|
PHP - Separating Data After Query?
I'm having trouble trying to separate the output into variables that I can use/echo on my page. when I do a print I see the two rows of data all grouped together how can I separate each result base on field and row? maybe something like $line['m_id'][0], $line['m_name'][0], $line['m_id'][1], $line['m_name'][1], etc...
$bio = mysql_query("SELECT * FROM soc_meminfo WHERE m_id = '".mysql_real_escape_string($en['mm_id'])."'"); if (mysql_num_rows($bio) == 0) call404(); while ($line = mysql_fetch_assoc($bio)) { foreach ($line as $key => $value) { $en['b'.$key] = str_replace("\n",'<br/>',stripslashes($value)); } echo '<pre>'; print_r($line); echo '</pre>'; } Similar TutorialsI was just wondering if it's possible to run a query on data that has been returned from a previous query? For example, if I do Code: [Select] $sql = 'My query'; $rs = mysql_query($sql, $mysql_conn); Is it then possible to run a second query on this data such as Code: [Select] $sql = 'My query'; $secondrs = mysql_query($sql, $rs, $mysql_conn); Thanks for any help Hi, I want to pull data from db, where sometimes all rows and sometimes rows matching given "username". Here is my code:
//Grab Username of who's Browsing History needs to be searched.
if (isset($_GET['followee_username']) && !empty($_GET['followee_username']))
{
$followee_username = $_GET['followee_username'];
if($followee_username != "followee_all" OR "Followee_All")
{
$query = "SELECT * FROM browsing_histories WHERE username = \"$followee_username\"";
$query_type = "followee_username";
$followed_word = "$followee_username";
$follower_username = "$user";
echo "$followee_username";
}
else
{
$query = "SELECT * FROM browsing_histories";
$query_type = "followee_all";
$followed_word = "followee_all";
$follower_username = "$user";
echo "all";
}
}
When I specify a "username" in the query via the url: browsing_histories_v1.php?followee_username=requinix&page_number=1 I see result as I should. So far so good.
Now, when I specify "all" as username then I see no results. Why ? All records from the tbl should be pulled! browsing_histories_v1.php?followee_username=all&page_number=1 This query shouldv'e worked:
$query = "SELECT * FROM browsing_histories";
I'm trying to separate the HTML(form) into a separate .html file (from the php file below). Someone suggested to "place the HTML in someFile.html and the PHP in someFile.php, and alter the <form> tag's 'action' element in the .html file to target someFile.php". But I don't know how to alter the <form> tag's 'action' element in the .html file to target this upload_file.php file. Any additional help will be appreciated.
<?php
session_start();
require_once 'phps3integration_lib.php';
$message = "";
if (@$_POST['submit'] != "") {
$allowed_ext = array("gif", "jpeg", "jpg", "png", "pdf", "doc", "docs", "zip", "flv", "mp4");
$extension = end(explode(".", $_FILES["file"]["name"]));
if (($_FILES["file"]["size"] < 32428800) && in_array($extension, $allowed_ext)) {
if ($_FILES["file"]["error"] > 0) {
//$message.="There is some error in upload, see: " . $_FILES["file"]["error"] . "<br>";//Enable this to see actual error
$message.="There is some error in upload. Please try after some time.";
} else {
$uploaded_file = uploaded_file_to_s3($_FILES["file"], "uploads", true);
if ($uploaded_file != FALSE) {
$user_name = @$_POST['user_name'] != "" ? @$_POST['user_name'] : "Anonymous";
$form_data = array(
'file' => $uploaded_file,
'user_name' => $user_name,
'type' => 'file'
);
mysql_query("INSERT INTO `phps3files` (`id`, `file`, `user_name`, `type`) VALUES (NULL, '" . $uploaded_file . "', '" . $user_name . "', 'file')") or die(mysql_error());
$message.= "File successfully uploaded in S3 Bucket.";
} else {
$message.="There is some error in upload. Please try after some time.";
}
}
} else {
$message.= "Invalid file, Please upload a gif/jpeg/jpg/png/pdf/doc/docs/zip file of maximum size 30 MB.";
}
}
?>
<?php
require_once 'header.php';
?>
<fieldset>
<legend>PHP AWS S3 integration library Demo1</legend>
Description: In this demo a file is being upload to an S3 bucket using "PHP AWS S3 integration library". After upload you can check the uploaded file in below table.
