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PHP - Php Array Into A Select List
nothing gets put into the select list
Similar TutorialsHi, In my mysql database i have a text input option, in the registration form and edit my details form i have a multiple select dropdown list, which user selects options to populate the text input box, which ultimately populates the text field in the mysql database. All works perfectly. The dropdownlist consists of 3 parts <optgroups> first is current selection (what is the usesr current selection)works fine, The second <optgroup> is existing words, what words we(the site) have given as options, and the third <optgroup> is the words that others have used. This is where im having a small problem. Because its a text field when i call the data from the database, it calls the entire text box as a single option in my select list.. I want to break the words in the text field (at the comma) and have them listed each one as an option in the select list. Example what i need: Words in text box:(my input allows the "comma") word1, word2, word3, word4, word5, word6, How i want them called/displayed: <option value=\"word1\">word1</option> <option value=\"word2\">word2</option> <option value=\"word3\">word3</option> <option value=\"word4\">word4</option> <option value=\"word5\">word5</option> <option value=\"word6\">word6</option> here's my code: $query = "SELECT allwords FROM #__functions_experience WHERE profile_id = '".(int)$profileId."' LIMIT 1"; $original_functionsexperience =doSelectSql($query,1); $query = "SELECT allwords FROM #__functions_experience WHERE profile_id = '".(int)$profileId."' LIMIT 1"; $functionsexperiencelist=doSelectSql($query); $funcexpList ="<select multiple=\"multiple\" onchange=\"setFunctionsexperience(this.options)\">"; foreach ($functionsexperiencelist as $functionsexperienceal) { $selected=""; if ($functionsexperienceals->allwords == $original_functionsexperience) $selected=' selected="selected"'; $allwords=$functionsexperienceal->allwords; $funcexpList .= "<optgroup label=\"Current selection\"> <option value=\"".$allwords."\" ".$selected." >".$allwords."</option> </optgroup> <optgroup label=\"Existing Words\"> <option value=\"existing1,\">existing1</option> <option value=\"existing2,\">existing2</option> <option value=\"existing3,\">existing3</option> <option value=\"existing4,\">existing4</option> <option value=\"existing5,\">existing5</option> <option value=\"existing6,\">existing6</option> </optgroup> <optgroup label=\"Others added\"> //heres problem <option value=\"".$allwordsgeneral."\">".$allwordsgeneral."</option> </optgroup>"; } $funcexpList.="</select>"; $output['FUNCEXPLIST']=$funcexpList; The result im getting for optgroup others added: word1, word2, word3, word4, word5, how can i get it like this: <option value=\"word1\">word1</option> <option value=\"word2\">word2</option> <option value=\"word3\">word3</option> <option value=\"word4\">word4</option> <option value=\"word5\">word5</option> <option value=\"word6\">word6</option> Hi, I use the following code to create a select menu from an array of options stored in LISTS.php: include 'LISTS.php'; print('<select id="from" name="from">'); foreach ($langList as $lang) {printf('<option %s>%s</option>', ($from1 == $lang ? 'selected="selected"' : ''), $lang); } echo '</select>'; where LISTS.php includes the following: $langList = array(' ','English', 'French', 'German', 'Dutch', 'Spanish'); This works great, but now I want to do something similar with a checkbox list, where each checkbox has an associated 'onchange' javascript function and I'm getting pretty stuck. My checkbox list is of the following form: Code: [Select] <html> <ul style="height: 95px; overflow: auto; width: 200px; border: 1px solid #480091; list-style-type: none; margin: 0; padding: 0;"> <li id="li1b"><label for="chk1b"><input name="chk1b" id="chk1b" type="checkbox" onchange="function1('chk1b','li1b')">Option1</label></li> <li id="li2b"><label for="chk2b"><input name="chk2b" id="chk2b" type="checkbox" onchange="function1('chk2b','li2b')">Option2</label></li> //etc. </ul> </html> What I want to do is have 'Option1', 'Option2', etc. stored in an array in LISTS.php and have a PHP script that populates the checkbox list accordingly, in a similar manner to my select menu above. I can't work out how to get the ID of the next <li> and the next <input> in the list to go up by one each time, e.g. 'li1b' then 'li2b', 'li3b', etc. Could someone pls help me out? Thanks! I have the following code currently: Code: [Select] <?php foreach ((array)$node->field_buy_at as $item) { ?> <?php print $item['view'] ?> <?php } ?> I would like to make the list a drop down with a link so that when a user selects, he goes to a new page. I tried the following: Code: [Select] <select name="select"> <?php foreach ((array)$node->field_buy_at as $item) { ?> <?php $url = $node->field_buy_at[0]['url']; $store = $item['view']; ?> <? echo "<option value='$url'>$store</option>";?> <?php } ?> </select> I'm pretty sure it's this "$url = $node->field_buy_at[0]['url'];" that I don't have correct. I want to make a list like this: I store the genres in a array like this: Code: [Select] $genre_list = array( array("ACTION", "First Person Shooter"), array("ACTION", "Third Person Shooter"), array("ACTION", "Tactical Shooter"), array("ACTION", "Fighting"), array("ACTION", "Arcade"), array("ADVENTURE", "Adventure"), array("ADVENTURE", "Platformer"), array("ADVENTURE", "Point and Click") ); This is my first attempt to only show the subgenre, i only get a blank list (apart from this <option value="genre">Genre</option> ). Code: [Select] <div class="forminput"> <select id="genres" name="genres"> <option value="genre">Genre</option> <?php for($i=0; $i<count($genre_list[0]); $i++) { ?