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PHP - Using Value From Database In If Statement
Hi Can you please help me with an issue I am having? I have three description fields that I am pulling from a database. I did this so when it displays it isn't one long string. However the issue I am running into is if the 3rd description box is empty I do not what to display it. I cant figure out why it is only hitting the first display option. I created this variable $dishdesc3 = $_POST["dish_desc_3"]. I would like to say when it is NULL display the only the first 2 description boxes. Otherwise have it display all three description boxes. Is there a way to do that?
<?php
// Include the database configuration file
require_once 'con_php.php';
//check connection
if ($con -> connect_error){
die ("connection failed: " . $con -> connect_error);
}
$chsql = "SELECT * FROM dish_pic where cat_name = 'Chicken'";
$chresult = $con ->query($chsql);
$dishdesc3 = $_POST["dish_desc_3"];
if ($chresult-> num_rows >0){
while ($row = $chresult -> fetch_assoc()){
if ($dishdesc3 = "3")
{
echo"<table border='0' cellpadding='4' cellspacing='0' width='100%'>";
echo" <tr><td class='style1' align='center' colspan='3' valign='top'>". $row["dish_name"] .".........................$" . $row["price"] . "<br><td></tr>
<tr>
<td class='style2' align='center' >" . $row["dish_desc"] . "<br>
" . $row["dish_desc_2"] . "<br>
<tr><td class='style3' align='center' >Serving " . $row["dish_size"] . " Calories " . $row["dish_cal"] . " Total Carbs " . $row["dish_tot_carbs"] . " Net Carbs " . $row["dish_net_carbs"] . " Fat " . $row["dish_fat"] . " Fiber " . $row["dish_fiber"] ." Protein " . $row["dish_protein"] . "<br></td></tr>
<tr><td class='style2' align='center' >Recommend Side " . $row["dish_recommend"] . "</td></tr>
<br>
";
echo"</table>";
}
Else
{
echo"<table border='0' cellpadding='4' cellspacing='0' width='100%'>";
echo" <tr><td class='style1' align='center' colspan='3' valign='top'>". $row["dish_name"] .".........................$" . $row["price"] . "<br><td></tr>
<tr>
<td class='style2' align='center' >" . $row["dish_desc"] . "<br>
" . $row["dish_desc_2"] . "<br>
" . $row["dish_desc_3"] . "<br></td></tr>
<tr><td class='style3' align='center' >Serving " . $row["dish_size"] . " Calories " . $row["dish_cal"] . " Total Carbs " . $row["dish_tot_carbs"] . " Net Carbs " . $row["dish_net_carbs"] . " Fat " . $row["dish_fat"] . " Fiber " . $row["dish_fiber"] ." Protein " . $row["dish_protein"] . "<br></td></tr>
<tr><td class='style2' align='center' >Recommend Side " . $row["dish_recommend"] . "</td></tr>
<br>
";
echo"</table>";
}
}
}
$con->close();
?>
Similar TutorialsSorry going to be a big post, but trying to query a sql database for a value Original code <?php if($this->item->params->get('catItemExtraFields') && count($this->item->extra_fields)): ?> <!-- Item extra fields --> <div class="catItemExtraFields"> <h4><?php echo JText::_('K2_ADDITIONAL_INFO'); ?></h4> <ul> <?php foreach ($this->item->extra_fields as $key=>$extraField): ?> <?php if($extraField->value): ?> <li class="<?php echo ($key%2) ? "odd" : "even"; ?> type<?php echo ucfirst($extraField->type); ?> group<?php echo $extraField->group; ?>"> <span class="catItemExtraFieldsLabel"><?php echo $extraField->name; ?></span> <span class="catItemExtraFieldsValue"><?php echo $extraField->value; ?></span> </li> <?php endif; ?> <?php endforeach; ?> </ul> <div class="clr"></div> </div> <?php endif; ?> Now I want to check the name field for a value and if it is Closing Date print out something different from the value in value, so I've changed the red code to <span class="catItemExtraFieldsLabel"><?php if $extraField->name ="Closing date"; ?