|
PHP - Duplicate Display From Select?
Hey guys, not sure what's gone wrong here, but this displays the select query twice on my page.
Code: [Select] <?php $result = mysql_query("SELECT * FROM members WHERE Username='$_SESSION[Username]'") or die(mysql_error()); while($row = mysql_fetch_array( $result )) { echo '<b>Resources</b><br />'; echo 'Wood: ' . $row['res1'] . ' <br />'; echo 'Iron: ' . $row['res2'] . ' <br />'; } ?> Displays as: Resources Wood: 2000 Iron: 1000 Resources Wood: 0 Iron: 0 The first set of numbers are correct, but I'm not sure where the second lot are been called from? Thanks in advance. Similar TutorialsHi everybody, I have a problem and was wondering if I could pick your brains. My situation is that I am making a fitness website and one feature is that the user can set themselves targets. Now each target has some fields that are mandatory (for example the date to be completed) and some fields that are unique to each target (for example if the target is weight loss I will need to grab the current weight/target weight fields). Just to make it clear I have all fields in the same database table. So I have written a script that displays that users current targets they have set. It grabs all the data correctly but it duplicates the display and although I am pretty new to php even I can tell the code if not best practise because I have a query inside a loop and a loop inside another loop. Here is the code I have done so far: Code: [Select] //Get target types from database for that particular user $get_user_targets = mysql_query("SELECT target_type FROM user_targets WHERE user_id='$id'") or die (mysql_error()); while ($row = mysql_fetch_array($get_user_targets)){ $target_type_display = $row['target_type']; //If the target type is equal to Weight Loss run this script to get all the information and make a display if($target_type_display == 'Weight Loss'){ //Search the database for all weight loass goals $weight_loss_display_sql = mysql_query("SELECT current_weight, target_weight, target_date, date_set, extra_info FROM user_targets WHERE user_id='$id' AND target_type='$target_type_display'") or die (mysql_error()); //Put the data from the query into page variables while($row1 = mysql_fetch_array($weight_loss_display_sql)){ $current_weight_loss_dis = $row1['current_weight']; $target_weight_loss_dis = $row1['target_weight']; $weight_loss_date_set_dis = $row1['date_set']; $weight_loss_target_date_dis = $row1['target_date']; $weight_loss_extra_info_dis = $row1['extra_info']; //Convert date_set into ago time $convertedTime = ($myObject -> convert_datetime($weight_loss_date_set_dis)); $weight_loss_set = ($myObject -> makeAgo($convertedTime)); //Construst a display out of the collected variables $weight_loss_display .= '<div class="blackText" style="width: 475px"> <table width="475px" class="blackText"> <tr><td class="profileUsername" colspan="2">'.$target_type_display.'</td></tr> <tr><td width="250px">Current Weight:</td><td width="225px">'.$current_weight_loss_dis.'</td></tr> <tr><td>Target Weight:</td><td>'.$target_weight_loss_dis.'</td></tr> <tr><td>Date for completion:</td><td>'.$weight_loss_target_date_dis.'</td></tr> <tr><td>'.$weight_loss_extra_info_dis.'</td><td><h6>Date set: '.$weight_loss_set.'</h6></td></tr> </table> </div>'; } } } So far it just checks for a weight loss goal and I was going to expand the code with more if statements for each of the other types of target. So my questions a 1)Can anybody see why the data is displaying twice? 2) Is there a better way of doing this? (NOTE: im not too worried that the code isn't best practise, I just want to get it functioning properly) Thanks in advance for any help Hi... I encountered problem in my query and while loop my data was duplicates when I join 2 tables, because the one field that I need to display is from another table. here is my code: Code: [Select] <div> <table> <thead> <th>Items</th> <th>Sub Items</th> <th>Item Code</th> <th>Demanded Qty</th> <th>UoM</th> <th>Class</th> <th>Description</th> <th>BIN Location</th> </thead> <?php $sql = "SELECT Items FROM bom_items ORDER BY Items"; $res_bom = mysql_query($sql, $con); while($row = mysql_fetch_assoc($res_bom)){ echo "<tr> <td style='border: none;font-weight: bold;'> $row[Items]</td> </tr>"; //$sql = "SELECT SubItems, ItemCode, UoM, Class, Description FROM bom_subitems WHERE Items = '$row[Items]' ORDER BY Items"or die(mysql_error()); //$sql = "SELECT DISTINCT bs.SubItems, bs.ItemCode, bs.UoM, bs.Class, bs.Description,w.BINLocation FROM bom_subitems bs LEFT JOIN wms w ON bs.Items = w.Items WHERE bs.Items = '$row[Items]' AND w.Items = '$row[Items]' ORDER BY bs.Items, w.Items"or die(mysql_error()); $sql = "SELECT DISTINCT bs.SubItems, bs.ItemCode, bs.UoM, bs.Class, bs.Description,w.BINLocation FROM bom_subitems bs, wms w WHERE bs.Items = '$row[Items]' AND w.Items = '$row[Items]' ORDER BY bs.Items, w.Items"or die(mysql_error()); $res_sub = mysql_query($sql, $con); while($row_sub = mysql_fetch_assoc($res_sub)){ echo "<tr> <td style='border: none;'> </td> <td style='border: none;'> $row_sub[SubItems]</td> <td style='border: none;'> $row_sub[ItemCode]</td> <td> </td> <td style='border: none;' size='3'> $row_sub[UoM]</td> <td style='border: none;'> $row_sub[Class]</td> <td style='border: none;'> $row_sub[Description]</td> <td style='border: none;'> $row_sub[BINLocation]</td> </tr>"; } } ?> I will attach my sample page. Thank you Hi, i got help earlier with the this code-- exploding a field and getting an array, then inserting the array into a multiselect box-- That part is easy and works fine. Now the explode function gives an array of words with duplicate words in it: I tried several ways most either give a blank option/value, or the array word.. here is the code: $query = "SELECT DISTINCT property_functionsexperience FROM #__users_profiles WHERE published = '1' ORDER BY property_functionsexperience ASC"; $functionsexperiencelistgeneral=doSelectSql($query); foreach($functionsexperiencelistgeneral as $words) $property_funcexperilist=$words->property_functionsexperience; $wording = explode(', ', $property_funcexperilist); foreach($wording as $funciex) $funcexi=array_unique($funciex); $funcexpList .= "<option value=\"".$funcexi.','."\">".$funcexi."</option>"; $funcexpList .= "</optgroup>"; } $funcexpList.="</select>"; $output['FUNCEXPLIST']=$funcexpList; this code is adapted slightly from the working original that gives the double words-- this code gives a blank, but i think its on the correct track!? Hi all, I am a php student and i am doing a project in php. I am doing the project of online attendance register. I am stuck with the part of reports. We need to select three different values from an HTML form and pick the selected rows and wants to display those rows in a table. I have done a code but it is not working. Please help me out of this issue. HTML Form: <!doctype html> <html> <head> <meta charset="utf-8"> <title>Untitled Document</title> </head> <body><form action="reportpro.php" method="get" name="services"> <table width="447" border="0" align="center" > <tr style="text-align: center"> <td width="441"><table width="481" border="1"> <tr> <td width="216"><div align="center"> SECTION </div></td> <td width="102"><div align="center"> MONTH </div></td> <td width="141"><div align="center"> YEAR </div></td> </tr> <tr> <td><select name="section" id="select"> <option>SERVICES-A</option> <option>SERVICES-B</option> <option>SERVICES-C</option> </select></td> <td><select name="month" id="select"> <option>01</option> <option>02</option> <option>03</option> <option>04</option> <option>05</option> <option>06</option> <option>07</option> <option>08</option> <option>09</option> <option>10</option> <option>11</option> <option>12</option> </select></td> <td><select name="year" id="select"> <option>2014</option> <option>2015</option> <option>2016</option> <option>2017</option> <option>2018</option> <option>2019</option> <option>2020</option> </select></td> </tr> <tr> <td colspan="4"> <center><input name="submit" type="submit" id="submit" formaction="reportpro.php" formmethod="get" value="Search"></center></td> </tr> </table> <p> </p> <p> </p> <p> </p></td> </tr> </table> </form> </body> </html> PHP Code:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<form action="" method="get">
<table border="1" align="center" >
<tr>
<td>BOOK NO</td>
<td>Name</td>
<td>Desigination</td>
<td>Section</td>
<td>Day</td>
<td>Month</td>
<td>Year</td>
<td>Attendance</td>
</tr>
<?php
$con=mysql_connect("localhost","root","");
mysql_select_db("niy",$con);
$m=$_REQUEST['month'];
$y=$_REQUEST['year'];
$sec=$_REQUEST['section'];
$sql="select month='$m', year='$y', sectn='$sec' from attendance ";
echo $sql;
$res=mysql_query($sql,$con);
while($row=mysql_fetch_array($res))
{
?>
<tr>
<td><?php echo $row[1];?></td>
<td><?php echo $row[2];?></td>
<td><?php echo $row[3];?></td>
<td><?php echo $row[4];?></td>
<td><?php echo $row[5];?></td>
<td><?php echo $m;?></td>
<td><?php echo $y;?></td>
<td><?php echo $row[8];?></td>
</tr>
<?php
}
?>
</table>
</body>
</html>