> <option value="<?=$i?>"<?=$genre_list[0][$i]?></option> <?php } ?> </select> </div> **Disclaimer: It's been a while since I last wrote any code. The quality of my code is likely to be sub-par. You've been warned.** I have a basic form that's meant to search flat files on our server. The "search engine" I created as two select lists: one for the file names and one for the customer site files come from. For a reason I can't figure out, whatever option I select from the second select list is never captured when I hit Submit. However, whatever option I select from the first select list is always captured. What am I missing? I am sure it's starting right at me.... Any hints welcome. Thank you. Here's my code: Code: [Select] <HTML> <head><title>SEARCH TOOL - PROTOTYPE</title></head> <body><h1>SEARCH TOOL - PROTOTYPE</h1> <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <fieldset> <legend>Filename (one item)</legend><select name="DBFilename" id="DBFilename" size="0"> <?php $con = mysql_connect("localhost", "user", "pass"); if (!$con) { die('Could not connect: ' . mysql_error());} mysql_select_db("dev", $con) or die(mysql_error()); $result = mysql_query("select distinct filename from search_test"); while ($row = mysql_fetch_array($result)) { ?> <option value="<?php echo $row['filename']; ?>"><?php echo $row['filename']; ?></option> <?php } mysql_close($con); ?> </select></fieldset> <fieldset> <legend>Site (one item)</legend><select name="DBSite" id="DBSite"> <?php $con = mysql_connect("localhost", "user", "pass"); if (!$con) { die('Could not connect: ' . mysql_error());} mysql_select_db("dev", $con) or die(mysql_error()); $result = mysql_query("select distinct site from search_test"); while ($row = mysql_fetch_array($result)) { ?> <option value="<?php echo $row['site']; ?>"><?php echo $row['site']; ?></option> <?php } mysql_close($con); ?> </select></fieldset> <input type="submit" name="submit" value="submit" > <input type="button" value="Reset Form" onClick="this.form.reset();return false;" /> </form> </body> </HTML> <?php if (isset($_POST['submit'])) { if (!empty($_POST['DBFilename'])) {doFileSearch();} elseif (!empty($_POST['DBSite'])) {doSite();} } function doFileSearch() { $mydir = $_SERVER['DOCUMENT_ROOT'] . "/filedepot"; $dir = opendir($mydir); $DBFilename = $_POST['DBFilename']; $con = mysql_connect("localhost", "user", "pass"); if (!$con) {die('Could not connect: ' . mysql_error());} mysql_select_db("dev", $con) or die("Couldn't select the database."); $getfilename = mysql_query("select filename from search_test where filename='" . $DBFilename . "'") or die(mysql_error()); echo "<table><tbody><tr><td>Results.</td></tr>"; while ($row = mysql_fetch_array($getfilename)) { $filename = $row['filename']; echo '<tr><td><a href="' . basename($mydir) . '/' . $filename . '" target="_blank">' . $filename . '</a></td></tr>'; } echo "</table></body>"; } function doSite() { $mydir = $_SERVER['DOCUMENT_ROOT'] . "/filedepot"; $dir = opendir($mydir); $DBSite = $_POST['DBSite']; $con = mysql_connect("localhost", "user", "pass"); if (!$con) {die('Could not connect: ' . mysql_error());} mysql_select_db("dev", $con) or die("Couldn't select the database."); $getfilename = mysql_query("select distinct filename from search_test where site='" . $DBSite . "'") or die(mysql_error()); echo "<table><tbody><tr><td>Results.</td></tr>"; while ($row = mysql_fetch_array($getfilename)) { $filename = $row['filename']; echo '<tr><td><a href="' . basename($mydir) . '/' . $filename . '" target="_blank">' . $filename . '</a></td></tr>'; } echo "</table></body>"; } ?> Okay, really newbie question, but for this code... Code: [Select] <!-- Gender --> <label for="gender">Gender:</label> <select id="gender" name="gender"> <option value="">--</option> <option value="F">Female</option> <option value="M">Male</option> </select> 1.) How do I assign a variable to this? 2.) How do I make this "sticky"? Here is how I have usually done other form types... Code: [Select] <!-- First Name --> <label for="firstName">First Name:</label> <input id="firstName" name="firstName" type="text" maxlength="30" value="<?php if(isset($firstName)){echo htmlentities($firstName, ENT_QUOTES);} ?>" /><!-- Sticky Field --> <?php if (!empty($errors['firstName'])){ echo '<span class="error">' . $errors['firstName'] . '</span>'; } ?> Oh, by the way, at the top of my PHP file I have this code... Code: [Select] if ($_SERVER['REQUEST_METHOD']=='POST'){ // Form was Submitted (Post). // Initialize Errors Array. $errors = array(); // Trim all form data. $trimmed = array_map('trim', $_POST); Thanks, Debbie could anyone please help me with the code which is i have already displayed data as a multi select list but now i need to select one or more from them and insert into another database table. would be appreciate your help. thanx Hi, I'm a php newbie, with some mysql experience. I have a mysql database as follows: Database=watch, Table=events - fields id, reportno, sdate, comments What I need is: 1. A dropdown list to display reportno from mysql database. 2. Depending on which reportno I choose, I'd like to open a popup(or separate) page to display the stored information. Tks in advance for any help Hi. I will try and explain simply.
