PHP - Select List Problem
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> 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> hirealimo.com.au/code1.php this works as i want it: Quote SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1 but apparelty selecting * is not a good thing???? but if I do this: Quote SELECT priceID, price FROM price INNER JOIN vehicle....etc it works but i lose the info from the vehicle table. but how do i make this work: Quote SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc so that i am specifiying which colums from which tables to query?? thanks nothing gets put into the select list 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. **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 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! 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... Hello all , here is another problem of my project. I need to create a textarea , drop down list and submit button . At first , I can type whatever I want in the textarea , but for certain part I can just choose the word I want from drop down list and click submit , then the word will appear in the textarea as my next word . But I have no idea how to make this works , is there any simple example for this function ? Thanks for any help provided . 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 have a series of steps a user goes through to add a product. I'm having a problem with the first going to second page. The first page is a form to input information which has a "add" php file that goes with it that basically Posts and Inserts the information into the table, then moves on to the next page. It also auto increments a new id for the product in the table during this process. The problem is I'm trying to get the product id to carry over to the next page so they can review the information they just input. Any ideas on how I can pull that new id by going from that one page to the next with that move (Form>Insert(form action)>EditPage)? I'm just having a hard time visualizing this because I'm inserting the new product then there's nothing to reference because I can't pinpoint on the id yet. i need help with this issue.i designed a comment page where people can comment on a topic but what i descovered is that once i comment on a topic it will be showed on the page i asked it to go but if i comment on thesame topic it will not show exceeptb the one that as been there before.even if i click on anotherr topic and i comment once i cannot comment again but if i remove the WHERE clause all the comments meant for other topics will just display which is not good.Please help me thanks. this is my code: Code: [Select] <?php include"header.php"; $topicid=$_GET['id']; ///the two tables POST and TOPICS share thesame id $sql="SELECT post_comments FROM post WHERE $topicid=topicsID"; $result=mysql_query($sql) or die(mysql_error()); while($row=mysql_fetch_array($result)) { echo"{$row['post_comments']}"; } ?> Hello everybody, I have a problem selecting rows from db. Copied error message: Quote Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\scriptforsite\index.php on line 223 I have searched for this problem, and I found tons of posts but I didn't do anything. Here is a code: Code: [Select] <?php include("include/functions.php"); db_connect(); lang_en(); lang_smiles(); lang_links(); $sql= mysql_query("SELECT msg_id,msg from messages order by msg_id desc"); while($row=mysql_fetch_array($sql)) { $msg=$row['msg']; $msg_id=$row['msg_id']; ?> <li class="bar<?php echo $msg_id; ?>"> <div align="left"> <span ><?php echo $msg; ?> </span> <span class="delete_button"><a href="#" id="<?php echo $msg_id; ?>" class="delete_update">X</a></span> </div> </li> <?php } db_close(); ?> This line 223 is this: $sql= mysql_query("SELECT msg_id,msg from messages order by msg_id desc"); ... I have tried to replace msg_id,msg with * but on this way I don't have dump from table... I have 5 entries in a table Code: [Select] $sql = "select count(distinct columnName) from table"; $result = mysql_result($sql); $count = mysql_fetch_array($result); echo $count[0];The output is 5 as expected. Code: [Select] $sql = "select distinct columnName from table"; $result = mysql_result($sql); $count = mysql_fetch_array($result); echo $count[0];the ouput is the first file name as expected, however Code: [Select] echo $count[1];gives undefined offset 1, which does not make any sense. Can anyone explain why the offset 1 is undefined if the count is 5? Hey guys,
My problem is, iam loading with mysql_fetch_array a table from my database and draw it into my website page.
The table is a user table which has an id, user, password and rights(Admin, Mod , User).
Only the rights are given out as <select> </select> boxes with the 3 options(Admin, Mod , User).
And i also have 1 more column which has for each row a href to a new site with the id of the array_index as paramter.
My Problem is i want to know which select box was changed and which option is selected so i can change the rights in my database of this user with the right id.
At the moment i cant post the code in here cause i have a problem with my laptop now.
I hope you can help me,
Thank you
Edited by Kandey, 30 October 2014 - 02:28 PM. Hi, i'm trying to select some entries from my DB, i don't quite understand what's the problem here, i'm using this: Code: [Select] //Customers Online $query = "SELECT COUNT(*) as Anzahl FROM customers WHERE status = 1"; $queryerg = mysql_query($query) OR die(mysql_error()); while($row = mysql_fetch_array($queryerg)){ $customers_online = $row[0]; } Which works fine, now i want to select by country and i'm doing this: Code: [Select] //Customers DE $query = "SELECT COUNT(*) as Anzahl FROM customers WHERE country = 'de'"; $queryerg = mysql_query($query) OR die(mysql_error()); while($row = mysql_fetch_array($queryerg)){ $customers_de = $row[0]; } Which doesn't work? If i print / echo $customers_de i just get a blank value, not even 0 or anything just blank. Any help ? I'm working on my website, and I'm having a bit of trouble in getting PHP to select the proper data. I'm trying to select usernames from my database where rank is equal to one (the highest, in my system). As such, I attempted this code; Code: [Select] // Connection info above this line... mysql_select_db("users"); $query = mysql_query ("SELECT displayname FROM login WHERE rank = 1"); $query_a = mysql_fetch_array($query); var_dump($query_a); The var_dump resulting from that is as follows; Code: [Select] array 0 => string 'Seltzer' (length=7) 'displayname' => string 'Seltzer' (length=7) Everything is working correctly, except for the fact that my database contains two displayname rows where rank is equal to one (EDIT: Two distinct rows). In fact, I can run a search of it in phpMyAdmin and get the two that my PHP code should be returning. phpMyAdmin generated the following query, which Inserted into my code in order to double-check things; Code: [Select] SELECT `displayname` FROM `login` WHERE `rank` =1 LIMIT 0 , 30 Even after I swapped my queries, the PHP code still returned the same var_dump as above. Complicating things further, I've noticed another function I've made, which queries "SELECT rank WHERE displayname = '$displayname'", functions perfectly. I've gotten rid of pretty much any source of the error I could think of; I'm testing the function on an otherwise empty page, I've removed any classes being used, and I've tried about a million different queries. Can someone help me out with this? I'm being held up by it and I'm sure that this is a simple fix. Thanks in advance, Dustin |