Okay so, I'm adding a reply feature to my text-board, and I'm wondering how I can select a single result from the DB I've tried this $board = mysql_real_escape_string($_GET['board']); $thread = mysql_real_escape_string($_GET['thread']); $result = mysql_query("SELECT subject, name, id, timestamp, body FROM $board LIMIT $thread,$thread"); while ($row = mysql_fetch_array($result, MYSQL_NUM)) { printf("<div id='html1' style='background-color:#F0E0D6;border:solid 1px #800000;font-family:arial; font-size: 12px; padding:2px; margin:2px;'><font color='#CC1105'>%s </font><font color='#117743'>%s</font> <font color='#800000'>Post # %s - %s<br />%s<br /></div>", $row[0], $row[1], $row[2], $row[3], $row[4], $row[5], $row[6]); mysql_free_result($result); } ?> Kinda works but if I pu ?thread=56 then it would show thread 57, which in this example does not exist. Guys, you are so going to be sick of me. But I really appreciate the help I am receiving here thus far. I have the code below sending the data from the form to the database. That's excellent. However, where I am stuck is doing the following two: 1. Updating currently inserted data 2. Displaying the data on the site. Here is settings.php: <?php include "../settings.php"; include "../lang/english.php"; include "../config/config.php"; $errors = array(); // condition and map inputs $fields = array('site_name'=>'sitename','site_email'=>'email','your_name'=>'name','meta_description'=>'meta-description','meta_keywords'=>'meta-keywords'); foreach($fields as $var => $field){ $$var = isset($_GET[$field]) ? trim($_GET[$field]) : ''; } // check if the form was submitted if(isset($_GET['submit'])){ // connect to/select database mysql_connect("$db_hostname", "$db_username", "$db_password") or die(mysql_error()); mysql_select_db("$db_database") or die(mysql_error()); // basic validation (not empty) and escape data foreach($fields as $var => $field){ if(empty($$var)){ $errors[] = "The form field: $field, is empty!"; } // escape the data $$var = mysql_real_escape_string($$var); } // if no validation errors, insert the data if(empty($errors)){ $insert = sprintf("INSERT INTO settings (site_name, description, keywords, email, name) VALUES ('%s','%s','%s','%s','%s')", $site_name, $meta_description, $meta_keywords, $site_email, $your_name ); if(!mysql_query($insert)){ // a query error occurred // check if due to duplicate data if(mysql_errno() == 1062){ // 1062 = duplicate primary key error (your error number might be different depending on your table definition) $errors[] = "The site name: $site_name, already exists and cannot be inserted!"; } else { // all other query errors - $errors[] = "A database error occurred and your query cannot be processed!"; trigger_error(mysql_error()); // use error_reporting/display_errors/log_errors to display/log the error condition } } else { echo "The data was successfully inserted!"; } } } //database_connect(); $sql = "UPDATE settings SET site_name='$site_name', description='$description', keywords='$keywords', email='$site_email', name='$your_name' WHERE id='$id'"; $query = mysql_query($sql)or die("There's a problem with the query: ". mysql_error()); if($query) echo "<br>The settings have been updated.<br>"; $_SESSION['tekst']=""; ?> <!DOCTYPE html> <html> <head> <title><?php echo $site_name; ?> :: :: Powered by osPHPSite</title> <meta http-equiv="content-type" content="text/html; charset=iso-8859-1" /> <link href="css/admin.css" rel="stylesheet" type="text/css" /> </head> <body> <div id="main"> <div id="header"> <h1><?php echo $sitename; ?></h1> <ul id="top-navigation"> <li><a href="index.php"><?php echo $lang_button_index; ?></a></li> <li><a href="settings.php" class="active"><?php echo $lang_button_settings; ?></a></li> <li><a href="pages.php"><?php echo $lang_button_pages; ?></a></li> <li><a href="gallery.php"><?php echo $lang_button_gallery; ?></a></li> <li><a href="/"><?php echo $lang_button_viewsite; ?></a></li> <li><a href="logout.php"><?php echo $lang_button_logout; ?></a></li> </ul> </div> <div id="middle"> <div id="left-column"> <h3><?php echo $eng_navigation; ?></h3> <ul class="nav"> <li><a href="index.php"><?php echo $lang_button_index; ?></a></li> <li><a href="settings.php"><?php echo $lang_button_settings; ?></a></li> <li><a href="pages.php"><?php echo $lang_button_pages; ?></a></li> <li><a href="gallery.php"><?php echo $lang_button_gallery; ?></a></li> <li><a href="/"><?php echo $lang_button_viewsite; ?></a></li> <li><a href="logout.php"><?php echo $lang_button_logout; ?></a></li> </ul> <a href="http://www.osphpsite.com" target="_blank" class="link">osPHPSite</a> <a href="http://www.osphpsite.com/forums/" target="_blank" class="link">Support Forums</a> <a href="http://www.vichost.com" target="_blank" class="link">VicHost.Com</a> </div> <div id="center-column"> <div class="table"> <?php if(!empty($errors)){ echo 'The following errors occurred:<br />'; foreach($errors as $error){ echo "$error<br />"; } } ?