></span> { <span class="catItemExtraFieldsLabel"><?php echo $extraField->name; ?></span> <span class="catItemExtraFieldsValue"><?php echo $this->item->publish_down, JText::_('K2_DATE_FORMAT_LC2')); ?></span> } else { <span class="catItemExtraFieldsLabel"><?php echo $extraField->name; ?></span> <span class="catItemExtraFieldsValue"><?php echo $extraField->value; ?></span> } <?php endif; ?> Doesn't work though, tried a heap of variations but can't figure it out. Thanks Marc I have database that is a list of categories that I would like to insert into a switch statement. This is the only way that I know how to do it right off hand and of course it doesn't work. I've looked on the internet and found examples but they are slightly over my head as ways to do it. Depending on $group which comes from the url a title an h1 would be assigned. Code: [Select] $q = "SELECT * FROM categories"; $r = @mysqli_query ($dbc, $q); switch ($group) { if ($r) { //If $r ran OK while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)){ case '$row['id_cat']': $h1 = '.$row['group']'; break; } } } Thanks S just a little confused where ( brackets go in this statement . im just wanting to to display the information differently depending on which message is returned from the database. thanks Code: [Select] if ($info['message'])="<a href=\"freindacept.php?username=$myusername\">Has sent you a friend request </a>";{ Echo "<a href='viewprofile.php?username={$info['username']}'><img src='http://datenight.netne.net/images/".$info['img'] ."' width='30' height='30''></a>".$info['from_user']." ".$info['message']; }elseif ($info['message'])="You have sent a contact request to"; Echo " .$info['message']<a href='viewprofile.php?username={$info['username']}'><img src='http://datenight.netne.net/images/".$info['img'] ."' width='30' height='30''></a>".$info['from_user']."; } } Hello I am creating a simple Discussion Forum, and I cant get past my IF Statement to verify topics exist? Please help? <?php //check for required info from the query string if (!$_GET[topic_id]) { header("Location: topiclist.php"); exit; } //connect to server and select database $link = mysql_connect('votpservicescom.ipagemysql.com', 'mantest', 'testman') or die(mysql_error()); mysql_select_db("learn2db",$link) or die(mysql_error()); Hi everyone, I am currently making a page for a friend to upload a bunch of photos at a time. I was quite pleased that after a bit of googling and trial and error, I figured out how to do this so that multiple records could be added to my table with one submit button. However, my form has 10 browse iconcs. A few tests have revealed that my problem is that if only one picture is uploaded, I still get 9 entries in my database, which I don't want. My question is how can I alter the code so that a row is only populated in the database if an image is uploaded. I guess something that sort of says : if($imgx!="") { populate that row in the table } else { don't } ...and the same for $imgx002 through to $imgx010 The current query is below. Any pointers are much appreciated. Code: [Select] $query = "INSERT INTO photo_uploads (date, photo_name)" . "VALUES (NOW(), '$imgx'), (NOW(), '$imgx002'), (NOW(), '$imgx003'), (NOW(), '$imgx004'), (NOW(), '$imgx005'), (NOW(), '$imgx006'), (NOW(), '$imgx007'), (NOW(), '$imgx008'), (NOW(), '$imgx009'), (NOW(), '$imgx010')"; I'm having issues with the following: Code: [Select] <?php session_start(); $_SESSION['username']=$_POST['username']; $_SESSION['password']=$_POST['password']; if($_SESSION['username']=="username" && $_SESSION['password']=="password"){ if($_GET['product']=="add"){ $content.