If you require some manipulation before uploading file to S3 then check <a href="upload_file_manually.php">Demo2</a> <br />
<br />
<form action="" method="post" enctype="multipart/form-data">
<div class="control-group">
<label for="file" class="control-label">Choose a file to upload: <span style="color:red">*</span></label>
<div class='controls'>
<input id="file" type="file" name="file" />
<?php //echo form_error('file'); ?> </div>
</div>
<div class="control-group">
<label for="user_name" class="control-label">Your name:</label>
<div class='controls'>
<input id="user_name" type="text" name="user_name" maxlength="255" value="" />
<?php //echo form_error('user_name'); ?> </div>
</div>
<div class="control-group">
<label></label>
<div class='controls'>
<input type="submit" name="submit" value="Submit" class="btn">
</div>
</div>
</form>
</fieldset>
<?php
if ($message != "" || @$_SESSION['message'] != "") {
?>
<div class="alert alert-success">
<?php echo $message; ?>
<?php
echo @$_SESSION['message'];
@$_SESSION['message'] = '';
?>
</div>
<?php
}
?>
<div>
<table class="table table-hover">
<caption>
<strong>Last 10 user uploaded files</strong>
</caption>
<?php
$files_result = mysql_query("SELECT * from `phps3files` WHERE type LIKE 'file' ORDER by id DESC LIMIT 10");
$i = 1;
while ($file = mysql_fetch_object($files_result)) {
?>
<tr>
<td><?php echo $i++; ?></td>
<td><a href="<?php echo site_url_s3("uploads/" . $file->file); ?>" target="_blank">View/Download</a> </td>
<td><a href="<?php echo site_url("delete_file.php?id=" . $file->id); ?>">Delete file from S3</a></td>
<td><?php echo "Uploaded by: " . $file->user_name; ?></td>
</tr>
<?php
}
if ($i == 1) {
?>
<tr>
<td colspan="2"> No files uploaded yet</td>
</tr>
<?php
}
?>
</table>
</div>
<h4>Source Code Part of Demo</h4>
<pre class="prettyprint lang-php linenums">
<?php
session_start();
require_once 'phps3integration_lib.php';
$message = "";
if (@$_POST['submit'] != "") {
$allowed_ext = array("gif", "jpeg", "jpg", "png", "pdf", "doc", "docs", "zip");
$extension = end(explode(".", $_FILES["file"]["name"]));
if (($_FILES["file"]["size"] < 32428800) && in_array($extension, $allowed_ext)) {
if ($_FILES["file"]["error"] > 0) {
//$message.="There is some error in upload, see: " . $_FILES["file"]["error"] . "<br>";//Enable this to see actual error
$message.="There is some error in upload. Please try after some time.";
} else {
$uploaded_file = uploaded_file_to_s3($_FILES["file"], "uploads", true);
if ($uploaded_file != FALSE) {
$user_name = @$_POST['user_name'] != "" ? @$_POST['user_name'] : "Anonymous";
$form_data = array(
'file' => $uploaded_file,
'user_name' => $user_name,
'type' => 'file'
);
mysql_query("INSERT INTO `phps3files` (`id`, `file`, `user_name`, `type`) VALUES (NULL, '" . $uploaded_file . "', '" . $user_name . "', 'file')") or die(mysql_error());
$message.= "File successfully uploaded in S3 Bucket.";
} else {
$message.="There is some error in upload. Please try after some time.";
}
}
} else {
$message.= "Invalid file, Please upload a gif/jpeg/jpg/png/pdf/doc/docs/zip file of maximum size 30 MB.";
}
}
?>
</pre>
<?php require_once 'footer.php'; ?>
Lets say I pull a value from a database field that has a value of 1,2 and what I want to do is seperate them and then run each of those values against a separate table how would that be accomplished. I'm going through my project trying to find a way of separating the backend (the query and php code) from the frontend (html/css). I want to bring the two together with the include function. I've done other but I'm getting confused because the while loops html code. I don't want someone to do this for me, I just what someone to highlight the most logical way. <?php include 'auth.inc.php'; include 'db.inc.php'; include 'buy.inc.php'; //connect to database. $db = mysql_connect(MYSQL_HOST, MYSQL_USER, MYSQL_PASSWORD) or die ('Unable to connect. Check your connection parameters.'); mysql_select_db(MYSQL_DB, $db) or die(mysql_error($db)); //select orders that this user has bought. $query2 = sprintf('SELECT d.name_id, d.order_id, d.order_qty, d.product_code, feedback, p.title, c.email FROM order_details d LEFT JOIN product p ON d.product_code = p.product_code LEFT JOIN contact c ON d.name_id = c.name_id WHERE d.buyer_id = "%u"', mysql_real_escape_string($_SESSION['name_id'])); $result2 = mysql_query($query2, $db) or die(mysql_error()); //echo out the information in a while loop, add pagination? $odd = true; while ($row2 = mysql_fetch_array($result2)) { echo ($odd == true) ? '<tr class="odd_row">' : '<tr class="even_row">'; $odd = !