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PHP - Only Displays One Row Of Data
Hi.
I am pretty new to this and am having difficulty looping through rows. I have the code below which first selects a name based upon the id sent from a GET on a separate page. It then has a second select query to get the data from all rows which have this name. I then want it to display the name, and city once but all of the comments. The code below only shows one of each though. Thanks in advance Code: [Select] <?php $id = $_GET['id']; // Retrieve data from database $sql="SELECT landlord_name AS name FROM $tbl_name WHERE id=$id"; $query=mysql_query($sql); //loop through the rows while ($rows = mysql_fetch_array($query)){ $name = $rows['name']; } $sql2="SELECT city, landlord_name, landlord_comments FROM $tbl_name WHERE landlord_name= '$name'"; $query2=mysql_query($sql2) or die(mysql_error()); while ($data = mysql_fetch_array($query2)){ ?> <table width="600" border="1" cellspacing="0" cellpadding="0" align="center"> <tr> <div id="name"><? echo $name ?></div> <div id="city"><? echo $data['city']; ?></div> <div id="comments"><? echo $data['landlord_comments']; ?></div> </tr> </table> <? } //End While ?> Similar TutorialsI must be mixing apples and oranges here or something trying to get two columns/fields of MySQL data to display.
The basic HTML page display OK and there are no MySQL Connection errors (finally resolved those).
<HTML snipped>
<?php
require '...<URL snipped>...';
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="SELECT `name`, `id` FROM `roster`";
$result = mysqli_query($con,$sql);
$num=mysqli_num_rows($result);
mysqli_close($con);
?>
<table border="0" cellspacing="2" cellpadding="2">
<tr>
<td>
<font face="Arial, Helvetica, sans-serif">NAME</font>
</td>
<td>
<font face="Arial, Helvetica, sans-serif">ID</font>
</td>
</tr>
<?php
function mysqli_result($res, $row, $field=0) {
$res->data_seek($row);
$datarow = $res->fetch_array();
return $datarow[$field];
}
$i=0;
while ($i < $num) {
$f1=mysqli_result($result,$i,$datarow[$field]);
$f2=mysqli_result($result,$i,$datarow[$field]);
?>
<tr>
<td>
<font face="Arial, Helvetica, sans-serif"><?php echo $f1; ?></font>
</td>
<td>
<font face="Arial, Helvetica, sans-serif"><?php echo $f2; ?></font>
</td>
</tr>
<?php
$i++;
}
?>
</table>
<?php
?>
<HTML snipped>
Any assistance is appreciated.
Thanks very much.
- FreakingOUT
Hi There, I have the following code: two arrays: $user_body_group & $DB_Seconday_muscles I want to look up the "$DB_Seconday_muscles" Array and search for the elements in the "$user_body_group" array The result would be displaying a checkbox list with only the items in the "$user_body_group" array checked: foreach($DB_Seconday_muscles as $value) { echo "<input name=\"colors[]\" type=\"checkbox\" value=\"$value\""; if (in_array($value,$user_body_group)) { echo "CHECKED"; } echo "> $value "; } BUT... When i run thi script i only get the first element in the searched array ticked. Can anyone help! Many Thanks! Leatfield I have set up two directories in www which reflect two domains I have using uniform server z which is a wamp package.
I have set up Vhosts also.
