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PHP - Execute Query From Listbox
Hi,
Im starting to write a simple (so I thought! ) project at work, I have no problems with the HTML, CSS or even the php mysql setup, but I am struggling with a simple piece of code! I have a list box, that populates from a table column using the following code: Code: [Select] <?php function select(){ $query="SELECT appname FROM kpe_apps"; $result = mysql_query($query); echo '<select name="item" onchange="this.form.submit()">'; while($nt=mysql_fetch_array($result)){ echo '<option value="' . $nt['id'] . '">' . $nt['appname'] . '</option>'; } echo '</select>'; } ?> I then called the function on the page I needed. which works fine, now all I need is whatever is selected in the listbox other columns in the same table relating to the selected item are seen. I just cant seem to do it, loads of people are using jquery and other code, surely this can be done in php? any help would really be appreciated.. Many Thanks MOD EDIT: code tags fixed, linefeeds and indenting added. Similar TutorialsUnable to execute query (SELECT * FROM lb-players) in the database : 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 '-players' at line 1 I don't know why i am getting this error. All i can think of that is taking away lb from "lb-players" when you add a " - "? Code: [Select] [m]<?php $conn = mysql_connect("23.23.23.23", "unknown", "itsAsecertxD");//ignore this xD if (!$conn) { echo "Unable to connect to the database : " . mysql_error(); exit; } if (!mysql_select_db("minecraft-rc")) { echo "Unable to select database mydbname : " . mysql_error(); exit; } $sql = 'SELECT * FROM lb-players'; $result = mysql_query($sql); if (!$result) { echo "Unable to execute query ($sql) in the database : " . mysql_error(); exit; } if (mysql_num_rows($result) == 0) { echo "No rows found, nothing to display."; exit; } while ($row = mysql_fetch_assoc($result)) { echo $row["playername"]; } mysql_free_result($result); ?>[/m] tell me if i posted this in the wrong place its been forever since i last been on this site :$ Hi. I have a query that determines if user is online/offline. Works perfectly when page loads. I want to change to auto run query every X seconds. NOTES: - The query is on a .php page with multiple other queries. -IF it is possible to still run the query with-in this page? or make a separate file and include in an iframe? i would like to avoid the iframe and just $output the result of query every X seconds. Here is my $query code: $query = "SELECT `manager_id` FROM #__profiles_xref WHERE profileid = '$profid'"; $onofflineid=doSelectSql($query,1); foreach ($onofflineid as $isonoffline){ $profmanagerid=$isonoffline->manager_id; } $query = "SELECT `session_id` FROM table_session WHERE userid = '$propmanagerid'"; $onofflinestatus=doSelectSql($query,1); if (count($onofflinestatus)>0) { $output['PROPSTATUS']='Online'; } else { $output['PROPSTATUS']='Offline'; } Hi all ! The following code has two similar queries. Query 1 and Query 2. While Query 1 works just fine, Query 2 simply fails persistently. I have not been able to figure out why?
I have checked every single variable and field names but found no discrepancy of any sort or any mismatch of spelling with the fields in the database.
I would be grateful if anybody can point out what is preventing the second query from returning a valid mysqli object as the first one does. Here is the code for the two.
<?php
$db_host = "localhost";
$db_user ="root";
$db_pass ="";
$db_database ="allscores";
//////////////// CHECKING VARIABLE FILEDS IN A UPDATE QUERY //////////////////////////////////
//////////////// QUERY 1 /////////////////////////////////////////////////////////////////////
/*
$db_database ="test";
$db_table ="users";
$RecNo = 1;
$field1 = 'name';
$field2 = 'password';
$field3 = 'email';
$field4 = 'id';
$val1 = "Ajay";
$val2 = "howzatt";
$val3 = "me@mymail.com";
$con = mysqli_connect($db_host,$db_user,$db_pass,$db_database) or die('Unable to establish a DB connection');
$query = "UPDATE $db_table SET $field1 = ?,
$field2 = ?, $field3 = ?
