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PHP - Relational List Submit Values Instead Of Text In Querry
I have a form where I have inserted 7 pre-populated relational lists. All of the information is pulling correctly from the databases, but when it posts, it's posting the value "ids" instead of the chosen text.
The files a www.kcwell.com/gcc_form.php and www.kcwell.com/gccsuccess_form.php How do I set up a query to obtain the data that I need? Help! 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> 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 . Hello!How can I echo this query? Code: [Select] SELECT i.internareID, CONCAT_WS( ' ', p.Nume, p.Prenume ) , p.cnp, m.nume, d.nume, ca.nume FROM internari i INNER JOIN medici m ON m.medicID = i.medicID INNER JOIN pacienti p ON p.pacientID = i.pacientID LEFT JOIN diagnostic d ON d.diagnosticID = i.diagnosticID LEFT JOIN case_asigurari ca ON ca.casaID = i.casaID ORDER BY `internareID` DESC I am trying to create seperate tables from this. Maybe a div class will work. I want each cat variable 'catname' to be in seperate tables at the top of each table with the subcats for that cat in 1 or 2 rows under the cat. I want to be able to also choose whether to use 1 or 2 table rows for the subcats under the cat. any ideas? Will appreciate your help. // Categories $sql = "SELECT catid, catname AS catname FROM $t_cats WHERE enabled = '1' $sortcatsql"; $rescats = mysql_query($sql) or die(mysql_error()); $catcount = @mysql_num_rows($rescats); $percol_short = floor($catcount/$dir_cols); $percol_long = $percol_short+1; $longcols = $catcount%$dir_cols; $i = 0; $j = 0; $col = 0; $thiscolcats = 0; while($rowcat=mysql_fetch_array($rescats)) { if ($j >= $thiscolcats) { $col++; $thiscolcats = ($col > $longcols) ? $percol_short : $percol_long; $j = 0; echo "<td valign=\"top\" width=\"$cell_width%\">"; } $i++; $j++; $catlink = buildURL("ads", array($xcityid, $rowcat['catid'], $rowcat['catname'])); $adcount = 0+$catadcounts[$rowcat['catid']]; ?> <table border="0" cellspacing="0" cellpadding="0" width="100%" class="dir_cat"> <tr> <th width="25" valign="top"><img src="images/category.gif" border="0" align="absmiddle"></th> <th><a href="<?php echo $catlink; ?>"><?php echo $rowcat['catname']; ?></a> <?php if($show_cat_adcount) { ?><span class="count">(<?php echo $adcount; ?>)</span><?php } ?> </th> </tr> <?php $sql = "SELECT scat.subcatid, scat.subcatname AS subcatname FROM $t_subcats scat WHERE scat.catid = $rowcat[catid] AND scat.enabled = '1' $sortsubcatsql"; $ressubcats = mysql_query($sql) or die(mysql_error()."<br>$sql"); $subcatcount = mysql_num_rows($ressubcats); while ($rowsubcat = mysql_fetch_array($ressubcats)) { if ($shortcut_categories && $subcatcount == 1 && $rowsubcat['subcatname'] == $rowcat['catname']) { continue; } $adcount = 0+$subcatadcounts[$rowsubcat['subcatid']]; $subcat_url = buildURL("ads", array($xcityid, $rowcat['catid'], $rowcat['catname'], $rowsubcat['subcatid'], $rowsubcat['subcatname'])); ?> <tr> <td> </td> <td> <a href="<?php echo $subcat_url; ?>"><?php echo $rowsubcat['subcatname']; ?></a> <?php if($show_subcat_adcount) { ?><span class="count">(<?php echo $adcount; ?>)</span><?php } ?> <br> </td> </tr> <?php } ?> </table> <br> <?php if($j==$thiscolcats || $i==$catcount) echo "</td>"; } ?> </tr></table> Hi Guys I have two table in mysql one is users and the second one is profiles my database name is movies users contains: users_id, username, password, email movies contains: id, user_id, name, movie_name, movielink I need users to login and then update their profile according to relational database setup so once user loggs in they can add the link of their movies to the database. Could some one help me with inserting data into relational database according to database setup I have introduced above? Appreciate your help in advance Hi Everyone, Well I have a bit of a problem... I have created a search page that I would like to return results from my MySQL database. However for some reason I can not get my page to display the results. Instead I am getting an error page. Below is the code I am using: In addition to this, should I wish for a user to be able to edit the returned results by clicking a link, how would I do that? <?php session_start(); ?