I have a list of item codes eg.
size1 Acolor1
size1 Bcolor3
size1 Bcolor1
size1 Acolor3
size1 Acolor4
size2 Acolor1
size3 Bcolor1
size3 Bcolor2
size3 Bcolor2
size3 Acolor1
I want to take these and split them into parts:
size1 Acolor2
size1 Bcolor3
etc.
Then I want to create a form with 2 dropdowns. The first option would just have all unique sizes so:
size1
size2
size3
Then when one is selected, option 2 would show the colours available in that size.
Is there a standard way of doing this? Can I use PHP to create the array and then jquery to display and hide the options?
I'm totally stuck and i'm not just looking for somebody to code it all for me. any help would be greatly appreciated
thanks
Sam
I have a search function in php where the text characters are matched to characters in a tables field--- works perfectly.... I need to make the input box have a droplist of words from database, this is also easy for me to do. the problem here is there is no definitive value! the options list always outputs a blank in the url--- its supposed to search a matching value and then output the matching value to url... Here is the droplist code: $output['RATESTITLE']='<input class="inputbox" type="text" size="24px" name="ratestitle" value="'.$sch->filter['ratestitle'].'" onfocus="if (this.value ==\''.$output['LANGUAGE_SEARCH_RATESTITLE'].'\') {this.value = \'\'}" />'; this outputs a input text box--- i want to have a droplist of options to populate this text box... If you must know this is the third day im at it... I need some help with the query for a form that selects values across multiple columns, and allow users to select multiple values in several columns. http://brinsterinc.com/tpa/tpasearch.php I assume you build the where clause for the sql by determining if there is an option value has been selected in the single-value select boxes. But how do I handle the multi-select list boxes? I'm desperate for help! Need to get somewhere with this over the weekend! TIA! How can I select a value in an array but not its key. In witch conditions you need to transfer the data from a select in array and after that to show from array in content of the page ? Is it possible to achieve a similar thing to this... SELECT FROM db WHERE value!=$array My example code is this: $browser_hide = array("1" => "YandexBot 0.0 for unknown"); $query = "SELECT * FROM hitlist_hits WHERE month='$month' AND browser!='$browser_hide' ORDER by id DESC"; Thanks Guys. Hi guyz, please help me I want to get an id from my database, I use a word to search it which is stored as keys of array. But all the keys can't be found in my database, even though it's there. Here some parts of my code Code: [Select] $arrKeys = array_keys($arrResult); foreach($arrKeys as $key) { //if($arrResult[$key]>1) { echo $key/*."=>". $arrResult[$key]*/."<br>"; //$q = mysql_query("select id_katadasar from tb_katadasar where katadasar='".$key."'"); $kon->query("select id_katadasar from tb_katadasar where katadasar='".$key."'"); var_dump($kon->query); //echo "select id_katadasar from tb_katadasar where katadasar='".$key."'"; //$jum = $kon->getJumlah(); //$row = mysql_fetch_array($q, MYSQL_ASSOC) or die(mysql_error() . ''. $q); //echo count($row); if($row = $kon->tampilkan()) //if($row = mysql_fetch_array($q, MYSQL_NUM)) { //$res = $kon->tampilkan(); echo $row[0]; } //else { //echo "tidak ada di db <br><br>"; } } } $kon->query is equal to mysql_query() $kon->tampilkan return $row=mysql_fetch_array(query) Thanks I have a Prepared Statement that runs a SELECT statement and returns 2 records, and I would like to store the Field Value for each Record into an Array. Here is the code I usually use for queries that return just a single value... // ****************** // Populate Form. * // ****************** // Build query. $q2 = "SELECT response FROM bio_answer WHERE member_id=?"; // Prepare statement. $stmt2 = mysqli_prepare($dbc, $q2); // Bind variable to query. mysqli_stmt_bind_param($stmt2, 'i', $memberID); // Execute query. mysqli_stmt_execute($stmt2); // Store results. mysqli_stmt_store_result($stmt2); // Check # of Records Returned. if (mysqli_stmt_num_rows($stmt2)>0){ // Details Found. // Bind result-set to variable. mysqli_stmt_bind_result($stmt2, $response); // Fetch record. mysqli_stmt_fetch($stmt2); // Close prepared statement. mysqli_stmt_close($stmt2); }else{ // Details Not Found. $_SESSION['resultsCode'] = 'DETAILS_NOT_FOUND_2133'; // Close prepared statement. mysqli_stmt_close($stmt2); // Set Error Source. $_SESSION['errorPage'] = $_SERVER['SCRIPT_NAME']; // Redirect to Display Outcome. header("Location: " . BASE_URL . "/members/results.php"); // End script. exit(); }//End of POPULATE FORM Can someone help me out with the syntax so that I get an end result like this... $answerArray[0] = 'I want to be my own boss!!' $answerArray[1] = 'Don't waste your time trying to do your own Taxes!' Thanks, Debbie Hi I need to do a SELECT query like $query = "SELECT name FROM myusers WHERE myidnum = ".$_SESSION['myid'].""; and put the results into an array like $myusers = array("David","John","Lucy","Sarah"); But I cannot figure out how to do this Can anyone help? Thanks hi i have this function for select / function select($table, $rows = '*', $where = null, $group = null, $order = null, $limit = null) this is what i do to display the records. $db->select('loan'); $records = $db->getResult(); foreach($records as $row) { $user = $row['id']; //$name = $row['firstname']; //print_r($row); echo $user; ?> the codes is running.but when i want to use WHERE id = $_POST['id']; this is what im doing. $db->select('member')->where(array('id' => $_POST['client'])); an error as occured where in function where is not existing. please help how to use the proper way of using select * $table where id=$id; in array. thanks hello, have a question. im trying to do a select * where id exists in array $id8 here is my array Code: [Select] $result8 = mysql_query("SELECT * FROM customers WHERE whosclient = 'Lansoft'"); while($row8 = mysql_fetch_array($result8)) { $id8=$row8['id']; } here is my select code. it is only returning results for the highest number in the array Code: [Select] $result = mysql_query("SELECT * FROM timesheets WHERE status = 3 AND billable = 1 AND client = '$id8' ORDER BY date, starttime"); while($row = mysql_fetch_array($result)) { Thanks for any help! I am writing a CRON job that will execute daily. First it will identify from a MySql table the date in a field 'FAPforSale_repost35' If the date is the today date it will then execute commands to delete photo images in a directory, delete the directory, and finally remove the record from the database.
I am on step one which is to build the array of records that match the days date. When I run the code, there are no errors but I am not getting results even though the records in the test table are set for today. Below is the select
<?php
define( "DIR", "../zabp_employee_benefits_processor_filesSm/", true );
require( '../zipconfig.php' );
require( DIR . 'lib/db.class.php' );
require_once( $_SERVER['DOCUMENT_ROOT'] . '/_ZABP_merchants/configRecognition.php' );
require_once( $_SERVER['DOCUMENT_ROOT'] . '/_ZABP_merchants/libRecognition/MailClass.inc' );
$todayRepost35 = date("Y-m-d");
echo $todayRepost35;
function repostEndSelect()
{
global $db;
$this->db = $db;
$data = $this->db->searchQuery( "SELECT `FAPforSale_IDnumber`, `FAPforSale_image1`, `FAPforSale_image2`, `FAPforSale_image3`, `FAPforSale_repost35` FROM `FAP_forSaleTest` Where `FAPforSale_repost35` = '$todayRepost35' ");
$this->FAPforSale_IDnumber = $data[0]['FAPforSale_IDnumber'];
$this->FAPforSale_image1 = $data[0]['FAPforSale_image1'];
$this->FAPforSale_image2 = $data[0]['FAPforSale_image2'];
$this->FAPforSale_image3 = $data[0]['FAPforSale_image3'];
$this->FAPforSale_repost35 = $data[0]['FAPforSale_repost35'];
echo $this->FAPforSale_IDnumber;
echo $this->FAPforSale_image1;
echo $this->FAPforSale_image2;
echo $this->FAPforSale_image3;
echo $this->FAPforSale_repost35;
} // ends function...
echo( ' Finished...' );
?>
Thanks in advance for any suggestions or direction. Chapter two will be when I start testing the commands to delete.
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