WHERE $field4 = ? "; // session terminated by setting the sessionstatus as 1
$stmt = $con->prepare($query);
var_dump($stmt);
$stmt->bind_param('sssi',$val1,$val2,$val3,$RecNo);
if($stmt->execute())
{
$count = $stmt->affected_rows;
echo $count;
}
//////////////// QUERY 1 END /////////////////////////////////////////////////////////////////////
*/
//////////////// QUERY 2 /////////////////////////////////////////////////////////////////////
$con = mysqli_connect($db_host,$db_user,$db_pass,$db_database) or die('Unable to establish a DB connection');
$table = 'scores';
$date = date('Y-m-d H:i:s');
/*
$prestr = "Mrt_M_";
$STATUS = $prestr."Status";
$S_R1 = $prestr."Scr_1";
$S_R2 = $prestr."Scr_2";
$PCT = $prestr."PPT";
$DPM = $prestr."DSP";
$TIMETAKEN = $prestr."TimeTaken";
*/
$STATUS = "Mrt_M_Status";
$S_R1 = "Mrt_M_Scr_1";
$S_R2 = "Mrt_M_Scr_2";
$PPT = "Mrt_M_PPT";
$DSP = "Mrt_M_DSP";
$TIMETAKEN = "Mrt_M_TimeTaken";
$TimeOfLogin = $date;
$no_of_logins = 10;
$time_of_nLogin = $date;
$m_scr_row1 = 5;
$m_scr_row2 = 5;
$m_ppt = 20;
$m_dsp = 60;
$m_time = 120;
$date = $date;
$RecNo = 24;
$query = "UPDATE $table SET
TimeOfLogin = ?,
no_of_logins = ?,
time_of_nLogin = ?,
$S_R1 = ?,
$S_R2 = ?,
$PPT = ?,
$DSP = ?,
$TIMETAKEN = ?,
$STATUS = '1',
TimeOfLogout = ?,
WHERE RecNo = ?";
$stmt = $con->prepare($query);
var_dump($stmt);
$stmt->bind_param('sisiiddssi',$TimeOfLogin,$no_of_logins,$time_of_nLogin,$m_scr_row1
$m_scr_row2,$m_ppt,$m_dsp,$m_time,$date,$RecNo);
if($stmt->execute()) echo " DONE !";
?>
Thanks to all
I have this piece of code: Code: [Select] if (isset($_POST['type1'])) { mysql_query("UPDATE tabke SET status = 1 WHERE id = $rid", $c2) or die(mysql_error()); } if (isset($_POST['type2'])) { mysql_query("UPDATE table SET status = 1 WHERE id = $rid", $c2) or die(mysql_error()); } if (isset($_POST['type3'])) { mysql_query("UPDATE table SET status = 1 WHERE id = $rid", $c2) or die(mysql_error()); } ?> <br /> <form method="post" action=""> <input type="hidden" name="pageid" value="plyrmgmt"> <input type="hidden" name="action" value="changeBan"> <input type="hidden" name="uid" value="<?php echo $hr_uid; ?>"> <input type="hidden" name="repid" value="<?php echo $rid; ?>"> <input type="submit" name="type1" value="Choice1" onClick="this.form.submit()"> </form> <form method="post" action=""> <input type="hidden" name="pageid" value="plyrmgmt"> <input type="hidden" name="action" value="changeMute"> <input type="hidden" name="uid" value="<?php echo $hr_uid; ?>"> <input type="hidden" name="repid" value="<?php echo $rid; ?>"> <input type="submit" name="type2" value="Choice2" onClick="this.form.submit()"> </form> <form method="post" action=""> <input type="hidden" name="pageid" value="plyrmgmt"> <input type="hidden" name="action" value="changeLock"> <input type="hidden" name="uid" value="<?php echo $hr_uid; ?>"> <input type="hidden" name="repid" value="<?php echo $rid; ?>"> <input type="submit" name="type3" value="Choice3" onClick="this.form.submit()"> </form> Now, for some reason when I click on Choice1, the query doesn't execute. Also, when I click on Choice2 or Choice3, the URL in my browser doesn't change for some odd reason...it stays the same as it was prior to clicking the Submit Button. Can anyone point out some errors I have? Thanks, Mark. So I am trying to execute an sql query by clicking a button or a link. Ultimately I want users to be able to click the "Fav" button and it then fades into "Added to favorites!" (I am building a favorite system which first collects data on page load like userid and itemid and then it needs to send it to the DB after a button is clicked) The fade and stuff is not the most important part now (although a tip on doing that would be awesome) but the most important part is how to get this functional: PHP Code: <?php // Start Favorite System! $user =& JFactory::getUser(); $userid = $user->id; if ($userid == 62) { echo "userID: ".$userid."<p />"; if ($user->guest) {echo "You are a guest<p />";} $itemid = $this->item->id; echo "itemID: ".$itemid."<p />"; $catid= $this->item->category->id; echo "catID: ".$catid."<p />"; $query = "INSERT INTO jos_k2_fav_xref (userID, itemID, catID) VALUES ('$userid', '$itemid', '$catid')"; // Need a way to make this into a button $run = mysql_query($query) or die(mysql_error()); } else {} ?> Currently all the values get stored in the DB as the page is loaded. I however want to be able to click on something to execute that query how do i do that? I read a bunch of stuff on javascript and ajax but it got me nowhere... Help is greatly appreciated! This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=355955.0 Hi,
I need help in populating data from DB in the 2nd listbox depending on 1st listbox data (where the data of 1st listbox is also fetched from DB), in the same form.