=' <p><label>Product Name:</label> <input type="text" name="product_name" size="30" /> <label>Product Price:</label> <input type="text" name="product_price" size="5" /> </p> <p><label>Product Category:</label> <input type="text" name="product_category" size="30" /></p> <p><label>Product Link:</label> <input type="text" name="product_link" size="30" /></p> <p><label>Product Image:</label> <input type="text" name="product_image" size="30" /></p> <p><label>Product Tag:</label> <input type="text" name="product_tag" size="30" /></p> <p><label>Product Keywords:</label> <input type="text" name="keyword" size="30" /></p> <p><label>Product Features:</label><br /> <textarea name="product_features" rows="10" cols="60"></textarea> </p> <p><label>Product Pros:</label><br /> <textarea name="product_pros" rows="5" cols="30"></textarea> </p> <p><label>Product Cons:</label><br /> <textarea name="product_cons" rows="5" cols="30"></textarea> </p> <p><label>Product Description:</label><br /> <textarea name="product_description" rows="10" cols="60"></textarea> </p> <p><label>Product Notes:</label><br /> <textarea name="product_notes" rows="5" cols="30"></textarea> </p> '; $logout='<div><a href="./acp_admincp.php?log-out">Log-Out</a></div>'; } elseif($_GET['product']=="view"){ } else{ $content.=' <a href="./admincp.php?product=add">Add New Product</a> <br /> <a href="./admincp.php?product=view">View Products</a> '; } } elseif(isset($_GET['log-out'])){ session_start(); session_unset(); session_destroy(); header("Location: ./admincp.php"); } else{ $content=' <form action="./admincp.php" method="post"> <p><label>Username:</label> <input type="text" name="username" size="30" />'; $content.='</p> <p><label>Password:</label> <input type="password" name="password" /></p>'; $content.='<p><input type="submit" value="Submit" name="Submit" /></p> </form>'; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en"> <head> <base href="http://ghosthuntersportal.com/" /> <title>Ghost Hunter's Portal - Admin Control Panel</title> <meta http-equiv="content-type" content="text/html; charset=utf-8" /> <meta name="verify-v1" content="" /> <meta name="keywords" content="ghost, hunters, hunter, ghosts, spirit, spirits, paranormal, investigation, investigator, investigators, k2, emf, meter, kii" /> <meta name="description" content="Ghost Hunters Potal. Parnormal research equipment store." /> <meta name="author" content="Andrew McCarrick" /> <meta name="robots" content="index, follow" <link rel="stylesheet" type="text/css" href="style.css" /> </head> <body> <img src="./logo.png" alt="Ghost Hunter's Portal Admin Control Panel" /> <br /> <div style="color: #AA44AA; font-size: 26px; margin-top: -30px; margin-left: 125px;">Admin Control Panel</div> <?php echo $logout; echo $content; ?> </body> </html> I can log-in, and get to the page with the two links on it. However, once I click one of the links it falls back to the log-in page, and it ends up being a never ending loop. It's doing this: Log-In --> Page with links ---> Log-In page again Should be doing this: Log-In --> Page with links --> Add Product page or View Products page I can never get into the the actual sub page. Just to be clear, the address bar actually shows product=add or product=view, but it still shows the log-in page. Hello Everyone, I have to change the if statements to a switch statement, in this game. Can anyone help? Thanks.
var checkKeyPressed = function (e) {
// Press key A or key D to make dog run to the left or right
// The running speed is specified by the step variable, 10 by default.
// If you replace 10 with a larger integer, the dog will run faster.