The sever is accessible from the internet The directories hold a simple hello world page ( one index HTML and one index.php) I set these up on uniform server z to test access as I have previously had problems getting virtual host to work on coral 6.8 I can test the IP address fine and ping the sever. I can access the two addresses from the net (not on my local network) How ever when I access either site externally I get the Index of/ My DirectoryIndex in the main apache config allows index.php & index.html. I also have that in the .htaccess and in the Vhosts ht access. The Index of however only shows a favicon and not the index.html/php Any help appreciated. Howdy! A client's web page is .shtml and includes various PHP files via standard virtual include: Code: [Select] <!--#include virtual="session.php" --> These PHP includes worked fine until recently, when PHP files that produce no visible output began displaying a 0 (zero). For instance, the session.php file above is here in it's entirety: Code: [Select] <?php session_start(); ?> The include is on the first line of the .shtml file, and includes that bit of code, but displays a 0 in the browser. If I echo anything at all the 0 goes away, but of course we don't want to echo anything before the <html> tag in the .shtml file. The short-term and unacceptable solution is to echo an empty HTML comment, as follows: Code: [Select] <?php session_start(); echo "<!-- -->"; ?> As noted, this causes the 0 to disappear but now we have <!-- --> before the HTML. Repeat: This was not happening when the .shtml and PHP was installed, and the files have not changed. Therefore, it seems that this problem is caused by a change in PHP and/or Apache configuration. Has anyone seen this and/or can anyone explain what's going on? Thanks! First, I would like to say when i tried to recover my account from this website, I took me 10 attempts to get the captcha right and then finally it said it sent me an email to my gmail account. I checked spam folder and everything there was no such email from this site. Then I decided to create a new account, well, it took me another 10 attempts to get the captcha right and finally when it was submitted, the page was loading for around 3 minutes before it signed me in.
My question is about the php date() function. It accepts a format to display a time. In the following example I use F for full representation of month, d for 2-digit day of month with leading zeros, Y for full year, g for 12-hour format without leading 0s, s for seconds and A for meridiem. It uses the correct format, but it gives me the wrong time. My local time is 4:43 and it prints out 4:12:
<?php echo "<p>order processed on " . date("F d, Y g:sA") . "</p>"; ?> Why is it 30 minutes behind?
Hi, I need code that reads from the roles database and then selects which file from these 3 which I want. For example, the user.php file would be loaded if the user has UName = user, Pass = 124, and Roles = User added to the database. But the admin.php and boss.php files would not appear to him.
<?php
session_start();
if(!(isset($_SESSION['User'])))
{
header("Location: index.php");
exit(0);
}
?>
<!DOCTYPE html>
<html>
<body>
<?php
include "config.php";
?>
<!--show for User-->
<?php include 'user.php';?>
<!--show for Admin-->
<?php include 'admin.php';?>
<!--show for Boss-->
<?php include 'boss.php';?>
</body>
</html>
Hi I have a list of states using the array method in a form. The drop down menu works fine. I want to save the user choice,if the form is re-displayed due to a blank field or pattern mismatch. I know I can use the selected=selected, but don't know wher to put the statement: My array is: state_province = array ("list of states", "provinces") Var in my labels array is "state"=>"state" Here is my code for the select/option statement: { if($field == "state") { echo "<div class='province_state'><label for='state' size='10'>* Province/State</label><select>"; foreach($state_province as $state) { echo "\n<option value='$state_province' /> "; echo $state ; echo "</option>"; } echo "</select></div>\n"; } ?Is this the correct code to add and