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PHP - Add A Default Select Option With Value "empty" This Dropdownlist
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; } 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[$key]);?>"></p> <?php } ?> <input name="Update" type="submit" value="Update"> </form> <?php } ?> </div> What I am looking for is that if any field in a database is empty, the form comes back and puts in the word "empty" into the web page, but not the database. I got this right now, it shows the first part fine, but the second part it doesn't show the "Empty" comment... I don't understand why... as far as I can tell the code is fine... Code: [Select] <?php //Nonconformity, Disposition, Comments and Comments & Additional Details echo '<div id="box3">'; if (!empty($row['Nonconformity']) || !empty($row['Disposition']) || !empty($row['Comments']) || !empty($row['CommentsAdditional_Details'])) { echo '<div id="non"><span class="b">Nonconformity: </span><br />' . $row['Nonconformity'] . '</div>'; echo '<div id="dis"><span class="b">Disposition: </span><br />' . $row['Disposition'] . '</div>'; echo '<div id="comm"><span class="b">Comments: </span><br />' . $row['Comments'] . '</div>'; echo '<div id="comma"><span class="b">Comments and/or Additional Details: </span><br />' . $row['CommentsAdditional_Details'] . '</div>';} else if (empty($row['Nonconformity']) || empty($row['Disposition']) || empty($row['Comments']) || empty($row['CommentsAdditional_Details'])) { echo '<div id="non"><span class="b">Nonconformity: </span><br />Empty</div>'; echo '<div id="dis"><span class="b">Disposition: <br /></span>Empty</div>'; echo '<div id="comm"><span class="b">Comments: <br /></span>Empty</div>'; echo '<div id="comma"><span class="b">Comments and/or Additional Details: </span><br />Empty</div>';} echo '</div>'; ?> 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 Is it possible to put a multiplication problem in the mysql "default value" field? In my script I need to take the unit amount * quantity, I know how to do this in php, but can I just put my equation in the "default value" field in MYSQL and make it equal to another field?... Like excell? Thanks Hello there, Could anyone tell me why you are not allowed to use "Default" as a function name in php? I searched around and couldn't get a straight answer. 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! Code: [Select] <?php if (isset($_POST['submit'])) { // Grab the selected value $optionValue = $_POST['subscriptionLevel']; // Grab the id of the selected value ** HOW DO I DO THIS? ** $optionID = ?????????????????????????; } ?> <html> <body> <form name="form" method="post" action=""> <table> <tr> <td> <label for="subscriptionLevel"></label> <select name="subscriptionLevel" id="subscriptionLevel"> <?php do { ?> <option id="<?php echo $dbRowResult['sub_level']; ?>" value="<?php echo $dbRowResult['sub_price']; ?>"><?php echo $dbRowResult['sub_desc']; ?></option> <?php } while ($dbRowResult = $db->dbGetRow($dbQueryResult)); ?> </select> </td> <td> <label> <input type="submit" name="submit" id="submit" value="submit"> </label> </td> </tr> </table> </form> </body> </html> How do I grab the id of the selected value? Thank you. Hello Everyone, I am having issues where people are submitting null characters on my website (I.e. %00). Currently, I check if their is a value by using the empty function, and I also am checking if the value after running the trim command is =="". How can I check for these special characters to make sure that whatever they are submitting actually contains some sort of legitimate value? Thanks! 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']; Hi guys , can someone help me with this code its giving me "query was empty Error" its connecting but...? and please find attached the entire php page for clarity. Thanks . <?php //connect database require_once "connect_to_mysqli.php"; //query the datbase $sqlcommand = "SELECT post_id, title, body From posts ORDER by post_id desc"; $query = mysql_query($sqlCommand) or die (mysql_error()); while($row = mysql_fetch_array($query)) { $title = $row["title"]; echo "<h3>$title</h3>"; } // Free the result set if it is large mysql_free_result($query); // close mysql connection mysql_close(); ?> 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. 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 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!!! 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 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 |
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