//press A, dog runs left
if (e.keyCode == "65") {
if (prex[0] > 5) {
prex[0] = prex[0] - step;
}
}
//press D, dog runs right
if (e.keyCode == "68") {
if (prex[0] < right) {
prex[0] = prex[0] + step;
}
}
};
At the moment I am creating a search function for my website. The approach I have in mind is a pseudo-PHP database. To give an example: A HTML form will submit the results to a PHP file. HTML FORM - Colour: Black PHP RESULT PAGE - if ($_POST['color'] == 'Black') {readfile("./products/black/*.html");} HTML FORM - Price: <$50 PHP RESULT PAGE - if ($_POST['Price'] == '<$50') {readfile("./products/less50/*.html");} The problem here is if there is an item that is black and costs less than $50, then its going to be listed twice. There is probably some code I can write to ommit the listing of duplicate entries, but it is probably going to be messy, so I am wondering if its better to use a centralized MySQL database, rather than a pseudo-PHP database? I've never used MySQL and don't know much about it and this is my first real attempt at using PHP. hello everyone, I am about to start coding my pages to display results from a database but before i do i want to know information about the following : Is it best to upload images to a database?and display them accordingly? or is it best to use images from a directory? What is most commonly used and or more reliable? Another topic i have trouble finding information on is actually positioning the output from you mysql database, is this practice done with tables?fields and rows? What is this method called? And is there more then one way to go about controlling result layout on your page? Sorry about the 1001 questions , but i am unable to find a clear answer on the topic ..especially question two . Thanks in advance. I have a very tricky php problem here. At least for me. On a page that is editing a job entry. I have a database generated group of checkboxes some of which have been checked when the job was entered. So the script has to do two things generate the complete list of checkboxes and check the ones that have already been selected in the job request. The below script is what I've done trying to figure it out. Description of what I'm trying to do: I'm generating the the boxes with a call to my database for the list of checkboxes. Then I'm building an array with all the possible items that could be checked to compare against. Ok here is where I think my first problem is. I need to make a second query to another table "worklog" in the same database. This is where the list of checked items are stored. What is the best way to do this? A second query seems wrong "or at least not efficient" Code: [Select] <div class="text">Job Type (<a class="editlist" href="javascript:editlist('listedit.php?edit=typelist',700,400);">edit list</a>):</div> <div class="field"> <?PHP $connection=mysql_connect ("localhost", "user", "password") or die ("I cannot connect to the database."); $db=mysql_select_db ("database", $connection) or die (mysql_error()); $query = "SELECT type FROM typelist ORDER BY type ASC"; $sql_result = mysql_query($query, $connection) or die (mysql_error()); $i=1; echo '<table valign="top"><tr><td>'; while ($row = mysql_fetch_array($sql_result)) { $type = $row["type"]; if ($i > 1 && $i % 26 == 0) echo '</td><td>'; else if ($i) echo ''; ++$i; $aTypes = array ("Spec Ad Campaign", "100 x 100 Logo", "100 x 35 (Featured Developer/Broker)", "100 x 40 (Featured Lender)", "120 x 180 (Home Page Auto)", "120 x 45 (Half Tile - Jobs)", "120x60 (Section Sponsor)", "125 x 40 (Profile Page logo)", "135 x 31 (Autos)", "135 x 60 (Autos)", "135 x 60 (Logotile - Jobs)", "150 x 40 (Auto Logo)", "160 x 240 (monster tile)", "160 x 400 (Skyscraper)", "160 x 600 (Tower)", "170x30 (Section Sponsor)", "240 x 180 JPG/GIF Auto Video", "240 x 180 size FLV video ", "300 x 250 (Story Ad)", "300 x 600 (Halfpage)", "468 