$odd; echo '<td style="text-align: center; width:100px;">' . $row2['title'] . ' <table> <tr> <th colspan="2">Shipping Information</th> </tr><tr> <td>First Name:</td> <td>' . $row2['title']. '</td> </tr><tr> <td>First Name:</td> <td>' . $row2['order_qty']. '</td> </tr><tr> <td>Last Name:</td> <td>' . $row2['product_code']. '</td> </tr><tr> <td>Billing Address:</td> <td>' . $row2['email'] . '</td> <td>' . $row2['order_id'] . '</td> </tr> </table> </tr>'; if ($row2['feedback'] == 0){ echo ' <form action="reputation.php" method="POST"><p>Please give the seller feedback on how you have received your product. <td><input type="hidden" name="name_id" value="' . $row2['name_id'] . '"></td> <td><input type="hidden" name="order_id" value="' . $row2['order_id'] . '"></td> <td><input type="submit" name="submit" value="Feedback"/></td> </form>'; } } ?> Hi there, I'm new to the board, so apologise if something like this has been asked before. I'm also quite new to php, so the disaster that is my code will probably make your stomach turn, but it works so far... I am using file_get_contents to grab a website I have access to (my work rota, incidentally), and parse a bit of the html. The aim is to be able to take my rota and do a few things with the times and dates, etc.. So far, I have got the file, parsed the crap out of the string that I don't need, and i'm left with something that looks like this: Code: [Select] <TR><TD ></TD></TR><TR bgcolor='ffff00'><TD>Date</TD><TD>Duty</TD><TD>Dep</TD><TD>Begin</TD><TD>End</TD><TD>Arr</TD></TR><TR><TD ></TD></TR><TR><TD>23 Apr 12, Mon</TD><TD>297</TD><TD>STN</TD><TD>15:55</TD><TD>17:10</TD><TD>DUB</TD></TR><TR><TD>23 Apr 12, Mon</TD><TD>288</TD><TD>DUB</TD><TD>17:35</TD><TD>18:50</TD><TD>STN</TD></TR><TR><TD>23 Apr 12, Mon</TD><TD>293</TD><TD>STN</TD><TD>19:15</TD><TD>20:30</TD><TD>DUB</TD></TR><TR><TD>23 Apr 12, Mon</TD><TD>298</TD><TD>DUB</TD><TD>20:55</TD><TD>22:05</TD><TD>STN</TD></TR><TR><TD ></TD></TR> The people who wrote my rota weren't very tidy, but in a nutshell, I want the information between the TD/TR brackets. Each Row contains cells with my date, time, destination etc in it. I can use striptags, but I end up with a long string that is difficult to split into useful information. I need something that parses the string like this.. "for every row grab information between <td> and </td> and load into $td[0]" etc.." How would I go about doing something like this? I'm a bit stumped. Thanks in advance Horgy Can someone give me some guidance of how to separate Form Display from Form Processing? I have always used forms that submitted back to themselves which isn't so bad, but then trying to cram in code to display the form, validation errors, and messages after the form is processed all in one file is insane?! Currently I am working on a simple "Add a Comment" form. It would be nice to have a separate form processing script, but I don't know where to begin... Debbie This is a LINK to the page -> http://shoutkey.com/carriage I have a problem of separating MySQL concatenated info, yet still maintaining their relationship to other arrays in the same order. First off here is the MySQL code: Code: [Select] $query ="SELECT report,"; $query.="GROUP_CONCAT(DISTINCT docID) AS docIDs, "; $query.="GROUP_CONCAT(DISTINCT analyst) AS analysts, "; $query.="GROUP_CONCAT(DISTINCT region) AS regions, "; $query.="GROUP_CONCAT(DISTINCT country) AS countries, "; $query.