where would I add it? if(@$_POST['state'] == $value) { echo "selected='selected' "; } Hi, I am not a PHP programmer. I took on a new client with a simple PHP site, without any databases. The site is up and running on the web. I would like to get it running on my local machine for further development. I have latest version of WAMP installed, running Apache version 2.2.11 and PHP version 5.3.0 I created a directory in the WAMP "www" project directory and it shows up there like it's supposed to when I browse to "localhost" Problem: The home page of website displays text but no, images, styles, footer, header, nav links, etc. Here is the code for the home page: <? define("NAV","home"); require_once('local/local.php'); ?> <!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> <title>TITLE</title> <meta name="keywords" content=""> <meta name="Description" content=""> <? include("common/dochead.php"); ?> </head> <body onLoad="<? include('common/preloads.php'); ?>"> <!-- ============================ main ============================= --> <div id="main-frame"><div id="main" class="noCollapse"> <? include("common/sign.php"); ?> <div id="right-frame"> <? include("common/navigation.php"); ?> <div id="content-frame"> <div id="content"> <h1>Welcome</h1> <p>This is the content area. This is the content area. This is the content area. </p> </div><!-- end content --> </div><!-- end content-frame --> </div><!-- end right-frame --> <div class="clearFloats"></div> </div><!-- end main --></div><!-- end main-frame --> <? include("common/footer.php"); ?> </body> </html> Any help would be greatly appreciated. I have spent many hours on this. Regards Below is my output on the browser: Student: Kevin Smith (u0867587) Course: INFO101 - Bsc Information Communication Technology Course Mark 70 Grade Year: 3 Module: CHI2550 - Modern Database Applications Module Mark: 41 Mark Percentage: 68 Grade: B Session: AAB Session Mark: 72 Session Weight Contribution 20% Session: AAE Session Mark: 67 Session Weight Contribution 40% Module: CHI2513 - Systems Strategy Module Mark: 31 Mark Percentage: 62 Grade: B Session: AAD Session Mark: 61 Session Weight Contribution 50% Now where it says course mark above it says 70. This is incorrect as it should be 65 (The average between the module marks percentage should be 65 in the example above) but for some stange reason I can get the answer 65. I have a variable called $courseMark and that does the calculation. Now if the $courseMark is echo outside the where loop, then it will equal 65 but if it is put in while loop where I want the variable to be displayed, then it adds up to 70. Why does it do this. Below is the code: Code: [Select] $sessionMark = 0; $sessionWeight = 0; $courseMark = 0; $output = ""; $studentId = false; $courseId = false; $moduleId = false; while ($row = mysql_fetch_array($result)) { $sessionMark += round($row['Mark'] / 100 * $row['SessionWeight']); $sessionWeight += ($row['SessionWeight']); $courseMark = ($sessionMark / $sessionWeight * 100); if($studentId != $row['StudentUsername']) { //Student has changed $studentId = $row['StudentUsername']; $output .= "<p><strong>Student:</strong> {$row['StudentForename']} {$row['StudentSurname']} ({$row['StudentUsername']})\n"; } if($courseId != $row['CourseId']) { //Course has changed $courseId = $row['CourseId']; $output .= "<br><strong>Course:</strong> {$row['CourseId']} - {$row['CourseName']} <strong>Course Mark</strong>" round($courseMark) "<strong>Grade</strong> <br><strong>Year:</strong> {$row['Year']}</p>\n"; } if($moduleId != $row['ModuleId']) { //Module has changed if(isset($sessionsAry)) //Don't run function for first record { //Get output for last module and sessions $output .= outputModule($moduleId, $moduleName, $sessionsAry); } //Reset sessions data array and Set values for new module $sessionsAry = array(); $moduleId = $row['ModuleId']; $moduleName = $row['ModuleName']; } //Add session data to array for current module $sessionsAry[] = array('SessionId'=>$row['SessionId'], 'Mark'=>$row['Mark'], 'SessionWeight'=>$row['SessionWeight']); } //Get output for last module $output .