x 60 (Banner Jpg/Gif)", "728 x 90 (Leaderboard)", "95 x 75 (Sweeps Logo) ", "Agent Profile Page", "Contest", "Creative Change", "Employer Profile", "Holiday Shop", "Home of the Week", "HP Half Banner (234 x 60 Static)", "Mobile Ads (320x53 *300x50*216x36*168x28)", "Newsletter", "Peelback", "Pencil Ad", "Print", "Real Deals", "Resize ad campaign", "Roll-Over Skyscraper", "Site Sponsor logo (170x30)", "Splash Page Production", "Travel Deals Update", "Video Ad", "Web Development", "Web Maintenance", "Web Quote (or Design)"); $dbTypes = explode(',',$row['type']); foreach ($aTypes as $type){ if(in_array($aType,$dbTypes)){ echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type' CHECKED><span style='color:#000;'>$type</span></input><br/>"; }else { echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type'><span style='color:#000;'>$type</span></input><br/>"; } } } echo '</td></tr></table>'; ?> </div> after cloasing connection of database i still got the values form database. Code: [Select] <?php session_start(); /* * To change this template, choose Tools | Templates * and open the template in the editor. */ require_once '../database/db_connecting.php'; $dbname="sahansevena";//set database name $con= setConnections();//make connections use implemented methode in db_connectiong.php mysql_select_db($dbname, $con); //update the time and date of the admin table $update_time="update admin set last_logged_date =CURDATE(), last_log_time=CURTIME() where username='$uname'limit 3,4"; //my admin table contain 5 colums they are id, username,password, last_logged_date, last_log_time $link= mysql_query($update_time); // mysql_select_db($dbname, $link); //$con=mysql_connect('localhost', 'root','ijts'); $result="select * from admin where username='a'"; $result=mysql_query($result); mysql_close($con); //here i just check after closing data baseconnection whether i do get reselts but i do, why? echo "after the cnnection was closed"; if(!$result){ echo "cont fetch data"; }else{ $row= mysql_fetch_array($result); echo "id".$row[0]."usrname".$row[1]."passwped".$row[2]."date".$row[3]."time".$row[4]; } // echo "<html>"; //echo "<table border='1' cellspacing='1' cellpadding='2' align='center'>"; // echo "<thead>"; // echo"<tr>"; // echo "<th>"; // echo ID; // echo"</th>"; // echo" <th>";echo Username; echo"</th>"; // echo"<th>";echo Password; echo"</th>"; // echo"<th>";echo Last_logged_date; echo "</th>"; // echo "<th>";echo Last_logged_time; echo "</th>"; // echo" </tr>"; // echo" </thead>"; // echo" <tbody>"; //while($row= mysql_fetch_array($result,MYSQL_BOTH)){ // echo "<tr>"; // echo "<td>"; // echo $row[0]; // echo "</td>"; // echo "<td>"; // echo $row[1]; // echo "</td>"; // echo "<td>"; // echo $row[2]; // echo "</td>"; // echo "<td>"; // echo $row[3]; // echo "</td>"; // echo "<td>"; // echo $row[4]; // echo "</td>"; // echo "</tr>"; // } // echo" </tbody>"; // echo "</table>"; // echo "</html>"; session_destroy(); session_commit(); echo "session and database are closed but i still get values from doatabase session is destroyed".$_SESSION['admin']; ?> session is destroyed but database connection is not closed. thanks I have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.
my form page test.php
<?PHP
require_once("./include/membersite_config.php");
if(!$fgmembersite->CheckLogin())
{
$fgmembersite->RedirectToURL("login.php");
exit;
}
?>
<form action="editform.php?id_user=<?= $fgmembersite->UserId() ?>" method="POST">
<input type="hidden" name="id_user" value="<?= $fgmembersite->UserId() ?>"><br>
Name:<br> <input type="text" name="name" size="40" value="<?= $fgmembersite->UserFullName() ?>"><br><br>
Email:<br> <input type="text" name="email" size="40" value="<?= $fgmembersite->UserEmail() ?> "><br><br>
Address:<br> <input type="text" name="address" size="40" value="<?= $fgmembersite->UserAddress() ?> "><br><br>
<button>Submit</button>