="GROUP_CONCAT(DISTINCT topic) AS topics, "; $query.="GROUP_CONCAT(DISTINCT date) AS dates, "; $query.="GROUP_CONCAT(DISTINCT event) AS events, "; $query.="GROUP_CONCAT(DISTINCT link) AS links, "; $query.="GROUP_CONCAT(DISTINCT province) AS provinces "; $query.="FROM reports GROUP BY report ORDER BY dates DESC, docIDs DESC"; The most important thing is to get all attributes about each report. The thing is, each report can have as many records as it must, in order to store various types of information, like, if multiple analysts are working on the report, then each one saves their own record, but when retrieving these records MYSQL concatenates the records, to make it look to the analysts like this Report 1 --- Josh, Rebecca --- Philippines However, my problem is trying to display the reports by country... as you can see http://shoutkey.com/carriage. I have no idea about how to split up the countries, while maintaining their relationship to the $rows pulled from MySQL. I am gonna post the entire code, and let me know if anybody has any ideas. Code: [Select] $result = mysql_query($query); $numrows = mysql_affected_rows(); $current = false; while($row=mysql_fetch_assoc($result)) { $dates = $row['dates']; $analysts = $row['analysts']; $provinces = $row['provinces']; $topics = $row['topics']; $regions = $row['regions']; $countries = $row['countries']; $reports = $row['report']; $links = $row['links']; $events = $row['events']; if (!$current){ //first call echo "<h3><a href=\"#\">{$countries}</a></h3>"; //echo first call $current = $countries; echo "<div>" ;//open content div } elseif($current != $countries){ // country has changed echo "</div>"; //close previous content div echo "<h3><a href=\"#\">{$countries}</a></h3>"; //echo subsequent countries $current = $countries; echo "<div><p>" ;//open content div } echo $analysts." ----- ".$provinces." ----- ".$reports; echo "<br /><br />"; } echo "</div>"; ?> </div> Hi Guys I'm returning to programming after 5 years, almost a newbie. I have a field(column) in a MySQL database table that has strings of characters separated by a delimiter (||). For example: (TOM||PAUL||HARRIS) I would like to separate the strings into an array. Something like this: ARRAY[0] = 'TOM'; ARRAY[1]='PAUL'; ARRAY[3]='HARRIS'; Can some one tell me how to do this? TIA. hi guys Ive written this php to take in two variables from the http POST, the idea is that I can multiple devices submit temperature readings to the php script, the script then append the unix time stamp and then the the 3 variables - device id, temp and unix time are then stored in a mysql DB. I can get the variables to present on a php page for debugging but I cant get the variables to be stored in the mysql DB. See the code: Code: [Select] <?php $unixtime = time(); // get device variables $device_id=$_GET['device']; $device_temp=$_GET['temp']; /* //for testing purposes echo "unixtime: " . $unixtime . "<br />"; echo "device id: " . $device_id . "<br />"; echo "device_temp: " . $device_temp . "<br />"; */ // Make a MySQL Connection mysql_connect("localhost", "username", "password") or die(mysql_error()); mysql_select_db("test") or die(mysql_error()); //mysql query $query = "INSERT INTO temperature VALUES ('',$device_id,$temp,$unixtime)"; // Insert a row of information into the relevant device table mysql_query($query); or die(mysql_error()); mysql_close(); echo "Data Inserted!"; ?> I cant see where Im going wrong to correct this, but as nothing is displayed on the page i believe I am not forming the query correctly? - any ideas would be much appreciated. Thank you Mathew Hello, I hope all of you are safe with your families. Currently I am starting with PHP Coding and I am trying to do a simple query to MySQL DB using PHP but even when is able to bring the number of rows is not displaying the values in the DB. This is my code:
<?php
$servername = "localhost";
$database = "mydbtest";
$username = "root";
$password = "root";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $database);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
