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PHP - How Do I Show Data From Sql In Php
i am doing a little project but I get stuck on a problem. How do I show the content of my SQL database in PHP or HTML. I want it to show the title at the top, the text in the middle and maybe a price at the bottom. How do I do that? I can't seem to find a tutorial which covers only that, and I don't have the time to search each tutorial for some usefull information right now. And a second thing I want is that it will show the database in one list on a specific webpage. So people can look trough the list and find the thing they want and than click on it so they awill get directed to that item from the database.
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Hi, I have this script that returns the entire table including column names and the row data. However I only want certain column names and the corresponding data to be returned. Code: [Select] <?php include_once("data/mysql.php"); $mysqlPassword = (base64_decode($mysqlpword)); $db = mysql_connect("$localhost", "$mysqlusername", "$mysqlPassword") or die ("Error connecting to database"); mysql_select_db("$dbname", $db) or die ("An error occured when connecting to database"); // sending query $result = mysql_query("SELECT * FROM members"); if (!$result) { die("Query to show fields from table failed"); } $fields_num = mysql_num_fields($result); echo "<h1>Table: members</h1>"; echo "<table border='1'><tr>"; for($i=0; $i<$fields_num; $i++) { $field = mysql_fetch_field($result); echo "<td>{$field->name}</td>"; } echo "</tr>\n"; while($row = mysql_fetch_row($result)) { echo "<tr>"; foreach($row as $cell) echo "<td>$cell</td>"; echo "</tr>\n"; } mysql_query($result); ?> Any help is always appreciated Hi all expert. I am a newbie in this PHP programming. I need your help or advise on the PHP. And question is, I have a list of data and the details are as below:
ID BILLNO DATE AMOUNT ITEM DESCRIPTION QTY UPRICE 1 IV001 01/01/2015 100.00 A1 Balloon 1 30.00 2 IV001 01/01/2015 100.00 A2 Bag 2 20.00 3 IV001 01/01/2015 100.00 A3 Pen 3 10.00 4 IV002 02/01/2015 20.00 A3 Pen 2 10.00 5 IV003 02/01/2015 50.00 A1 Balloon 1 30.00 6 IV003 02/01/2015 50.00 A2 Bag 1 20.00 How can I make the output in xml by using PHP to output as below: <RECORD> <HEADER BILLNO="IV001" DATE="01/01/2015" AMOUNT="100.00> <DETAIL ITEM="A1" DESCRIPTION="Balloon" QTY="1" UPRICE="30.00"> </DETAIL> <DETAIL ITEM="A2" DESCRIPTION="Bag" QTY="2" UPRICE="20.00"> </DETAIL> <DETAIL ITEM="A3" DESCRIPTION="Pen" QTY="3" UPRICE="10.00"> </DETAIL> </HEADER> <HEADER BILLNO="IV002" DATE="02/01/2015" AMOUNT="20.00> <DETAIL ITEM="A3" DESCRIPTION="Balloon" QTY="2" UPRICE="10.00"> </DETAIL> </HEADER> </RECORD> Your feedback is highly appreciated. Thank you. I want to show data for logged in user, i am using sessions to login. This is the code i already have: // Connect to server and select database. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); //this selects everything for the current user, ready to be used in the script below $result = mysql_query("SELECT id, points, ingame_points, ingame_money, ingame_items FROM members; WHERE username = $_SESSION['myusername']"); //this function will take the above query and create an array while($row = mysql_fetch_array($result)) { //with the array created above, I can create variables (left) with the outputted array (right) $points = $row['points']; $id = $row['id']; $ingame_points = $row['ingame_points']; $ingame_money = $row['ingame_money']; $ingame_items = $row['ingame_items']; } Help ? I'm trying to show the same data, but updated right away. For example. I want to update my coords on a map and refresh a div to show the new data, but as the code now, it keeps the same data until I reload the page. Here is the code I have now. if ($north) { $ylocation = $users['y'] + 1; if ($ylocation > 5) { $ylocation = 0; } $locationyupdate = ("UPDATE players SET y = '$ylocation' WHERE name='$users[name]'"); mysql_query($locationyupdate) or die("could not register");?> <script type="text/javascript"> $('#npc').load('npc.php'); $('#description').load('description.php'); </script><?} The update code is before the reload script for the two div's. The data DOES change in the database, but the two div's won't display the new data until it is refreshed again. Do I need to reactivate fetch to get the new data? not sure if this will find an answer but I am posting anyway. Any help is appreciated....my coder has left the team and no matter how I invite her to come and do some bug squashing, her new schedules won't fit a re-visit into my little aplication...I am eager to delve into the codes at my very low knowledge of php. I have initiated reading and is still trying to teach myself...but of course not as fast as the app needs the patch...here's the meat: I want to show a cookie of a referral's username on a sign up page. The link is like this, www.mysite.com/signup?ref=johnsmith. The cookie doesn't show if I go to that url page. But it does show up once I reload the page. So I'm wondering if it's possible to show the cookie the first time around, instead of reloading the page? Here is my code.