> <link rel="stylesheet" type="text/css" href="css/layout.css"/> <html> <?php $record = $_POST['record']; echo "<p>Search results for: $record<br><BR>"; $host = "localhost"; $login_name = "root"; $password = "P@ssword"; //Connecting to MYSQL MySQL_connect("$host","$login_name","$password"); //Select the database we want to use mysql_select_db("schedules_2010") or die("Could not find database"); $result = mysql_query("SELECT * FROM schedule_september_2010 WHERE champ LIKE '%$record%' ") or die(mysql_error()); // keeps getting the next row until there are no more to get while($row = mysql_fetch_array( $result )) { // Print out the contents of each row echo "<br><p>Your Schedule<BR></p><br>"; echo "<table border=1>\n"; echo "<tr> <td bgcolor=#444444 align=center><p><b>Champ</p></td> <td bgcolor=#444444 align=center><p><b>Date</p></td> <td bgcolor=#444444 align=center><p><b>Start Time</p></td> <td bgcolor=#444444 align=center><p><b>End Time</p></td> <td bgcolor=#444444 align=center><p><b>Department</p></td> <td bgcolor=#444444 align=center><p><b>First Break</p></td> <td bgcolor=#444444 align=center><p><b>Second Break</p></td> <td bgcolor=#444444 align=center><p><b>Login ID</p></td> </tr>\n"; do { printf("<tr> <td><p>%s</p></td> <td><p>%s</p></td> <td><p>%s</p></td> <td><p>%s</p></td> <td><p>%s</p></td> <td><p>%s</p></td> <td><p>%s</p></td> <td><p>%s</p></td> </tr>\n" , $row["champ"] , $row["date_time"] , $row["start_time"] , $row["end_time"] , $row["department"] , $row["first_break"] , $row["second_break"] , $row["login_id"] ); } while ($row = mysql_fetch_array($result)); echo "</table>\n"; } else { echo "$champ No Records Found"; } mysql_free_result($result); mysql_close($con); ?> </html> I've got a form that submits values to itself and the user input values are saved into the database in a table with an auto incrementing ID. After SUBMIT is clicked, it reloads the page and all the values are gone, I'm wondering how I can get it to keep the values so that the user can make a single change and resubmit again. I know I can save all the values into the session or in the URL or pass them through $_POST in an array but I'm wondering what is the proper way to do this. Thanks. I have a relational table call sales with prodcut_id and custmer_id tables. I also have a product and customer tables. products table with product_id as an auto increment and customer with custumer_id I want to join through the sales tables all t he fields of row 1 from tables customer and products and pull it at once. This is a member login script and I want to display the products by members. So far I have this query to display the products once the member is login in. $userid is the id of the customer coming from the $userid= $_SESSION['customer_id']; $mysqlSales="SELECT products.* FROM products JOIN sales ON (products.product_id = procuct_id ) WHERE sales.customer_id = '$userid'"; so far that statement is not working where should I have some type of incoherance with the english statement above expressing what I want the query to do. I am creating a simple social network, and i want the post visible only on its circle of friends but the problem is... let say user_a, user_b, user_c already registered and user_a and user_c connected/friends already and all their posts and comments are visible on their circle but when user_b write a post oh his wall, it's also visible to user_a and user_c which i dont want to happen. I dont know what was wrong on codes below. CREATE TABLE IF NOT EXISTS `user`( `uid` INT(11) AUTO_INCREMENT PRIMARY KEY NOT NULL, `uname` VARCHAR(25) NOT NULL, `pword` CHAR(60) NOT NULL, `fullname` VARCHAR(30) NOT NULL, INDEX(`uname`) ) Engine = InnoDB DEFAULT CHARSET=utf8 COLLATE utf8_general_ci; CREATE TABLE IF NOT EXISTS `friend`( `fid` INT(11) AUTO_INCREMENT PRIMARY KEY NOT NULL, `friend_id` INT(11) NOT NULL, `my_id` INT(11) NOT NULL, `stat` ENUM('0','1') NOT NULL, INDEX(`friend_id`, `my_id`), FOREIGN KEY(`friend_id`) REFERENCES `user`(`uid`) ON UPDATE CASCADE ON DELETE CASCADE ) Engine = InnoDB DEFAULT CHARSET=utf8 COLLATE utf8_general_ci; public function viewFriendIfExistOnTbl($uid) { $query = $this->mysqli->query("SELECT `friend_id` FROM `friend` WHERE `my_id` = '$uid' LIMIT 1"); if ($query->num_rows > 0) { return true; } } I am just new to relational algebra probably a pre-step before learning SQL queries. Can you help me make the expressions of relational algebra expression for each of the following queries. This is the table contained inside a bus driver database.
