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PHP - Need Help With Exporting Data To Csv
Hello,
I created a web page that executes request from mysql and prints it on web page. I also created a export submit button to export mysql query in csv format file. When I select values and then click on export buttons, I get nothing and page gets refreshed.
Here is the link of my site: rahil.me/index.php
Here is the code:
These are the buttons:
<input type="submit" name="search" value="Search"> <input type="submit" name="export" value="Export" />PHP code:
if (isset($_POST['export'])) {
if (empty($_POST['service'])) {
echo "Please select service in dropdown" . "</br>";
}
else {
$service = $_POST['service'];
}
if (empty($_POST['environment'])) {
echo "Please select Environment in dropdown" . "</br>";
}
else {
$env = $_POST['environment'];
}
if ((!empty($service)) && (!empty($env))) {
$sql="SQLQUERY REMOVE";
if (!mysqli_query($con,$sql)) {
die('Error: ' . mysqli_error($con));
}
$result = mysqli_query($con,$sql);
/*
* send response headers to the browser
* following headers instruct the browser to treat the data as a csv file called export.csv
*/
header('Content-Type: text/csv');
header('Content-Disposition: attachment;filename=export.csv');
/*
* output header row (if atleast one row exists)
*/
$row = mysqli_fetch_assoc($result);
if ($row) {
echocsv(array_keys($row));
}
/*
* output data rows (if atleast one row exists)
*/
while ($row) {
echocsv($row);
$row = mysqli_fetch_assoc($result);
}
/*
* echo the input array as csv data maintaining consistency with most CSV implementations
* - uses double-quotes as enclosure when necessary
* - uses double double-quotes to escape double-quotes
* - uses CRLF as a line separator
*/
function echocsv($fields)
{
$separator = '';
foreach ($fields as $field) {
if (preg_match('/\\r|\\n|,|"/', $field)) {
$field = '"' . str_replace('"', '""', $field) . '"';
}
echo $separator . $field;
$separator = ',';
}
echo "\r\n";
}
}
}
Edited by raymak, 25 May 2014 - 02:04 PM. Similar TutorialsHi all, this is the code i'm using at the moment which is at the bottom of a webpage Code: [Select] // Open file export.csv. $f = fopen("export.csv", "w"); // Put all values from $out to export.csv. fputs($f, $out); fclose($f); $filename = "StatisticalAnalysis-" . gmdate("Y-m-d-H-i-s") . ".csv"; header("Content-type: application/csv"); header("Content-Disposition: attachment; filename=" . $filename); readfile("export.csv"); The variable named $out is a long line of text in csv format. I've echoed that variable and it's all good. When a checkbox in a form further up the page is checked the code creates and fills in the $out variable and then carries out the code above. The problem i have is that when the csv save dialog comes up and the file is saved, all it contains is a copy of the html code which is on that page, not the info stored in the $out variable. Can anyone offer any advice on what the problem is? Cheers Hi, this is my first time here. i am trying to create an xls file from the data i get from mysql. here is the code i tried but i am not able to use it correctly. Code: [Select] ob_start(); include('config'); $datacat = mysql_query("SELECT * FROM `leads`") or die(mysql_error()); $name = mysql_fetch_assoc($datacat); $line1="Industry,Company Name,Officials Name,Job Title,Country,Direct Number Mobile Number,Switch Board Number,E-mail,Executive NamePitch Date,Call Back Date,Comments\t"; while($row = mysql_fetch_array($datacat, MYSQL_ASSOC)) { $line2= $name['industry'].",".$name['company_name'].",".$name['officials_name'].",".$name['job_title'].",".$name['country'].",".$name['direct_number_mobile_number'].",".$name['switch_board_number'].",".$name['email'].",".$name['executive_name'].",".$name['pitch_date'].",".$name['call_back_date'].",".$name['comment']."\t"."\n"; } $data="$line1\n$line2\n"; header("Content-type: application/x-msdownload"); header("Content-Disposition: attachment; filename=extraction.xls"); header("Pragma: no-cache"); header("Expires: 0"); print "$header\n$data"; i get my webpage title in the top of xls file and all the fields in xls file are separated by commas, whereas i want them to be in proper tables. Help would be much appreciated. I have a script that will export data to excel. After the export, I need the page to reload showing status information, but it does not reload. What could be causing an issue like this? I think that because the export forced the download file dialogue to open, those headers may be preventing the form from reloading properly, but how, why, and what can be done to resolve the issue? Hi people. Need some help if anyone could give a hand! I'm currently trying to incorporate an export to Excel feature on search returns on my web page. Here is how my searches are returned, so what I want now is when the user clicks on the 'Export to Excel' button it downloads the search results and opens it in the same format in Excel. I've tried using http://phpexcel.codeplex.com/ but can't understand how to use it properly, does anyone have any suggestions? I don't have any code written for the Export to Excel yet as it's just a submit button. How I see it in my head is the user clicks the button, it goes to a new page and performs a new query then downloads that but I don't no if thats right or not. Hello, guys i got some issue in my script for exporting data to CSV (exel for view) Its export all data in one collumn, but i need to export each data in separeta column like. a B a1 ID ID2 But i got there this: a B a1 ID""ID2 Here is the script: Code: [Select] <?php class WDealAdminExport extends UWidgetWorklet { public function accessRules() { return array( array('allow', 'roles' => array('company')), array('deny', 'users'=>array('*')) ); } public function taskConfig() { $list = wm()->get('deal.admin.coupon'); $_GET[$list->modelClassName] = $_POST[$list->modelClassName]; $list->init(); $dataProvider = $list->dataProvider(); $dataProvider->pagination = false; $csv = array( array( $this->t('ID'), $this->t('Order ID'), $this->t('Deal ID'), $this->t('User'), $this->t('Status'), ) ); foreach($dataProvider->data as $data) { $row = array( "#".