// This is in the header
$url_ref_name = (!empty($_GET['ref']) ? $_GET['ref'] : null);
if(!empty($url_ref_name)) {
$number_of_days = 365;
$date_of_expiry = time() + 60 * 60 * 24 * $number_of_days;
setcookie( "ref", $url_ref_name, $date_of_expiry,"/");
} else if(empty($url_ref_name)) {
if(isset($_COOKIE['ref'])) {
$user_cookie = $_COOKIE['ref'];
}
} else {}
// This is for the sign up form
if(isset($_COOKIE['ref'])) {
$user_cookie = $_COOKIE['ref'];
?>
<fieldset>
<label>Referred By</label>
<div id="ref-one"><span><?php if(!empty($user_cookie)){echo $user_cookie;} ?></span></div>
<input type="hidden" name="ref" value="<?php if(!empty($user_cookie)){echo $user_cookie;} ?>" maxlength="20" placeholder="Referrer's username" readonly onfocus="this.removeAttribute('readonly');" />
</fieldset>
<?php
}
I am displaying rows from a database onto a page using: while($row=mysql_fetch_array($query)){ echo $row['name']; } I need to figure out how to limit the rows shown to the page to 100. And if there are 100 rows on the page, a link will be displayed at the bottom, that says "Next 100". Then this will display the next 100 rows. Can you give an example how to do this please? Thanks Hi, I would like to do the following but not sure how. If the you/user is on index.php of http://www.domain.com/ show one page If not show another How would I do this? Thanks Hi , I know my code sucks but i'm learning fast!! I'm trying to show a form if the qty value in a database == 10 or a different form if the value ==20. I tried but failed. Any help really appreciated. Code: [Select] <?php require_once('Connections/book.php'); ?> <?php $colname_cardpayment = "-1"; if (isset($_GET['orderid'])) { $colname_cardpayment = (get_magic_quotes_gpc()) ? $_GET['orderid'] : addslashes($_GET['orderid']); } mysql_select_db($database_book, $book); $query_cardpayment = sprintf("SELECT * FROM cards WHERE orderid = '%s' ORDER BY qty ASC", $colname_cardpayment); $cardpayment = mysql_query($query_cardpayment, $book) or die(mysql_error()); $row_cardpayment = mysql_fetch_assoc($cardpayment); $totalRows_cardpayment = mysql_num_rows($cardpayment); // Database connect $con = mysql_connect("mysql1.myhost.ie","admin_book","root123"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("book_test", $con); //Parse Values from Coupon.php Form $orderid = mysql_real_escape_string(trim($_POST['orderid'])); $name = mysql_real_escape_string(trim($_POST['name'])); $surname = mysql_real_escape_string(trim($_POST['surname'])); $add1 = mysql_real_escape_string(trim($_POST['add1'])); $add2 = mysql_real_escape_string(trim($_POST['add2'])); $town = mysql_real_escape_string(trim($_POST['town'])); $county = mysql_real_escape_string(trim($_POST['county'])); $postcode = mysql_real_escape_string(trim($_POST['postcode'])); $phone = mysql_real_escape_string(trim($_POST['phone'])); $email = mysql_real_escape_string(trim($_POST['email'])); $letterstyle = mysql_real_escape_string(trim($_POST['letterstyle'])); $sql="INSERT INTO custdetails (orderid, name, surname, add1, add2, town, county, postcode, phone, email, letterstyle) VALUES ('$orderid','$name','$surname','$add1','$add2','$town','$county','$postcode','phone','$email','$letterstyle')"; if (!mysql_query($sql)) { die('Error: ' . mysql_error()); } ?> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <title>Digital Scribe Books</title> <link href="style.css" rel="stylesheet" type="text/css" /> <script type="text/javascript"> function MM_preloadImages() { //v3.0 var d=document; if(d.images){ if(!d.MM_p) d.MM_p=new Array(); var i,j=d.MM_p.length,a=MM_preloadImages.arguments; for(i=0; i<a.length; i++) if (a[i].indexOf("#")!