driver ( driver_id, driver_name, age, rating ); bus ( bus_id, bus_name, color); reserves ( driver_id, bus_id, date);a. Find the names of drivers who have reserved at least three busses. b. Find the names of drivers who have reserved all busses. c. Find the names of drivers who have reserved all busses called Shuttle. d. Find the IDs of drivers whose rating is better than some driver called Paul. I would be grateful if somebody can help me here. I'm simply trying to set up a form where, if when a user clicks 'Submit', and then 'Back', the values from the form are preserved. My question is, how do I preserve the values of drop down menus. The following is a snippet of my code: Code: [Select] <select name="dropdown_dept" id="dept_list"> <option value=0><?php echo "Please select one..."?></option> <?php $dropdown_dept = "select dept_name from departments"; $result_dept = $db_conn->query($dropdown_dept); if (!$result_dept) { echo '<p>Unable to get department data.</p>'; return false; } for($i=0; $i<$result_dept->num_rows; $i++) { $app_name_row = $result_dept -> fetch_array(); ?> <option><?php echo($app_name_row[0]); ?></option> <? } ?> </select> Above is where I have set up a drop down menu of departments. Given that code, how can I preserve the department name after a user clicks 'Submit'? I have a page that contains a List box containing a list of categories taken from a mysql database, 4 iframe elements and 4 submit button (code to follow)
When an item is selected from the List box the onchange event submits the page to jobs.php and loads it into the joblist iframe. Once the iframe is loaded and visible it makes the "Add Job To Category" submit button visible. This all works great
When I click the "Add Job To Category" Submit button it loads Jobnew.php into the jobed iframe this is working but I can not seem to figure out how to pass the selected item from the List box using this submit button
I hope I am clear in my question if not please advise any help is very much appreciated....
Jobtask.php
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Job Configuration</title>
<head>
</head>
<body>
<div style="position:absolute;left:5px;top:0px;width:1345px;height:50px;z-index:0;">
<img src="images/img0001.png" id="Shape1" alt="" style="border-width:0;width:1345px;height:50px;"></div>
<span style='position:fixed;left:525px;top:8px;font-family:Arial;font-size:32px;width:275px;'>Job Configuration</span>
<form action='index.html' target='' method='post'>
<div style="position:absolute;left:205px;top:55px;width:1345px;height:50px;z-index:0;">
<Input type="submit" action="index.html" Target="_top" method='post'Value="Main Menu">
</div>
</form>
<form style=position:fixed;left:200px;top:82px;>
<iframe name='joblist' id='joblist' style=position:fixed;left:200px;top:px;visibility:hidden;z-index:10 src='' height="475" width="1150" scrolling='yes' frameBorder='0' ></iframe>
<iframe name='job' id='job' style=absolute:fixed;left:200px;top:0px;visibility:hidden;z-index:0 src='' height="475" width="1150" scrolling='yes' frameBorder='0' ></iframe>
<iframe name='jobed' id='jobed' style=position:fixed;left:460px;top:150px;visibility:hidden;z-index:10 src='' height="217" width="350" scrolling='no' frameBorder='0' ></iframe>
<iframe name='newcat' id='newcat' style=position:fixed;left:460px;top:150px;visibility:hidden;z-index:10 src='' height="125" width="350" scrolling='no' frameBorder='0' ></iframe>
</form>
<?php
error_reporting(E_ALL);
include('dbcon/dbconnect.php');
$con=mysqli_connect($host,$user,$password,$db);
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if (isset($_POST['id'])) {
$insql= "INSERT INTO catagory (Catagory) VALUES ('$_REQUEST[newcat]')";
$con->query($insql);
}
$Catresult = mysqli_query($con, "SELECT CatID, Catagory FROM catagory");
echo "<div id='wb_CatSel' style='position:absolute;z-index:0;text-align:center;'bgcolor='#00b0e6';>";
echo "<span style='position:fixed;left:10px;top:82px;font-family:Arial;font-size:15px;background-color:#00b0e6;width:175px;'>Select Category</span></div>";
echo "<form action='jobs.php' target='joblist' style='position:fixed;left:10px;top:101px;'></td>";
echo "<select name='cat' size='11' style='width: 175px;' onchange='this.form.submit()'>";
while($row = mysqli_fetch_array($Catresult))
{
echo "<option value=\"".$row['CatID']."\">".$row['Catagory']."</option>\n ";
}
echo "</select>";
echo "</form>";
echo "<form action='Jobnew.php' target='jobed' method='post'>";
echo "<div style='position:absolute;left:10px;top:285px;width:1345px;height:50px;z-index:0;'>";
echo "<Input type='hidden' id='id1'><Input type='submit' id='sbtn' style='width:175px;visibility:hidden;' Value='Add Job To Category'></div></form>";
echo "<form action='Jobtask.php' target='' method='post'>";
echo "<div style='position:absolute;left:10px;top:310px;width:1345px;height:50px;z-index:0;'>";
echo "<Input type='submit' style='width:175px;' action='Jobtask.php' Target='_top' method='post' Value='Reset Form'></div></form>";
echo "<form action='newcat.php' target='newcat' method='post'>";
echo "<div style='position:absolute;left:10px;top:335px;width:1345px;height:50px;z-index:0;''>";
echo "<Input type='submit' style='width:175px;'' action='newcat.php' Target='newcat' method='post' Value='Add New Category'></div></form>";
?>
</body>
</html>
<?php
// Daniel URL duplicator.