my editform.php
<?php
$con = mysqli_connect("localhost","root","user","pass");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"UPDATE fgusers3 SET name = '".$_POST['name']."', email= '".$_POST['email']."', address= '".$_POST['address']."' WHERE id_user='".$_POST['id_user']."'");
header("Location: test.php");
?>
Hi guys, I was just wondering if anyone could help me. I've got a My_SQL database containing articles, a summary for the article and a date. I have a basic CMS system set-up, but I want to create a script that when users sign up to a mail list it forwards the summary and dates of the articles database. If that makes sense? But I only want it to forward the most recent 5 rows. I'm pretty new to PHP and I've been mostly following tutorials thus far, but this is quite specific. Thanks in advance! hello I want query from one table and insert in another table on another domain . each database on one domain name. for example http://www.site.com $con1 and http://www.site1.com $con. can anyone help me? my code is : <?php $dbuser1 = "insert in this database"; $dbpass1 = "insert in this database"; $dbhost1 = "localhost"; $dbname1 = "insert in this database"; // Connecting, selecting database $con1 = mysql_connect($dbhost1, $dbuser1, $dbpass1) or die('Could not connect: ' . mysql_error()); mysql_select_db($dbname1) or die('Could not select database'); $dbuser = "query from this database"; $dbpass = "query from this database"; $dbhost = "localhost"; $dbname = "query from this database"; // Connecting, selecting database $con = mysql_connect($dbhost, $dbuser, $dbpass) or die('Could not connect: ' . mysql_error()); mysql_select_db($dbname) or die('Could not select database'); //query from database $query = mysql_query("SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1"); while($row=mysql_fetch_array($query)){ $result=$row[0]; $text=$row[1]."</br>Size:(".$row[4].")"; $alias=$row[2]; $link = '<a target="_blank" href='.$row[3].'>Download</a>'; echo $result; } //insert into database mysql_query("SET NAMES 'utf8'", $con1); $query3= " INSERT INTO `jos_content` (`id`, `title`, `alias`, `) VALUES (NULL, '".$result."', '".$alias."', '')"; if (!mysql_query($query3,$con1)) { die('Error: text add' . mysql_error()); } mysql_close($con); mysql_close($con1); ?> the second database found on the cloud
i try to get JSON data but how to insert and update them to another online database with the same table my php script to return json data <?php include_once('db.php'); $users = array(); $users_data = $db -> prepare('SELECT id, username FROM users'); $users_data -> execute(); while($fetched = $users_data->fetch()) { $users[$fetchedt['id']] = array ( 'id' => $fetched['id'], 'username' => $fetched['name'] ); } echo json_encode($leaders);
i get
{"1":{"id":1,"username":"jeremia"},"2":{"id":2,"username":"Ernest"}} Edited March 24 by mahenda Good morning,
I am trying to return specific values depending on the value of the WJCStatusID, for example if WJCStatusID = < 4 but > 2 return "In production"
SQL QUERY:
SELECT tblWJCItem.AddedDescription,
tblWJC.WJCPrefix + Convert(Varchar(10),tblWJC.WJCNo) AS OurRef,
tblWJCItem.MaterialName, tblStockFamily.StockFamily,
tblWJCItem.WeightToSend,
tblWJC.DateCreated,
tblWJC.WJCStatusID
FROM tblWJC
INNER JOIN tblWJCItem ON tblWJC.WJCID = tblWJCItem.WJCID
INNER JOIN tblStockFamily ON tblWJCItem.ProductFamilyID = tblStockFamily.StockFamilyID
IF (tblWJC.WJCStatusID) < 2 THEN 'Pre Production'
ELSE IF (tblWJC.WJCStatusID) < 4 THEN 'In Production'
ELSE IF (tblWJC.WJCStatusID) > 4 THEN 'Ready To Ship'
ELSE 'Awaiting Lab Results';