= outputModule($moduleId, $moduleName, $sessionsAry); //Display the output echo $output; I think the problem is that it is outputting the answer of the calculation only for the first session mark. How in the while loop can I do it so it doesn't display it for the first mark only but for all the session marks so that it ends up showing the correct answer 65 and not 72? Am new here - looks like a great foru! I would sincerely appreciate any help anyone can give me. I have been trying to solve my problem for hours and I am not having any luck, so I thought I would post and see if anyone can help. I am very stuck and am not making much progress on this project, and I am certain the answer is very simple. I am constructing a form to collect data for a specialized purpose. The form and program actually work for its intended function, but I am trying to enhance the user experience by preventing customers from having to reenter all of their data should there be a problem with any of the data submitted. I have been able to do that with the contact form portion, but what I am having trouble with is the portion which has as many as 400 possible entries. So, in a nutshell, if the customers contact data is incomplete or in error, the form will ask them to return to the page and correct things. The previous data entered has been saved in the session and the input value will equal the previous entry. i.e. <tr> <td align="right" class="infoBox"><?php echo ENTRY_EMAIL_ADDRESS; ?></td> <td align=left><?php echo "<input type=text name='cemail' value=\"$cemail\" size=35 maxlength=35>" ?></td> </tr> Works perfectly, all well and good there. On the other 400 more or less entries, I am having a difficult time tweaking the string concatenation to work to achieve similar results. There are 4 columns each with $points entries asking for a dimension in either feet or inches. The <input name=> is one of ptaf,ptai,ptbf,ptbi, appended programatically with the corresponding row number or data point. i.e. "ptaf1", "ptai1", etc... This is produced by the example below and works perfectly also. <?php { $points=100; $i=1; while ($i <= $points) {echo ' <tr><td align="center" width="6"><b> ' .$i . '</b></td> <td align="right" NOWRAP>A' .$i . ' (ft) <input type="text" name="ptaf'.$i.'" size=4 maxlength=3> </td> <td align="right" NOWRAP>A' .$i . ' (in) <input type="text" name="ptai'.$i.'" size=4 maxlength=4> </td> <td align="right" NOWRAP>B' .$i . ' (ft) <input type="text" name="ptbf'.$i.'" size=4 maxlength=3> </td> <td align="right" NOWRAP>B' .$i . ' (in) <input type="text" name="ptbi'.$i.'" size=4 maxlength=4> </td> '; $i++; } } ?> I am trying to add <input value=$ptai.$i> for each field but as I mentioned I am not having any luck. It seems as if I have tried every combination imagineable, but still no luck. My head is spinning! The closest I seem to have gotten was with this: <td align="right" NOWRAP>A' .$i . ' (ft) <input type="text" size=6 maxlength=3 name="ptaf'.$i.'" value="' . "$ptaf" . $i . '" ></td> But line 17 for example returns this: <input type="text" value="17" name="ptaf17" maxlength="3" size="6"> To recap, I am trying to have the value set to whatever the customer may have entered previously. Again, I would most appreciate any help anyone can give me. If you need clarification on anything please let me know. Thanks AJ Hello to all, I have problem figuring out how to properly display data fetched from MySQL database in a HTML table. In the below example I am using two while loops, where the second one is nested inside first one, that check two different expressions fetching data from tables found in a MySQL database. The second expression compares the two tables IDs and after their match it displays the email of the account holder in each column in the HTML table. The main problem is that the 'email' row is displayed properly while its while expression is not nested and alone(meaning the other data is omitted or commented out), but either nested or neighbored to the first while loop, it is displayed horizontally and the other data ('validity', 'valid_from', 'valid_to') is not displayed.'