// Input links list.
$linksList = "links.txt";
// How many times to duplicate the url?
$manyTimes = 1000;
// Read in the list.
for ($x = 0; $x <= $manyTimes; $x++)
{
$handle = fopen($linksList, "r");
$line = fgets($handle);
echo $line;
fclose($handle);
}
?>
Hey Guys,
I'm stuck on this simple bit of code lol what I'm trying to do is load in a list of urls:
site1.com
site2.com
site3.com
etc
For each site that is read, I'm trying to duplicate it X times, above would print to screen the same url 1000 times, then move onto the next print it 1000 times etc until the list is done (or how ever many times I select)
I can't think of the best way to do it!
any help would be appreciated guys!
Graham
Hi i'm new to php. I want to get all the values of dropdown list on another page wether selected or not. Hi all !
I wish to post the value of an auto submit dropdown value to another page, be redirected to that page, and use it on that page. I am unable to achieve this in php and so I tried as :-follows:-
dropdown.php
<form method="post" action = dropdownaction.php>
<select name="myselect" onchange="this.form.submit();">
<option>blue</option>
<option>red</option>
</select>
</form>
and now I need the equivalent of
dropdownaction.php
<?php if(isset($_POST(['myselect']))) echo " I am selected".$_POST['myselect']; ?>Please can someone tell me how I may retrieve the value of 'myselect' in the dropdownaction.php after being redirected to it. Thanks loads. Hi all, i want my list/menu field values to come from my database. how can i accomplish that? thanks i did Code: [Select] <select name="select"> <option value="0">--select below--</option> <option value="1">Me</option> <?php require_once '../konnect/konex.php'; $result = mysql_query("SELECT * FROM is_clients"); while($row = mysql_fetch_array($result)) { echo "<option value ='2'>".$row"['name']</option>"; echo "<br />"; } ?> </select> Hi all.
how can i make the values show like a list. I tried html line break "<br>" and php \n but all to no avail. It just show all the values in one straigth line.
example of what i want is for the values to appear like this:
1234567890
0987654345
4567890675
instead of :
1234567890 0987654345 4567890675
Thanks
<form data-abide method="post" action="">
<div>
<select name="">
<option value="name">
<?php
$stmt = $pdo->query("SELECT acct_num FROM table order by id desc");
while ( $row = $stmt->fetch(PDO::FETCH_ASSOC) ) {
echo $row['acct_num'];
}
?>
</option>
</select>
</div>
<div>
<label>New Password <small>required</small></label>
<input type="password" name="password" id="password" required>
<small class="error">New password is required and must be a string.</small>
</div>
<div>
<label>Confirm New Password <small>required</small></label>
<input type="password" name="password2" id="password2" required>
<small class="error">Password must match.</small>
</div>
<input name="submit" type="submit" class="button small" value="Change Password">
</form>
Edited by Mr-Chidi, 13 November 2014 - 01:34 AM. Hi everyone, I am having trouble passing/displaying the values inside of a selected list. I created a add/remove list using Jquery and I tried to display the values passed using foreach and for loops but it is still not working. The values I am trying to get are $existing_mID[$j], which is inside of the option value attribute. Please kindly let me know what should I do in order to get the values and I really appreciate your help.
<?php
$selected = $_POST['selectto'];
if(isset($selected))
{
echo "something in selected<br />";
for ($i=0;$i<count($selected);$i++)
echo "selected #1 : $selected[$i]";
foreach ($selected as $item)
echo "selected: item: $item";
}
?>
This is the formHi i have this drop down list for date which contain 3 selects DAY MONTH YEAR hwo can i make so that when update form select keeps the same value has before help please <?php $months = array('','January','February','March','April','May','June','July','August','September','October','November','December'); echo '<select name="month_of_birth">'; for ($i=1;$i<13;++$i) { echo '<option value="' . sprintf("%02d",$i) . '">' . $months[$i] . '</option>'; } echo '</select>'; echo '<select name="day_of_birth">'; for ($i=1;$i<32;++$i) { echo '<option value="' . sprintf("%02d",$i) . '">' . $i . '</option>'; } echo '</select>'; echo '<select name="year_of_birth">'; $year = date("Y"); for ($i = $year;$i > $year-50;$i--) { $s = ($i == $year)?' selected':''; echo '<option value="' . $i . '" ' . $s . '>' . $i . '</option>'; } echo '</select>'; ?> |
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