=0){ d.MM_p[j]=new Image; d.MM_p[j++].src=a[i];}} } function MM_swapImgRestore() { //v3.0 var i,x,a=document.MM_sr; for(i=0;a&&i<a.length&&(x=a[i])&&x.oSrc;i++) x.src=x.oSrc; } function MM_findObj(n, d) { //v4.01 var p,i,x; if(!d) d=document; if((p=n.indexOf("?"))>0&&parent.frames.length) { d=parent.frames[n.substring(p+1)].document; n=n.substring(0,p);} if(!(x=d[n])&&d.all) x=d.all[n]; for (i=0;!x&&i<d.forms.length;i++) x=d.forms[i][n]; for(i=0;!x&&d.layers&&i<d.layers.length;i++) x=MM_findObj(n,d.layers[i].document); if(!x && d.getElementById) x=d.getElementById(n); return x; } function MM_swapImage() { //v3.0 var i,j=0,x,a=MM_swapImage.arguments; document.MM_sr=new Array; for(i=0;i<(a.length-2);i+=3) if ((x=MM_findObj(a[i]))!=null){document.MM_sr[j++]=x; if(!x.oSrc) x.oSrc=x.src; x.src=a[i+2];} } </script> </head> <body onload="MM_preloadImages('images/buttons/home_over.png','images/buttons/books_over.png','images/buttons/cards_over.png','images/buttons/letters_over.png')"> <div id="snow"> <div id="wrapper"> <div id="header"> <div id="logo"><img src="images/digital_scripe.png" width="218" height="91" /></div> <div id="menu"><a href="index.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Home','','images/buttons/home_over.png',1)"><img src="images/buttons/home_act.png" name="Home" width="131" height="132" border="0" id="Home" /></a><a href="books.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Books','','images/buttons/books_over.png',1)"><img src="images/buttons/books_act.png" name="Books" width="131" height="132" border="0" id="Books" /></a><a href="cards.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Cards','','images/buttons/cards_over.png',1)"><img src="images/buttons/cards_act.png" name="Cards" width="131" height="132" border="0" id="Cards" /></a><a href="letters.php" onmouseout="MM_swapImgRestore()" onmouseover="MM_swapImage('Letters','','images/buttons/letters_over.png',1)"><img src="images/buttons/letters_act.png" name="Letters" width="131" height="132" border="0" id="Letters" /></a></div> </div> <div id="content"> <?php echo 'Order ID is : '. $orderid . '.<br />'; if ($row2['qty'] == 10) echo "<div> <form action="https://www.paypal.com/cgi-bin/webscr" method="post"> <input type="hidden" name="cmd" value="_xclick"> <input type="hidden" name="business" value="accounts@agraphics.ie"> <input type="hidden" name="lc" value="IE"> <input type="hidden" name="item_name" value="10 Christmas Cards"> <input type="hidden" name="item_number" value="<? echo $orderid; ?>"> <input type="hidden" name="amount" value="12.99"> <input type="hidden" name="currency_code" value="EUR"> <input type="hidden" name="button_subtype" value="services"> <input type="hidden" name="shipping" value="2.99"> <input type="hidden" name="return" value="http://www.digitalscribe/thanks.php"> <input type="hidden" name="bn" value="PP-BuyNowBF:btn_buynowCC_LG.gif:NonHosted"> <input type="image" src="https://www.paypalobjects.com/en_US/i/btn/btn_buynowCC_LG.gif" border="0" name="submit" alt="PayPal - The safer, easier way to pay online!"> <img alt="" border="0" src="https://www.paypalobjects.com/en_US/i/scr/pixel.gif" width="1" height="1"> </form> </div>"; if ($row2['qty'] == 20) echo "<div> <form action="https://www.paypal.com/cgi-bin/webscr" method="post"> <input type="hidden" name="cmd" value="_xclick"> <input type="hidden" name="business" value="accounts@agraphics.ie"> <input type="hidden" name="lc" value="IE"> <input type="hidden" name="item_name" value="20 Christmas Cards"> <input type="hidden" name="item_number" value="<? echo $orderid; ?