//echo "Connected successfully";
$myquery = "SELECT * FROM Country";
$result = $conn->query($myquery);
$numf = $result->num_rows;
echo "Number of rows " . $numf . "<br>";
if($numf >0){
while($row = $result->fetch_object()){
echo "Code" . $row->countrycode . "<br>";
echo "Country" . $row->countryname . "<br>";
}
}else{
echo '0 results';
}
mysql_free_result($myout);
mysqli_close($conn);
?>
What is failing?
Thanks in advance for the assistance. I've only been studying PHP for a week now and have come across this problem which I'm sure there is a simple answer to, but I just can't figure it out and would appreciate some help. I've spent far too long on this minor issue already! I have a table which contains a list of products, in this case books, which stores the date when each new item is added. I have a query that then searches through this table and extracts the 6 most recent additions. Here is the code I have so far: Code: [Select] $sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC LIMIT 6"); $productCount = mysql_num_rows($sql); // count the output amount if ($productCount > 0) { while($row = mysql_fetch_array($sql)){ $id = $row["id"]; $title = $row["title"]; $author = $row["author"]; $price = $row["price"]; $date_added = strftime("%b %d, %Y", strtotime($row["date_added"])); $dynamiclist .= //My table showing the products } } else { $dynamicList = "There are currently no Books listed in this store"; } This works well when I need to display the 5 most recent products in a normal table fashion, one below another, on the page. However, I want to display the products in a more personalised order. For example display the newest item in one section of the page and 3rd newest in another. What is the best way to select an individual row from a query? To extract the details of the 2nd newest item to display in the header, for example. Thanks for your help. I am at a loss why my query is not inserting values into db. Even if I do echo $query or var_dump($query) there is nothing printed at all. All values are being passed successfully just not being inserted. I am getting 'Could not connect' but I do not know why. All connections are established and as a test I took this code and ran it on it's own with dummy data and it inserted the data fine. I can only think it has something to do with the $response_array. Where am I going wrong. and would appreciate any help. Thanks Code: [Select] <?php require_once('Connections/sample.php'); ?> <?php session_start(); $new = 1; $activity = 'General Contact Enquiry'; $mobile = 'Submitted from mobile'; $name = mysql_real_escape_string($_POST['GC_name']); $department = mysql_real_escape_string($_POST['GC_department']); $message = mysql_real_escape_string($_POST['GC_message']); $email = mysql_real_escape_string($_POST['GC_email']); $company = mysql_real_escape_string($_POST['GC_company']); $position = mysql_real_escape_string($_POST['GC_position']); //response array with status code and message $response_array = array(); //validate the post form //check the name field if(empty($name)){ //set the response $response_array['status'] = 'error'; $response_array['message'] = 'Name cannot be blank'; //check the name field } elseif(empty($company)) { //set the response $response_array['status'] = 'error'; $response_array['message'] = 'You must enter a company name'; //check the position field }elseif(empty($position)) { //set the response $response_array['status'] = 'error'; $response_array['message'] = 'You must enter a position'; //check the email field } elseif(empty($email)) { //set the response $response_array['status'] = 'error'; $response_array['message'] = 'You must enter a valid email address'; //check the dept field }elseif($department=="Choose Department") { //set the response $response_array['status'] = 'error'; $response_array['message'] = 'You must select a department'; //check the message field }elseif(empty($message)) { //set the response $response_array['status'] = 'error'; $response_array['message'] = 'You must enter a message'; //check the dept field } else { //set the response $response_array['status'] = 'success'; $response_array['message'] = 'Your enquiry has been sent succesfully'; $flag=1; } //send the response back echo json_encode($response_array); if($flag == 1) { mysql_select_db($database_sample, $sample); $query = 'INSERT INTO feedback (company, department, name, email, position, feedback, date, new) VALUES (\''.