I am quite new to the world of mssql so I maybe doing something wrong which is quite simply to fix. Can any body help?Hello I'm hoping someone here can help me with this. I know very little PHP I'm afraid but I think what I want to do is fairly basic and should be relatively simple. I have a footer which is referenced on each page of the site with <?php include($DOCUMENT_ROOT."/includes/php/footer.php"); ?> Within the footer I want it so that if it's the home page (default.php) it displays just the footer but for any other page I want it to display a Div and then the footer. Here is what I have so far but it isn't working: <?php if ($_SERVER['REQUEST_URI'] == "default.php") { ?> <!--Home Page Footer--> <div class="footer"> <div class="footer1"> </div> <div class="footer2"> </div> <div class="footer3"> </div> <div class="footerLinks"> <?php include($DOCUMENT_ROOT."/includes/php/bottomlinks.php"); ?> </div> <div class="footer4"> </div> </div> <?php } else { ?> <!--Blue Footer Image--> <div class="footer2"> </div> <!--Footer--> <div class="footer"> <div class="footer1"> </div> <div class="footer2"> </div> <div class="footer3"> </div> <div class="footerLinks"> <?php include($DOCUMENT_ROOT."/includes/php/bottomlinks.php"); ?> </div> <div class="footer4"> </div> </div> <?php } ?> I don't know that the request uri address is correct but when I've tried to use 'echo $_SERVER['REQUEST_URI']' to find out what it should be I can't get that to work either. I'm new here, this is my first post. Sorry if it's too basic or I've made a forum faux pas. Any help would be extremely appreciated. Hi All, I have an SQL query that returns a date (31/01/2011 - for example) and this get's placed into a variable - $req_date. I want to be able to display an image when the required date is within 5 days of the current date. Normally in an if statement I would use: <?php if ($ticket_no == 100) { echo "This is ticket 100" } else { echo "This is not ticket 100" ?> However I am struggling to work out the difference between the 2 dates. Any ideas? Cheers Matt Hello all, I need your help. I have two pages. This is the 1st page: <?php include "include/dbc.php"; include "include/header.inc"; ?> <style type="text/css"> .mydate{ color:#00F; text-decoration:underline; cursor:pointer; } </style> <script type="text/javascript"> function displayDate(d){ var date=new Date(); var D=date.getDate(); date.setDate(D+d); var YYYY=date.getFullYear(); var MM=date.getMonth()+1; MM<10?MM='0'+MM:null; var DD=date.getDate(); DD<10?DD='0'+DD:null; var span=document.getElementById('date'); span.innerHTML= 'Entries for '+MM+'/'+DD+'/'+YYYY; } onload=function(){displayDate(0)}; </script> <h1>Food Diary</h1> <div class="full"> <center><div><span class="mydate" onclick="displayDate(-1)"><img src="images/left_arrow.png" border="0">Yesterday</span> <span id="date" style="font-size:2em;"></span> <span class="mydate" onclick="displayDate(1)">Tomorrow<img src="images/right_arrow.png" border="0"></span></div><br /> <a href="#" onclick="displayDate(0);return false;">Today</a> </center> <div class="full"> <form name="exercise" id="exercise" method="GET" action=""> <center><table> <tr> <td><h3>Add an Activity</h3></td> </tr> <tr> <td><input name="NewSearchString" style="width: 100px" type="text"/> <input type="submit" value="Search" /> </td> </tr> <tr> <td> <select name="activity"> <option value="_">Activity Browse...</option> <option value="all">All Activities</option> <option value="biking">Biking</option> <option value="condition">Conditioning</option> <option value="dancing">Dancing</option> <option value="fish">Fishing & Hunting</option> <option value="Home">Home Activities</option> <option value="misc">Miscellaneous</option> <option value="music">Music Playing</option> <option value="occupation">Occupation</option> <option value="running">Running</option> <option value="sports">Sports</option> <option value="walking">Walking</option> <option value="water">Water Activities</option> <option value="winter">Winter Activities</option> </select> <input type="submit" value="Submit" /></td></tr></table></center></form> </td> </tr> </table> </center> <table width="100%"> <tr bgcolor="#66CC33"> <td><div>Activity</div></td> <td><div>Specific Activity</div></td> <td><div>Time (hh:mm)</div></td> <td><div>Distance</div></td> <td><div>Units</div></td> </tr> <tr bgcolor="#66CC33"> <td><div></div></td> <td><div></div></td> <td><div></div></td> <td><div class="Float"></div></td> <td class="cp_Distance"><div></div></td> </tr> <?php if(isset($_GET[activity])) { $category=$_GET[activity]; $result = mysql_query("SELECT * FROM exercise WHERE type='$category'"); ?