> <form action="" id="settings" name="settings"> <table class="listing form" cellpadding="0" cellspacing="0"> <tr> <th class="full" colspan="2">Site Configuration</th> </tr> <tr> <th colspan="2">From the options below, define the default settings for your website.</th> </tr> <tr> <th colspan="2"></th> </tr> <tr> <td>Site name: </td> <td><input type="text" name="sitename" value="<?php echo $site_name; ?>" width="172" /> <em>Site name for logo</em></td> </tr> <tr> <td>Email: </td> <td><input type="text" name="email" value="<?php echo $site_email; ?>" width="172" /> <em>Your email address</em></td> </tr> <tr> <td>Name: </td> <td><input type="text" name="name" value="<?php echo $your_name; ?>" width="172" /> <em>Your own name</em></td> </tr> <tr> <td>Meta Description: </td> <td><input type="text" name="meta-description" value="<?php echo $meta_description; ?>" width="172" /> <em>SEO</em></td> </tr> <tr> <td>Meta Keywords: </td> <td><input type="text" name="meta-keywords" value="<?php echo $meta_keywords; ?>" width="172" /> <em>Separate with Commas</em></td> </tr> <tr> <td><input type="submit" class="button" name="submit" value="Submit"></td> </tr> </table> </form> </div> </div> </div> </div> </body> </html> This will be a huge hurdle if I can get passed this one, I am well and truly on my way. Any help, advice etc that you can give, and what needs to go where, I would really appreciate it and I promise not to annoy you guys unless absolutely necessary. Hello: I am trying to select all the records from a table, and display them. Having an issue... This is the code: Code: [Select] <ul> <?php $query = mysql_query("SELECT id,mySectionTitle FROM myWebSiteData"); while($menuData = mysql_fetch_array($query)) { <li><a href="a_websiteData.php?id=" .echo $menuData['id'] ."">echo $menuData['mySectionTitle']</a></li> } ?> </ul> This is the error: Code: [Select] Parse error: syntax error, unexpected '<' in /html/admin/a_websiteData.php on line 91 line 91 is the one that starts with "<li>" What am I missing, or what would be the best way to do this? Hey guys I'm really confused at the minute, I have products in my database (It's MySQL) I want to display attributes from the products in a filtering form.
I currently have select elements in my template and I want to display all current manufacturers that are listed in the manufacturers column of my database.
How can I display the manufacturers once in the select element without displaying them multiple times and in alphabetical order? For example I might have 4 products that are Sony but I don't want the brand to display 4 times.
Any resources or examples on how to do this?
Thanks
I have the following html code and php code. I would like clients to select which fields in the mysql database they would like to have appear in the resultant table by checking the text box next to the field in the html form. For instance they should be able to select first name and city and have only those columns appear in the resultant table. # <html> # <head> # <title>Search Clients Database</title> # </head> # # <body> # <h1>Clients Database Search Page</h1> # <form action="searchclients.php" method="post"> # <table width="100%" border="1" cellspacing="1" cellpadding="1"> # <tr> # <th width="37" scope="col"> </th> # <th width="114" scope="col">Fields</th> # <th width="169" scope="col">Filter Value</th> # <th width="1157" scope="col"> </th> # </tr> # <tr> # <td><input type="checkbox" name="idc" id="idc"></td> # <td>ID</td> # <td><input type="text" name="ID" id="ID"></td> # <td> </td> # </tr> # <tr> # <td><label> # <input type="checkbox" name="fnc" id="fnc"> # </label></td> # <td>First Name</td> # <td><input type="text" name="FirstName" id="FirstName"></td> # <td> </td> # </tr> # <tr> # <td><label> # <input type="checkbox" name="lnc" id="lnc"> # </label></td> # <td>Last Name</td> # <td><input type="text" name="LastName" id="LastName"></td> # <td> </td> # </tr> # <tr> # <td><label> # <input type="checkbox" name="cc" id="cc"> # </label></td> # <td>City</td> # <td><input type="text" name="City" id="City"></td> # <td> </td> # </tr> # <tr> # <td><label> # <input type="checkbox" name="pc" id="pc"> # </label></td> # <td>Province</td> # <td><select name="Province" id="Province"> # <option selected> </option> # <option>KZN</option> # <option>North West Province</option> # <option>Gauteng</option> # <option>Free State</option> # <option>Mpumalanga</option> # <option>Eastern Cape</option> # <option>Limpopo Province</option> # <option>Northern Cape</option> # <option>Western Cape</option> # </select></td> # <td> </td> # </tr> # </table> # <p> # <input type="submit" name="submit" value="Search" /> # </p> # </form> # # </body> # </html> # <?php # mysql_connect ("localhost", "username","password") or die (mysql_error()); # mysql_select_db ("clients"); # # $fn = $_POST['FirstName']; # $ln = $_POST['LastName']; # $city = $_POST['City']; # $prov = $_POST['Province']; # # $idc = $_POST['idc']; # $fnc = $_POST['fnc']; # $lnc = $_POST['lnc']; # $cc = $_POST['cc']; # $pc = $_POST['pc']; # # ?