Thanks,
Jessentha
optionsT Does not show any results.Dont know where is the problem.
<script type="text/javascript" language="javascript">
function AddItemInList(fromLeftToRight, isAll){
var list1 = document.getElementById('listBoxF');
var list2 = document.getElementById('listBoxT');
if(Boolean(fromLeftToRight) == true){
MoveItems(list1,list2,isAll);
}else{
MoveItems(list2,list1,isAll);
}
return false;
}
function MoveItems(listFrom, listTo, isAll){
var toBeRemoved = "";
if(listFrom.options.length > 0){
for (i=0; i<listFrom.length; i++){
if (listFrom.options[i].selected || (isAll == true)){
if(Exist(listTo, listFrom.options[i].value) == 0){
listTo[listTo.length] = new Option(listFrom.options[i].text, listFrom.options[i].value, true);
toBeRemoved = toBeRemoved + listFrom.options[i].value + ';';
}
}
}
ClearSelection(listTo);
RemoveFromList(listFrom, toBeRemoved);
}else{
alert('Unable to Move Items. List is Empty!');
}
}
function RemoveFromList(listFrom, items){
var toBeRemoved = items.split(';');
for (var i=0; i < toBeRemoved.length; i++){
for (var j = 0; j < listFrom.length; j++){
if (listFrom.options[j] != null && listFrom.options[j].value == toBeRemoved[i]){
listFrom.options[j] = null;
}
}
}
}
function ClearSelection(list){
list.selectedIndex = -1;
}
function Exist(list, value){
var flag = 0;
for (var i=0; i < list.length; i++){
if (list.options[i].value == value){
flag = 1;
break;
}
}
return flag;
}
</script>
<?php
$opt = isset($_POST['optionsT']) ? $_POST['optionsT'] : '';
if(!empty($_POST['submit'])){
print $opt;
}
print"
<form method=\"POST\">
<table style=\"width:100%\">
<tr valign=\"top\">
<td>Options</td>
<td>
<select multiple size=\"5\" name=\"optionsF\" id=\"listBoxF\" style=\"width:350px\">
<option value=\"1\">1</option>
<option value=\"2\">2</option>
<option value=\"3\">3</option>
<option value=\"4\">4</option>
</select>
<div align=\"center\">
<input type=\"button\" onclick=\"return AddItemInList(true,true)\" value=\"+ All\">
<input type=\"button\" onclick=\"return AddItemInList(true,false)\" value=\"+\">
<input type=\"button\" onclick=\"return AddItemInList(false,false)\" value=\"-\">
<input type=\"button\" onclick=\"return AddItemInList(false,true)\" value=\"- All\">
</div>
<select multiple size=\"5\" name=\"optionsT[]\" id=\"listBoxT\" style=\"width:350px\">
</select>
</td>
</tr>
<tr>
<td></td>
<td><input type=\"submit\" name=\"submit\" value=\"Submit\"></td>
</tr>
</table>
</form>";
?>
Hi! every body. This is my first post in this form. Hope u people will help me. I'm using PHP 5.3. I've the following listboxes. From "emp_list" i've double click the value, which goes to "sel_list". I'm selecting multiple values. How can i get these values on the target page using PHP. <html><head> <script language="javascript"> function moveItem( box1, box2 ){ for ( var i=0; i < box1.options.length; i++ ){ if ( ( box1.options.selected == true ) ){ strItemToAdd = box1.options.text; box2.options[box2.length] = new Option(strItemToAdd); box1.options = null; i--; } } } </script> </head><body> <form id="project" name="project" action="" method="POST"> <select id='emp_list' name='emp_list' size='5' STYLE="width:170;" multiple onDblclick="moveItem( project.emp_list, project.sel_list)"> <option value='a1'>a1</option> <option value='a2'>a2</option> <option value='a3'>a3</option> </select> <select id='sel_list' name='sel_list' size='5' STYLE="width:170;" multiple onDblclick="moveItem(project.sel_list, project.emp_list )"> </select> <input type='submit' name='submit' value='submit'> </form> </body></html> Hi guys, I have a list box. say the code is Code: [Select] <select name="drop1" id="Select1" size="4" multiple="multiple"> <option value="1">item 1</option> <option value="2">item 2</option> <option value="3">item 3</option> <option value="4">item 4</option> <option value="0">All</option> </select> The user will select few items. How do i capture the data through the "POST" method. Code: [Select] $data = $_POST['drop']; would not work coz there is an array if data. Can any one help or point me in the right direction? Thanks in advance needing some help on this, i have a form with a dynamic listbox and and textfields, what i need to do is when i select item from listbox i need it to fetch the data from DB and fill out rest of fields in form. i 'm using dreamweaver and can post code if required kenny Hello, I'm using PHP and I have a list box where I should be able to select multiple values, that I then submit to run multiple reports. However, when I select two values I only get results for the last selected value instead of both value a & value b. Can anyone help me with my problem? Code: [Select] <td style="text-align: right">Job Type:</td> <td> <select name="job_type" multiple="multiple" size="4"> <option value="0" selected>ALL</option> {section name=jt loop=$job_type_list} <option value="{$job_type_list[jt][0]}">{$job_type_list[jt][1]}</option> {/section} </select> </td> I've read that I can change name="job_type" to name="job_type[]", but I'm not getting any luck with that...