> <form action="add_activity.php" method="POST"> <?php while($row = mysql_fetch_array($result)) { echo '<tr><td><div>'.$row[Type].'</div></td>'; echo '<td><div>'.$row[Name].'<input type="hidden" name="exerciseid[]" value="'.$row[Name].'"></div></td>'; echo '<td><div><input type="text" name="duration['.$row['Name'].']" value=""></div></td>'; echo '<td><div><input type="text" name="distance['.$row['Name'].']" value ""></div></td>'; echo '<td><div><select name="metric[]"> <option value="mile" name="mile">mile</option> <option value="Km" name="Km">km</option> <option value="M" name="M">m</option> <option value="Yard" name="yard">yrd</option> <option value="Feet" name="feet">ft</option> </select></div></td></tr>'; } mysql_close(); ?> <tr><td colspan="6" align="center"><input type="submit" name="submit" value="Add Activities" onClick="return confirm( 'Are you sure you want to submit the activities?');"></td></tr> </form> <?php } ?> <tr bgcolor="#66CC33"> <td><div></div></td> <td><div></div></td> <td><div></div></td> <td><div class="Float"></div></td> <td class="cp_Distance"><div></div></td> </tr></table> This page contains a table that is pulling info from a database stored in phpmyadmin. The values entered in that table are passed to the second page: <?php include "include/dbc.php"; include "include/header.inc"; $exerciseid = $_POST["exerciseid"]; $duration = $_POST["duration"]; $distance = $_POST["distance"]; $metric = $_POST["metric"]; echo'<h1>Added Activities</h1>'; // name of array echo '<h1>Exercise</h1>'; if (is_array($exerciseid)) { foreach ($exerciseid as $key => $value) { echo $key .' : '. $value .'<br />'; } } echo '<h1>Duration</h1>'; if ($duration == ""){ echo "No value"; } // name of array echo '<h1>Distance</h1>'; if (is_array($distance)) { foreach ($distance as $key => $value) { echo $key .' : '. $value .'<br />'; } } // name of array echo '<h1>Metric</h1>'; if (is_array($metric)) { foreach ($metric as $key => $value) { echo $key .' : '. $value .'<br />'; } } ?> The problem is that ALL the values, empty or w/content, are being passed. So I was wondering if there was a way to pass only user input with an if/else statement. Something like: if ($duration == ""){ echo "No value"; else echo duration; } This, however, is not working and I would like to know what possibly could work that would be structured similarly. Thanks in advance. Hi I am trying to perform an echo yes if field from query is equal to one or no if anything else. See the code below, Now its erroring saying unexpected t variable in string and its also saying expecting ' It seems to be erroring at the start of my IF statement. So can anyone please help with what I have done wrong. This will be my first personal written IF statement so if it is miles off i apologise. <?php $qGetCat = ("SELECT * FROM category ORDER BY `id`"); $rGetCat = mysql_query($qGetCat) or die(mysql_error()); while($Cat = mysql_fetch_assoc($rGetCat)) { ?> <td height="31" align="left" nowrap="nowrap"><?= $Cat['id'] ?></td> <td align="left" nowrap="nowrap"><?= $Cat['cat1'] ?></td> <td align="left" nowrap="nowrap"> <?php if $Cat['top']='1' { echo "Yes";} else { echo "No";} ?> </td> <td><a href="updatetopcategory.php?=<?= $Cat['id'] ?>"><img src="../images/Edit.png" width="29" height="29" /></a></td> <td><img src="../images/delete.jpg" width="29" height="29" /></td> <? } ?> |
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