> # # <html> # <body> # # <table border=1> # <tr> # <th>ID</th> # <th>First Name</th> # <th>Last Name</th> # <th>City</th> # <th>Province</th> # </tr> # # <?php # $sql = mysql_query("select ID, FirstName, LastName, City, Province from clients where FirstName like '%$fn%' and LastName like '%$ln%' and City like '%$city%' and Province like '%$prov%'"); # # while ($row = mysql_fetch_array($sql)){ # # $id = $row['ID']; # $fname = $row['FirstName']; # $lname = $row['LastName']; # $city = $row['City']; # $prov = $row['Province']; # # ?> # # <tr> # <th><?php echo $id;?></th> # <th><?php echo $fname;?></th> # <th><?php echo $lname;?></th> # <th><?php echo $city;?></th> # <th><?php echo $prov;?></th> # </tr> # # <?php } //this ends the if?> # # </table> # </html> I wrote the page below to display a member's area of a website. The page is a mix of html and php code. the first part of is a block of php, which authenticates the user and displays the header. The middle part is a block of html which contains the bulk of the page. Embedded in this block of html are two sections of php, the first of which is supposed to display a user uploaded picture, enclosed in div tags, and the second of which, prints out the user's first name, from the database. The end of the script is a line of php that displays the footer. Now the page displays correctly when the section of php that contains the select query for displaying the picture(starting line 104) is omitted. But once I include that section, all i see is a blank page. Why is that section of php problematic? What can be done to fix it? Here is the data for the full page. Thanks for any help. <?php //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //authenticate user require('auth.php'); //define title define('TITLE' , 'Members'); require ('header.html'); //need the header ?> <div id="main" style="background-color: #FFFFFF; height:71%; width:101%; border:0px none none; margin:auto; "> <!-- --> <div id="main_left" style="float:left; height:100%; width:20%; border:0px none none;"> <!--opens main left--> <div id="main_left_top" style="float:left; position:relative;bottom:5px;right:5px; height:31.25%; width:100%; background-color: #FFFFFF; border:1px solid #c0c0c0; margin:1px;"> <!--opens main left top--> </div> <!-- closes main left top--> <div id="main_left_center" style="float:left; background-color: #FFFFFF; height:33%; width:100%; border-color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <!--opens the white content area--> </div> <!-- closes main left center--> <div id="main_left_bottom" style="float:left; background-color: #FFFFFF; height:33%; width:100%; border-color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <!--opens the white content area--> </div> <!-- closes main left bottom--> </div> <!-- closes main left--> <div id="main_center" class="content_text" style="float:left; height:100%; width:58%; background-color: #FFFFFF; border:1px solid #c0c0c0;"> <!--opens main center--> <div id="image_box" style="float:left; background-color: #c0c0c0; height:150px; width:140px; border-color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <?php //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); $query = "SELECT* FROM images WHERE member_id ='{$_SESSION['id']}' AND cartegoty 'main' "; $result = mysql_query($query); $result_data = mysql_fetch_array($result); header("Content-type: image/jpeg") ; echo $result_data['image']; ?> </div> <a href="upload_image_page.php">click here to uplaod a picture</a> <h1>Welcome <?php echo ucfirst($_SESSION['firstname']);?></h1> <a href="member_profile.php">My Profile</a> | <a href="logout.php">Logout</a> <p>This is a password protected area only accessible to members. </p> <a href="blog_entries.php">Add to Hahap Tok Library</a> </div> <!-- closes main center--> <div id="main_right" style="float:left; background-color: #FFFFFF; height:100%; width:20%; border-color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <!--opens the white content area--> <div id="main_right_top" style="float:left; background-color: #FFFFFF; height:33%; width:100%; border-color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <!--opens the white content area--> </div> <!-- closes main left top--> <div id="main_right_center" style="float:left; background-color: #FFFFFF; height:33%; width:100%; border-color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <!--opens the white content area--> </div> <!-- closes main left center--> <div id="main_right_bottom" style="float:left; background-color: #FFFFFF; height:34%; width:100%; border-color:#a0a0a0;border-style:outset;border-width:1px; margin:auto; "> <!--opens the white content area--> </div> <!-- closes main left bottom--> </div> <!-- closes main right--> </div> <!-- closes main--> <?php require('footer.html'); ?> Hi... I have query in highlighting null data using this code: Code: [Select] <?php include 'config.php'; $currentEmpID = $_SESSION['empID']; if(!isset($_POST['Regsubmit_'])){ $DATE1 = $_GET['Regfirstinput']; $DATE2 = $_GET['Regsecondinput']; $sql = "SELECT DISTINCT IF(ISNULL(a.LOG_IN), 'rdc', '') AS LOGIN_CLASS, IF(ISNULL(a.LOG_OUT), 'rdc', '') AS LOGOUT_CLASS, a.EMP_ID, CONCAT(LASTNAME, ', ' , FIRSTNAME) AS FULLNAME, a.LOG_IN, a.LOG_OUT FROM $ATTENDANCE.attendance_build AS a JOIN $ADODB_DB.employment em ON (a.EMP_ID = em.EMP_NO AND em.STATUS IN ('Reg Operatives', 'Reg Staff')) WHERE LOG_IN BETWEEN '$DATE1' AND '$DATE2' OR ISNULL(LOG_IN) OR ISNULL(LOG_OUT)"; $DTR = $conn3->GetAll($sql); $smarty->assign('attendance', $DTR); } $smarty->display('header_att.tpl'); $smarty->display('RegAttendance.tpl'); $smarty->display('footer.tpl'); ?> and here is the tpl code: Code: [Select] {section name=att loop=$attendance} <tr> <td colspan="2">{$attendance[att].EMP_ID}</td> <td colspan="2">{$attendance[att].FULLNAME}</td> <td colspan="2" class="{$attendance[att].LOGIN_CLASS}">{$attendance[att].LOG_IN|date_format:"%d-%m-%Y %I:%M %p"}</td> <td colspan="2" class="{$attendance[att].LOGOUT_CLASS}">{$attendance[att].LOG_OUT|date_format:"%d-%m-%Y %I:%M %p"}</td> </tr> {sectionelse} <tr><td colspan="1">No DATA</td></tr> {/section} this code highlight the null value of login or logout or both. this is the css: Code: [Select] .rdc {background-color:#ff0000;} Now, I need to revised my query statement, because i have separate code for adding attendance if the employee has no attendance or no login or no logout. I just want to happen is if the employee is already add his attendance in NRS table or should I said if the LOG_IN in attendance table is equal to TIME_IN in NRS table the data will have a color yellow. For Example: I have this data in attendance table: EMP_ID = 012012 LOG_IN = NULL LOG_OUT = 2011-12-12 13:35:00 I will his attendance in NRS table to have his attendance: EMP_NO = 012012 TIME_IN = 2011-12-12 05:35:00 TIME_OUT = 2011-12-12 13:35:00 In my above query the LOG_IN has a background color of RED. I want to happen is if I add his attendance in NRS the EMP_NO, LOG_IN, LOGOUT will have a color to notice that it is already have in NRS. Because theirs a scenario that the employee has no login or no logout or both. Feel free to ask me if my explanation is not clear to you. Thank you in advance I had this working, but when I try and get fancy and use AJAX the data doesn't display. I think this is a PHP problem though. My code for the select form including AJAX code Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en-GB"> <head> <title>AJAX Example</title> <link rel="stylesheet" type="text/css" href="Form.css" media="screen" /> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script> <script type="text/javascript"> $(document).ready(function(){ $("tr:odd").addClass("odd"); }); </script> <script type="text/javascript"> function showPlayers(str) { var xmlhttp; if (str=="") { document.getElementById("DataDisplay").innerHTML=""; return; } if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); else {// code for IE6, IE5 } xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("DataDisplay").innerHTML=xmlhttp.responseText; } } xmlhttp.open("GET","Query.php?category_id="+str,true); xmlhttp.send(); } </script> </head> <body> <h1">AJAX Example</h1> <?php #connect to MySQL $conn = @mysql_connect( "localhost","username","pw") or die( "You did not successfully connect to the DB!" ); #select the specified database $rs = @mysql_SELECT_DB ("MyDB", $conn ) or die ( "Error connecting to the database test!"); ?> <form name="sports" id="sports"> <legend>Select a Sport</legend> <select name="category_id" onChange="showPlayers(this.value)"> <option value="">Select a Sport:</option> <?php $sql = "SELECT category_id, sport FROM sports ". "ORDER BY sport"; $rs = mysql_query($sql); while($row = mysql_fetch_array($rs)) { echo "<option value=\"".