$company.'\', \''.$department.'\', \''.$name.'\', \''.$email.'\', \''.$position.'\', \''.$message.'\', NOW() , \''.$new.'\')'; mysql_query($query) or die("Could not connect"); } ?> I know that I can fairly easily pull my data from the database and view it in a browser. I can also 'polish' it with some HTML or put it into a table. Can I get an item from a given VARIABLE to appear inside of an INPUT box, so that it looks the same as when it was initially submitted? Can it be done with a multiple choice SELECT dropdown, so that the item chosen is viewable again? Alright, I've spent over a week trying to fix this now - And Im getting frustrated! I asked at other forums, I asked co-workers and I asked friends-of-friends, and nobody can explain what happens. Let's take a look at this first: $name = mysql_real_escape_string($_POST['name']); mysql_query(sprintf("UPDATE em_users SET name='%s' WHERE id='" . $in_user['id'] . "'", $name)); This will insert NO data on the Name field in the database. Obviously, I thought the $_POST variable wasn't passed correctly, but echo'ing it just before the query WILL show data. And as I said, I tried everything possible for the last week. Switching variables, adding static text on the $name variable instead of using the $_POST content (this does work). I used very very simple test data on the form, such as my name "Mark" or "test" and "hey". The query is correctly executed everytime. The truely WEIRD thing is, if I ensure there is content in $name before executing the query it will work as expected everytime. Like this: $name = mysql_real_escape_string($_POST['name']); $name && mysql_query(sprintf("UPDATE em_users SET name='%s' WHERE id='" . $in_user['id'] . "'", $name)); Of course I could do this, but I want to know why my code does or doesn't work + it's a lot of work to do for something that worked fine a week ago. It has spread to a lot of forms on my website that $_POST variables aren't processed correctly - and it happened out of nowhere. Even on codes that havnt changed in months. I really need help on fixing this! This project has been in development for nearly two years, and without a fix it's pretty much lost Hi guys, im trying to connect to a database and get the value for the user in the row called 'user_credit', if it equals 1 or more then i want to show the ''You have £ ....'' bit in the script. Problem is nothing shows at all, even without the if statement. I have changed the value for me in the database so in user_credit the value is 100, which is more than 1 so it should appear. I have probably done something wrong. Any ideas? Code: [Select] <? include '../admin/database/membership_dbc.php'; $r = mysql_query("SELECT * FROM users WHERE user_name='".safe($_SESSION['user_name'])."'") or die ("Cannot find table"); while( $cred = mysql_fetch_array($r) ) { if ($cred >= '1' ) { ?> <p>You have £<? echo $cred['user_credit']; ?> available on you account, would you like to use it on this order?<br> <label for="credit"></label> <select name="credit" id="credit"> <option value="Y" selected>Yes, use credit</option> <option value="N">No, save credit</option> </select> </p> <? } } ?