Can someone help me on this, I guess the problem lies in the while loop? <thead> <tr> <th data-column-id="id" data-type="numeric">ID</th> <th data-column-id="email">Subscriber's Email</th> <th data-column-id="validity">Validity</th> <th data-column-id="valid_from">Valid From</th> <th data-column-id="valid_to">Valid To</th> </tr> </thead> Here is part of the PHP code:
<?php
while($row = $stmt->fetch(PDO::FETCH_ASSOC))
{
echo '
<tr>
<td>'.$row["id"].'</td>
';
while ($row1 = $stmt1->fetch(PDO::FETCH_ASSOC)) {
echo '
<td>'.$row1["email"].'</td>
';
}
if($row["validity"] == 1) {
echo '<td>'.$row["validity"].' month</td>';
}else{
echo '<td>'.$row["validity"].' months</td>';
}
echo ' <td>'.$row["valid_from"].'</td>
<td>'.$row["valid_to"].'</td>
</tr>';
}
?>
Thank you. Hi, after banging my head against the wall for a while thinking this would be a simple task, I'm discovering that this is more complicated than I thought. Basically what I have is a link table linking together source_id and subject_id. For each subject there are multiple sources associated with each. I had created a basic listing of sources by subject... no problem. I now need a way of having a form to create an ordered list in a user-specified way. In other words, I can currently order by id or alphabetically (subject name lives on a different table), but I need the option of choosing the order in which they display. I added another row to this table called order_by. No problem again, and I can manage all of this in the database, however I want to create a basic form where I can view sources by subject and then enter a number that I can use for sorting. I started off looping through each of the entries and the database (with a where), and creating a foreach like so (with the subject_id being grabbed via GET from the URL on a previous script) Code: [Select] while($row = mysqli_fetch_array($rs)) { //update row order if (isset($_POST['submit'])) { //get variables, and assign order $subject_id = $_GET['subject_id']; $order_by = $_POST['order_by']; $source_id = $row['source_id']; //echo 'Order by entered as ' . $order_by . '<br />'; foreach ($_POST['order_by'] as $order_by) { $qorder = "UPDATE source_subject set order_by = '$order_by' WHERE source_id = '$source_id' AND subject_id = '$subject_id'"; mysqli_query($dbc, $qorder) or die ('could not insert order'); // echo $subject_id . ', ' . $order_by . ', ' . $source_id; // echo '<br />'; } } else { $subject_id = $_GET['subject_id']; $order_by = $row['order_by']; $source_id = $row['source_id']; } And have the line in the form like so: Code: [Select] echo '<input type="text" id="order_by" name="order_by[]" size="1" value="'. $order_by .'"/> (yes I know I didn't escape the input field... it's all stored in an htaccess protected directory; I will clean it up later once I get it to work) This, of course, results in every source_id getting the same "order_by" no matter what I put into each field. I'm thinking that I need to do some sort of foreach where I go through foreach source_id and have it update the "order_by" field for each one, but I must admit I'm not sure how to go about this (the flaws of being self-taught I suppose; I don't have anyone to go to on this). I'm hoping someone here can help? Thanks a ton in advance This could be PHP or MySql so putting it in PHP forum for now... I have code below (last code listed) which processes a dynamically created Form which could have anywhere from 0 to 6 fields. So I clean all fields whether they were posted or not and then I update the mySQL table. The problem with this code below is that if, say, $cextra was not posted (i.e. it wasnt on the dynamically created form), then this code would enter a blank into the table for $cextra (i.e. if there was already a value in the table for $cextra, it gets overwritten, which is bad). What is the best way to handle this? I'm thinking i have to break my SQL query into a bunch of if/else statements like this... Code: [Select] $sql = "UPDATE cluesanswers SET "; if (isset($_POST['ctext'])){ echo "ctext='$ctext',"; } else { //do nothing } and so on 5 more times.... That seems horribly hackish/inefficient. Is there a better way? Code: [Select] if (isset($_POST['hidden']) && $_POST['hidden'] == "edit") { $cimage=trim(mysql_prep($_POST['cimage'])); $ctext=trim(mysql_prep($_POST['ctext'])); $cextra=trim(mysql_prep($_POST['cextra'])); $atext=trim(mysql_prep($_POST['atext'])); $aextra=trim(mysql_prep($_POST['aextra'])); $aimage=trim(mysql_prep($_POST['aimage'])); //update the answer edits $sql = "UPDATE cluesanswers SET ctext='$ctext', cextra='$cextra', cimage='$cimage', atext='$atext', aextra='$aextra', aimage='$aimage'"; $result = mysql_query($sql, $connection); if (!$result) { die("Database query failed: " . mysql_error()); } else { } Say there is a complex opt in process where people start to enter their data but certain questions stop them where they close out of the page. They already entered their data and I feel there is a way to grab it and post it to mysql even though they do not click submit.
How would this be done?
A super simple example (proof of concept) or a link to a tutorial would be very useful.