$row['category_id']."\">".$row['sport']."</option>\n "; } ?> </select> </form> <br /> <div id="DataDisplay"></div> </body> </html> Query.php <?php #get the id $id=$_GET["category_id"]; #connect to MySQL $conn = @mysql_connect( "localhost","username","pw") or die( "Error connecting to MySQL" ); #select the specified database $rs = @mysql_SELECT_DB ("MyDB", $conn ) or die ( "Could not select that particular Database"); #$id="category_id"; #create the query $sql ="SELECT * FROM sports INNER JOIN players ON sports.category_id = players.category_id WHERE players.category_id = '".$id."'"; echo $sql; #execute the query $rs = mysql_query($sql,$conn); #start the table code echo "<table><tr><th>Category ID</th><th>Sport</th><th>First Name</th><th>Last Name</th></tr>"; #write the data while( $row = mysql_fetch_array( $rs) ) { echo ("<tr><td>"); echo ($row["category_id"] ); echo ("</td>"); echo ("<td>"); echo ($row["sport"]); echo ("</td>"); echo ("<td>"); echo ($row["first_name"]); echo ("</td>"); echo ("<td>"); echo ($row["last_name"]); echo ("</td></tr>"); } echo "</tr></table>"; mysql_close($conn); ?> I think the problem is either with this part in the AJAX Code: [Select] xmlhttp.open("GET","Query.php?category_id="+str,true); or most likely in my Query.php code when I wasn't using AJAX and using POST it worked fine, but adding the AJAX stuff and GET it doesn't work. When I echo out the SQL the result is Code: [Select] SELECT * FROM sports INNER JOIN players ON sports.category_id = players.category_id WHERE players.category_id = '' so the category_id is not being selected properly and that is the primary key/foreign key in the MySQL table which connects the JOIN. I am working on a project where I want a select form to display information from a MySQL table. The select values will be different sports (basketball,baseball,hockey,football) and the display will be various players from those sports. I have set up so far two tables in MySQL. One is called 'sports' and contains two columns. Once called 'category_id' and that is the primary key and auto increments. The other column is 'sports' and contains the various sports I mentioned. For my select menu I created the following code. <?php #connect to MySQL $conn = @mysql_connect( "localhost","uname","pw") or die( "You did not successfully connect to the DB!" ); #select the specified database $rs = @mysql_SELECT_DB ("test", $conn ) or die ( "Error connecting to the database test!"); ?> <html> <head>Display MySQL</head> <body> <form name="form2" id="form2"action="" > <select name="categoryID"> <?php $sql = "SELECT category_id, sport FROM sports ". "ORDER BY sport"; $rs = mysql_query($sql); while($row = mysql_fetch_array($rs)) { echo "<option value=\"".$row['category_id']."\">".$row['sport']."</option>\n "; } ?> </select> </form> </body> </html> this works great. I also created another table called 'players' which contains the fields 'player_id' which is the primary key and auto increments, category_id' which is the foreign key for the sports table, sport, first_name, last_name. The code I am using the query and display the desired result is as follows <html> <head> <title>Get MySQL Data</title> </head> <body> <?php #connect to MySQL $conn = @mysql_connect( "localhost","uname","pw") or die( "Err:Db" ); #select the specified database $rs = @mysql_SELECT_DB ("test", $conn ) or die ( "Err:Db"); #create the query $sql ="SELECT * FROM sports INNER JOIN players ON sports.category_id = players.category_id WHERE players.sport = 'Basketball'"; #execute the query $rs = mysql_query($sql,$conn); #write the data while( $row = mysql_fetch_array( $rs) ) { echo ("<table border='1'><tr><td>"); echo ("Caetegory ID: " . $row["category_id"] ); echo ("</td>"); echo ("<td>"); echo ( "Sport: " .$row["sport"]); echo ("</td>"); echo ("<td>"); echo ( "first_name: " .$row["first_name"]); echo ("</td>"); echo ("<td>"); echo ( "last_name: " .$row["last_name"]); echo ("</td>"); echo ("</tr></table>"); } ?> </body> </html> this also works fine. All I need to do is tie the two together so that when a particular sport is selected, the query will display below in a table. I know I need to change my WHERE clause to a variable. This is what I need help with. thanks Hi, I have come up with the following code, I need it to get the details of several scattered products and echo the results, the trick is I don't want it to echo the results one after the other... I want to have the products scattered between unique text on the page but don't want to run the query several times for performance reasons. E.g.