"; echo "<img sro=\"$img_url\" width=50 height=50>"; echo "</td>"; echo "<td>"; echo "<a href=profile.php?member="; echo $row['commentor']; echo ">"; echo $row['commentor']; echo "</a>"; echo "</td>"; echo "<td>"; echo $row['date']; echo "</td>"; echo "</tr>"; echo "<tr>"; echo "<td cospan=3>"; echo $row['comment']; echo "</td>"; echo "</tr>"; echo "</table>"; echo "<br />"; } } As you can see I've thrown the "$img_url" in there as a placeholder for the time being. I've tried a couple of ideas and while none of them threw any errors it, of course, didn't return a value so the img was broken and the HTML source was just empty. Any and all ideas are greatly appreciated. Hi there, Does anyone know how to select from a mutlidimensional array, and put the results in a new array? I figured it out but it takes a long time. My array is huge (over 1000+ items) and this search is performed about 50 times before loading the page. My array is called $tasks and it looks like this Code: [Select] $tasks=Array( [0]=>array([id]=>"12"; [owner]=>"nancy"; [task]=>"clean the house"); [1]=>array([id]=>"23"; [owner]=>"toby"; [task]=>"do homework"); [2]=>array([id]=>"43"; [owner]=>"dan"; [task]=>"take out trash"); [3]=>array([id]=>"32"; [owner]=>"nancy"; [task]=>"cook dinner"); ) I want to be able to select from $tasks where owner="nancy", so i end up with this Code: [Select] $nancy_tasks=array( [0]=>array([id]=>"12"; [owner]=>"nancy"; [task]=>"clean the house"); [1]=>array([id]=>"32"; [owner]=>"nancy"; [task]=>"cook dinner");) )Thank you!!! This topic has been moved to Linux. http://www.phpfreaks.com/forums/index.php?topic=318175.0 Why is this not working? I can not seem to incorporate and into the statement $query = "SELECT * FROM Project_Registrations WHERE (First_Name = '$_POST[name_1]' AND Teacher = '$_POST[teacher]' AND School = '$_POST[school]') OR (Second_Name = '$_POST[name_1]' AND Teacher = '$_POST[teacher]' AND School = '$_POST[school]') OR (First_Name = '$_POST[name_2]' AND Teacher = '$_POST[teacher]' AND School = '$_POST[school]') OR (Second_Name = '$_POST[name_2]' AND Teacher = '$_POST[teacher]' AND School = '$_POST[school]') "; Thanks Ok, I am not sure if I can explain it correctly, but here it goes. I have a form where user is asked to enter up to three name. Let's say that the fields are called [name1], [name2] and [name3] and they are called the came in the table. Now I have a code which allows me to compare what names are in the database versus the name that was entered by the user. $query = "SELECT * FROM Project_Registrations WHERE name1= '$_POST[name1]' "; $result2 = mysql_query($query); if (mysql_numrows($result2) > 0) { while($row = mysql_fetch_array($result2)) {echo" ...error code... ";} } else { ...ok code... }; This works well for one name field but I have three in the form and three in the table. How do I define it so all three name fields in the form are compared to all three in the table and if just on matches I can give them the ...error code...? Oh, and they can enter up to three names in the form, but it could be two or one. Can I code it such: $query = "SELECT * FROM Project_Registrations WHERE name1= '$_POST[name1]' OR name1= '$_POST[name2] OR name1= '$_POST[name2]' OR ...'"; Thanks Hi, I currently have an if, elseif, else program that starts off with Code: [Select] if( $v_name == "" || $v_msg == "" ) echo "something" how do I turn this into a switch? but the very next elseif I have Code: [Select] elseif( strcspn( $_REQUEST['msg'], '0123456789' ) == strlen( $_REQUEST['msg'] ) ) echo "something" is it possible to turn this into a switch with 2 different strings like that to evaluate? This topic has been moved to PHP Installation & Configuration. http://www.phpfreaks.com/forums/index.php?topic=347922.0 |
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