> i want the name of a picture stored in my db after i upload it the data is not stored in the db after i run this script, but i dont get errors either i print the two vars before sending them, and they get printed fine any help on this would be greatly appreciated thanks ! <?php error_reporting(E_ALL); ini_set("display_errors", 1); // INCLUDE THE CLASS FILE include('ImageLib.Class.php'); include("./includes/egl_inc.php"); $displayMessage = ''; if($_POST){ if(isset($_FILES['image_file'])){ // SEE THE MAGIC HAPPEN $destination_path = 'uploads/'; $post_file_name = 'image_file'; $width = 600; $height = 400; $scale = false; $trim = true; $uniqueName = true; $img = ImageLib::getInstance()->upload($post_file_name, $destination_path, $uniqueName)->resize($width, $height, $scale, $trim)->save(); $imgstr = mysql_real_escape_string ($img); $fileName = $_FILES['image_file']['name']; $displayMessage = '<div class="image"><img src="'.$destination_path.$fileName.'" /><br />Uploaded And Resized...With new file name : "'.$img.'"</div><br /><br />'; $playerid=$_SESSION['tid']; $matchdetails = mysql_fetch_array(mysql_query("SELECT id FROM ffa_matches WHERE status=2 and admin=$playerid")); $id = $matchdetails[id]; print $img;print $imgstr; print $id; mysql_query(" INSERT INTO ffa_screens (imgname,match) VALUES( '" . mysql_real_escape_string($imgstr) . "', '" . mysql_real_escape_string($id) . "' )"); }} ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html> <head> <title>ImageLib Samples By Rahul Kate</title> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <style> body{font-family: arial; font-size:12px; color:#444444; padding:20px;} li{margin-top:10px;} .image{color:green;} .image img{margin-bottom:5px;} </style> </head> <body> <h1>ImageLib | Upload Image, move it to Uploads folder and Resize it and Trim</h1> <?php echo $displayMessage; ?> <form method="post" enctype="multipart/form-data"> Select Image<br /> <input type="file" name="image_file" id="image_file" /> <br /> <br /> <input type="submit" name="submit" value="Submit" /> <br /> <br /> <a href="index.html">Back TO Home</a> </form> </body> </html> Hi. I am trying to get a PHP Query to refresh every 10 seconds. I have scoured the internet for days and could not find anything of much use. Plenty of Ajax going on (whatever that is) but the scripts were immense. I only want to refresh 7 lines of PHP Query Code. Can you please tell me if this is possible? Thanks in Advance. I have a text box that I use to post comments and save them to my database. I can go to the database and the data I entered looks perfect but when I pull it out it all runs together.
This is how I entered it and the way it is in the database:
I have a text box that I use to post comments and save them to my database. This is how it came out: I have a text box that I use to post comments and save them to my database. I can go to the database, the data I entered looks perfect but when I pull it out and display it it all runs together. No new lines just all one paragraph. Can someone help? I got it indenting the paragraphs, but I get all the data back I entered all in 1 paragraph, is there something I am missing to be able to recognize the new lines?? Hi, I want to develop array like following Code: [Select] $BCD=array('type' =>'TYPE1', array( 0=>array('column1'=>'value1','column2'=>'value1','column3'=>'value1','column4'=>'value1','column5'=>'value1','column6'=>'value1','column7'=>0), 1=>array('column1'=>'value1','column2'=>'value1','column3'=>'value1','column4'=>'value1','column5'=>'value1','column6'=>'value1','column7'=>0), 2=>array('column1'=>'value1','column2'=>'value1','column3'=>'value1','column4'=>'value1','column5'=>'value1','column6'=>'value1','column7'=>0), 3=>array('column1'=>'value1','column2'=>'value1','column3'=>'value1','column4'=>'value1','column5'=>'value1','column6'=>'value1','column7'=>0), ) )); I have written following code to achieve the same but not getting result. Code: [Select] $sql = "select * from tablename "; $result = mysql_query($sql); $k=0; while ($row = $db->mysql_fetch_array($result)) { // array_push($BCD['type'],$row['type']); $BCD1=array('type' =>$row['type'], array( $k=>array('column1'=>$row['column1'],'column2'=>$row['column2'],'column3'=>$row['column3'],'column4'=>$row['column4'], 'column5'=>$row['column5'],'column6'=>$row['column6'],'column7'=>$row['column7']) )); $k++; } |
|||||||