- PAGE to look like this: $Product_1 unique text/images $Product_2 $Product_3 unique text/images $Product_4 Current Code: Code: [Select] <? $result = mysql_query("SELECT * FROM products where Product_ID IN (475, 465, 234, 567, 845)"); while($row = mysql_fetch_array($result)) { $x = "1"; while ($x<=3) { echo $x; $Product = "Product_"; $Product = $Product.$x; echo $Product; $Product = $row['Product_ID']; echo $Product; $x++; echo $x; } } At the moment it returns the following results: Quote 1 Product_1 465 2 2 Product_2 465 3 3 Product_3 465 4 1 Product_1 475 2 2 Product_2 475 3 3 Product_3 475 4 A few problems... In Blue... it duplicates for product 465 In Red... It repeats again for 475 Also.... it starts with 465, but I want it to go in order as how it appears - $result = mysql_query("SELECT * FROM products where Product_ID IN (475, 465, 234, 567, 845)"); so should start with 475 I want to get the following result: Quote 1 Product_1 475 2 2 Product_2 465 3 3 Product_3 234 4 4 Product_4 567 4 (and so on.....) If anyone could provide me assistance with my troubled 'while loop' statement that would be much appreciated! Would like to be able to click on a radio button that represents an image. Once selected and submitted, have that image display on another page. I have an idea, but need some guidance. BTW, is using php only doable? Is there a simpler or more elegant way to do this? Thanks all! hirealimo.com.au/code1.php this works as i want it: Quote SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1 but apparelty selecting * is not a good thing???? but if I do this: Quote SELECT priceID, price FROM price INNER JOIN vehicle....etc it works but i lose the info from the vehicle table. but how do i make this work: Quote SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc so that i am specifiying which colums from which tables to query?? thanks I am inserting the two records below simultaneously (one after the other), but what I want to do write the second ONLY if the first isn't a duplicate. Help? Code: [Select] //FIRST INSERT mysql_query("INSERT INTO spam (id, scm_mem_id) VALUES('', '$social_mem_id' ) ON DUPLICATE KEY UPDATE scm_mem_id=$social_mem_id") or die(mysql_error()); //SECOND INSERT mysql_query("INSERT INTO sc_messages (smg_from, smg_to, smg_subject, smg_body, smg_sent_del, smg_postdate) VALUES ('$social_mem_id','1','$subject','$body','1','$time')") or die(mysql_error()); Hi there, I know I'm doing something stupid, but i've tried for hours to try and figure this out and I just cant seem to figure out why tutorials i've seen seem to work yet my own code doesnt. basically i've got two functions: getAll() and get($id) inside a class, calling getAll, returns an array with one of the same record as opposed to all the records (which is my intended result). Can anyone help? thanks. Code: [Select] class Test{ protected $_db; protected $_table = "table"; protected $_friends = array(); public function __construct($db) { $this->_db = $db; } public function getAll() { $db = $this->_db; $sql = "SELECT * FROM $this->_table"; $db->query($sql); while($data = $db->fetch_array()) { $this->_friends[$data['id']] = $this->get($data['id']); } return $this->_friends; } public function get($id) { $db = $this->_db; $sql = "SELECT * FROM $this->_table WHERE id='$id'"; $db->query($sql); $data = $db->fetch_array(); return (object)$data; } } $test = new Test(new mysqlConnClass()); $arrayOfAllAsObjects = $test->getAll(); Code: [Select] foreach ($items as $product ) { echo $product; ?><input type="text" size="2" value="<?php \\display quantity here ?>" name="quantity" /><?php } $items is an array. It has multiple values, some duplicate. How do i count the duplicate values in the array and display in this foreach loop? i know array_count_values($product) but putting that in the foreach loop won't accomplish what i want. another thing. i can get the foreach loop to not display duplicates by doing this Code: [Select] foreach (array_unique($items) as $product ) { echo $product; ?><input type="text" size="2" value="<?php \\display quantity here ?>" name="quantity" /><?php } how would i accomplish both. basically i want it to display the quantity without displaying duplicate rows. aka shopping cart. Hey guys just wondering if this piece of code is correct because i get a mysql error saying Duplicate entry '*********' for key 2 this is the code i have and table structure. $sql = " INSERT INTO `IP_Address` VALUES(NULL,'".$ip."','".$date."','".$time."') ON DUPLICATE KEY UPDATE `date` = `".$date."`, `time` = `".$time."` "; |
|||||||