Edited by brentman, 23 September 2014 - 10:42 AM. Here's the code that deals with the client side:
<?php
session_start();
if(!isset($_SESSION['Logged_in'])){
header("Location: /page.php?page=login");
}
?>
<!DOCTYPE Html>
<html>
<head>
<!--Connections made and head included-->
<?php require_once("../INC/head.php"); ?>
<?php require_once("../Scripts/DB/connect.php"); ?>
<!--Asynchronously Return User Names-->
<script>
$(document).ready(function(){
function search(){
var textboxvalue = $('input[name=search]').val();
$.ajax(
{
type: "GET",
url: 'search.php',
data: {Search: textboxvalue},
success: function(result)
{
$("#results").html(result);
}
});
};
</script>
</head>
<body>
<div id="header-wrapper">
<?php include_once("../INC/nav2.php"); ?>
</div>
<div id="content">
<h1 style="color: red; text-align: center;">Member Directory</h1>
<form onsubmit="search()">
<label for="search">Search for User:</label>
<input type="text" size="70px" id="search" name="search">
</form>
<a href="index.php?do=">Show All Users</a>|<a href="index.php?do=ONLINE">Show All Online Users</a>
<div id="results">
<!--Results will be returned HERE!-->
</div>
search.php
<?php //testing if data is sent ok echo "<h1>Hello</h1><br>" . $_GET['search']; ?>This is the link I get after sending foo. http://www.family-li...php?&search=foo Is that mean it was sent, but I'm not processing it correctly? I'm new to the whole AJAX thing. I have two tables. Table Name:Users Fields: User_name user_email user_level pwd 2.Reference Fields: refid username origin destination user_name in the users table and the username field in reference fields are common fields. There is user order form.whenever an user places an order, refid field in reference table will be updated.So the user will be provided with an refid Steps: 1.User needs to log in with a valid user id and pwd 2.Once logged in, there will be search, where the user will input the refid which has been provided to him during the time of order placement. 3.Now User is able to view all the details for any refid 3.Up to this we have completed. Query: Now we need to retrieve the details based on the user logged in. For eg: user 'USER A' has been provided with the referenceid '1234' during the time of order placement user 'USER B' has been provided with the referenceid '2468' during the time of order placement When the userA login and enter the refid as '2468' he should not get any details.He should get details only for the reference ids which is assigned to him. <?php session_start(); if (!$_SESSION["user_name"]) { // User not logged in, redirect to login page Header("Location: login.php"); } $con = mysql_connect('localhost','root',''); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("login", $con); $user_name = $_POST['user_name']; $refid = $_POST['refid']; $query = "SELECT * from reference,users WHERE reference.username=users.user_name AND reference.refid='$refid' AND "; $result = mysql_query($query) or trigger_error('MySQL encountered a problem<br />Error: ' . mysql_error() . '<br />Query: ' . $query); while($row = mysql_fetch_array($result)) { echo $row['refid']; echo $row['origin']; echo $row['dest']; echo $row['date']; echo $row['exdate']; echo $row['username']; } echo "<p><a href=\"logout.php\">Click here to logout!</a></p>"; ?> <html> <form method="post" action="final.php"> Ref Id:<input type="text" name="refid"> <input type="submit" value="submit" name="submit"> </html> hi guys, im new to this forum I'm new also to php, I need help from you guys: I want to display personal information from a certain person (the data is on the mysql database) using his name as a link: example: (index.php) names 1. Bill Gates 2. Mr. nice Guy i want to click Bill Gates (output.php) Name: Bill Gates Country:xxxx Age: xx etc. How can i make this or how to learn this? I can add and delete data from my table. Now I need to be able to change one or more fields in an entry. So I want to retrieve a row from the db, display that data on a form where the user can change any field and then pass the changed data to an update.php program. I know how to go from form to php. But how do I pass the data from retrieve.php to a form so it will display? Do I use a URL and Get? Can I put the retrieve and form in the same program? |
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