<script type="text/javascript">
function listBoxSearch(){
var input = document.getElementById('searchBox'),
output = document.getElementById('listBoxF');
if(input.value != ''){
for (var i = 0; i < output.options.length; i++){
if(output.options[i].text.substring(0, input.value.length).toUpperCase() == input.value.toUpperCase()){
output.options[output.options.length] = new Option(output.options[i].innerHTML, output.options[i].text);
}
}
if (output.options.length == 1) {
output.selectedIndex = 0;
}
}
}
</script>
It dosen`t work. It should work like this[DEMO in attached filed]
Attached Files
listbox_with_keybord_search.htm 7.9KB
3 downloadsHi all, I have a listbox on a form that gets it's data from a MySQL query, all of which is working fine. I make a selection in my listbox and then submit the form which is reloading itself. However on the reload I would like the listbox to display the selection made by the user, instead it reverts to the first item in the list. I have included the code, if someone could point me in the right direction that would be great. I have searched these forums and tried a few things that are suggested but I still can't get it to work. Code: [Select] <?php include("dbconn.php"); $query = "SELECT location.location_name, printer.model_number, location_printer.serial_number" . " FROM location, printer, location_printer" . " WHERE location.location_id = location_printer.location_id" . " AND printer.printer_id = location_printer.printer_id"; $result = mysql_query($query); echo "<table>"; echo "<th>Location</th>"; echo "<th>Printer</th>"; echo "<th>Serial No</th>"; while(($row = mysql_fetch_array($result))) { echo "<tr>"; echo "<td>".$row['location_name']."</td>"; echo "<td>".$row['model_number']."</td>"; echo "<td>".$row['serial_number']."</td>"; echo "</tr>"; } echo "</table>"; if(isset($_POST['location'])) { $loc = $_POST['location']; echo "<p>Location = <b> $loc </b>"; } ?> <html> <head> <title>Loop Results</title> </head> <body> <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" > <table> <tr> <td> <?php $query = "SELECT * FROM location ORDER BY location_name"; $result = mysql_query($query); $default = $_POST['location']; echo "<p>Location: "; echo "<select name='location'>"; while(($row = mysql_fetch_array($result))) { echo "<option value='{$row['location_id']}"; //Selected value if($row['location_id'] == $default) echo " selected "; echo "'>{$row['location_name']}</option>\n"; } echo "</select>"; ?> </td> </tr> <tr> <td> <input type="submit" /> </td> </tr> </table> </form> </body> </html> Hi, I'm coding a page that starts with a select list loaded from a database and I want to make a different query depending on the item selected, I have been searching around and I can't find a php version of the selectedIndex of a listbox in javascript, any idea? thanks.
Hi everyone,
I'm a complete newbie here. I've looked around for help and tried a few solutions without success, so because time is short I'm asking for help.
I have a registration form which includes address details. The form holds a country field which is a listbox, the values being selected from a MySQL table. How do I assign the selected country option to a field list in the insert statement?
Here's what I have so far:
Registration page form:
<div class="form-group"> How can I get this form to execute? Do I need to use a hidden input or something? Code: [Select] <form action="partnerRequest.php" method="post"> <a href="" title="Become Associates" id="becomeassoc" class="mll"> <span class="f_right">Become Associates</span><img src="../assets/img/becomeassoc.jpg" class="mrs vmiddle" alt="Become Associates" /> </a> </form> hi, I am having problems setting a session from a SQL query, please help... here is my code: Code: [Select] $query1 = "SELECT * FROM members_copy WHERE rsUser = '$username' AND rsPass = '$password'"; $result1 = mysql_query($query1); echo "<br>".$result1['USERID']; $_SESSION['s_logged_n'] = 'true'; $_SESSION['s_username'] = $username; $_SESSION['USERID']=$result1['USERID']; $_SESSION['RSTOWN']=$result1['RSTOWN']; $_SESSION['RSEMAIL']=$result1['RSEMAIL']; $_SESSION['RSUSER']=$result1['RSUSER']; no sessions are currently being set?! |
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