>"> <input type="hidden" name="amount" value="21.99"> <input type="hidden" name="currency_code" value="EUR"> <input type="hidden" name="button_subtype" value="services"> <input type="hidden" name="shipping" value="2.99"> <input type="hidden" name="return" value="http://www.digitalscribe/thanks.php"> <input type="hidden" name="bn" value="PP-BuyNowBF:btn_buynowCC_LG.gif:NonHosted"> <input type="image" src="https://www.paypalobjects.com/en_US/i/btn/btn_buynowCC_LG.gif" border="0" name="submit" alt="PayPal - The safer, easier way to pay online!"> <img alt="" border="0" src="https://www.paypalobjects.com/en_US/i/scr/pixel.gif" width="1" height="1"> </form> </div>"; ?> </div> <div id="footer" class="clear"><div id="sign"><div id="sign_text">Personalised<br /> Books</div> </div></div> </div></div> </body> </html> <?php mysql_free_result($cardpayment); ?> 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>'.$row["valid_to"].'</td>
</tr>';
}
?>
Thank you. I have two tables. Table Name:Users Fields: User_name user_email user_level pwd 2.Reference Fields: refid username origin destination user_name in the users table and the username field in reference fields are common fields. There is user order form.whenever an user places an order, refid field in reference table will be updated.So the user will be provided with an refid Steps: 1.User needs to log in with a valid user id and pwd 2.Once logged in, there will be search, where the user will input the refid which has been provided to him during the time of order placement. 3.Now User is able to view all the details for any refid 3.Up to this we have completed. Query: Now we need to retrieve the details based on the user logged in. For eg: user 'USER A' has been provided with the referenceid '1234' during the time of order placement user 'USER B' has been provided with the referenceid '2468' during the time of order placement When the userA login and enter the refid as '2468' he should not get any details.He should get details only for the reference ids which is assigned to him. <?php session_start(); if (!$_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> 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. 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="'. $order_by .'"/> (yes I know I didn't escape the input field... it's all stored in an htaccess protected directory; I will clean it up later once I get it to work) This, of course, results in every source_id getting the same "order_by" no matter what I put into each field. I'm thinking that I need to do some sort of foreach where I go through foreach source_id and have it update the "order_by" field for each one, but I must admit I'm not sure how to go about this (the flaws of being self-taught I suppose; I don't have anyone to go to on this). I'm hoping someone here can help? Thanks a ton in advance Here's the code that deals with the client side:
<?php
session_start();
if(!isset($_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. 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? Hi I am trying to select and order data/numbers from a colum in a mysql data base however i run the code and it returns no value just a blank page no errors or any thing so i think the code is working right but then it returns no result? Please help thanks Here is the code: <?php $host= "XXXXXX"; $mysql_user = "XXXXXX"; $mysql_password = "XXXXXX"; $mysql_database = "XXXXXXX"; $connection = mysql_connect("$host","$mysql_user","$mysql_password") or die ("Unable to connect to MySQL server."); mysql_select_db($mysql_database) or die ("Unable to select requested database."); $row = mysql_fetch_assoc( mysql_query( "SELECT XP FROM Game ORDER BY number DESC LIMIT 1" ) ); $number = mysql_result(mysql_query("SELECT XP FROM Game ORDER BY number DESC LIMIT 1"), 0); echo "The the highest XP is $number"; ?> 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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