$data->couponId(), $data->orderId, $data->dealId, $data->user->email." [".$data->user->getName(true)."]", $data->status==1 ? ($data->deal->isExpired() ? $this->t('Expired') : $this->t('Available')) : $this->t('Used') ); if($_POST['charset'] != 'utf-8') { $row[3] = iconv('utf-8', $_POST['charset'], $row[3]); $row[4] = iconv('utf-8', $_POST['charset'], $row[4]); } $csv[] = $row; } $contents = $this->generateCSV($csv); $this->send($contents); } public function taskGenerateCSV($src) { $csv = ''; foreach($src as $row) { foreach($row as $item) $csv.= $this->escape($item).$this->param('delimiter'); $csv = rtrim($csv,$this->param('delimiter'))."\r\n"; } return $csv; } public function taskEscape($value) { return '"'.str_replace('"','""',$value).'"'; } public function taskSend($contents) { $content_type = "application/octet-stream"; $filename = 'export.csv'; // disable browser caching header('Cache-control: private'); header('Pragma: private'); header('Expires: Mon, 26 Jul 1997 05:00:00 GMT'); header('Content-Type: '.$content_type); header('Content-Transfer-Encoding: binary'); header('Content-Length: '.strlen($contents)); header('Content-Disposition: attachment;filename="'.$filename.'"'); echo $contents; app()->end(); } } hello guys.. i want help from u . i am displaying a table of 5 rows and 5 columns . i want to get a pdf copy of it dynamically. i want to re-use it again for my invoice also . covert in into pdf and then print that file. can u provide a sample code of how to covert it into pdf . (same as exporting into excel) How can I use PHP to export my database into a .sql file? Hi everyone, I got a bit of code off the web and did some tweaking. I want to use it to export data from a mysql table to excel. I do a website for a friend and have been exporting it myself and emailing it to him. However, I thought it'd be easier if I could just direct him to a link so he could get it whenever. Anyway, the code below works pretty well...the only problem being that it exports all data to the first column of the excel file. There are commas between all fields, so it is possible to use the 'text to columns' feature in excel to separate teh data into separate columns. However, if there is a way of fixing the code that would negate the need to have to do this, it would be much appreciated. Here goes: <?php $db = mysql_connect("localhost", "MYUSERNAME", "MYPASSWORD"); mysql_select_db("MYDATABASE",$db); $query="SELECT * FROM MYTABLENAME"; $result=mysql_query($query); $num=mysql_numrows($result); mysql_close(); $csv_output = "firstname,lastname,emailaddress,crew_name_001"; $csv_output .= "\r\n"; $i=0; while ($i < $num) { $id=mysql_result($result,$i,"id"); $firstname = mysql_result($result,$i,"firstname"); $lastname = mysql_result($result,$i,"lastname"); $emailaddress = mysql_result($result,$i,"emailaddress"); $crew_name_001 = mysql_result($result,$i,"crew_name_001"); $csv_output .= "$firstname,$lastname,$emailaddress,$crew_name_001\n"; ++$i; } $mode="xls"; $type="excel"; header("Content-type: text/x-csv"); header("Content-disposition: attachment; filename=".date("d-m-Y")."-export.xls"); print $csv_output; exit; ?> Thanks for your time, Dave hello I wrote a php code which will get the records from database and save them in excel. <?php $t = strtotime($date1); $t1 = strtotime($date2); $t2 = $t1 - $t; $t3 = floor($t2 /60/60/24); $j = 0; while ($j <= $t3 ) { if ($j == 0) { $t = $t - 86400; } $t = $t + 86400; $today = date('Y-m-d',$t); mysql_select_db("reporting", $connect); $result1 = mysql_query("SELECT * from report where date = '$today' "); $row1 = mysql_fetch_assoc($result1); ?><tr><?php foreach($row1 as $k=>$v){ ?> <td><?php echo $v; $op1 = $row[$v]; ?></td> <?php } ?></tr><?php $j++; } ?> </table> <body> </html> <?php mysql_close($connect); header("Content-type: application/vnd.ms-excel"); header('Content-disposition: attachment; filename = "report" '); exit; ?> Problem is I am able to retrieve only 2 rows to excel file. If the number of rows are more than 2 it displays a table. can any one help me Hey guys! I have the following script to create an excel file, the thing is that I dont want to be asked if I want to open or save the file when accessing the php file...I just want the php file directly to save the excel file. Heres the code: $filename = "test.xls"; $contents = "testdata1 \ntestdata2 \ntestdata3 \n"; header('Content-type: application/vnd.ms-excel'); header('Content-Disposition: attachment; filename='.$filename); echo $contents; Is there a way instead of using the header function, use something like: fopen, fwrite, fclose ? Thanks in advance! Cheers, I am trying to export multiple csv files to download from 2 separate tables in my database. Some background: I have a web app that links to a phpmyadmin database. There are 2 tables in the database (entering and exiting). These tables hold the phone inventory information of employees currently entering or exiting the organization. The fields in my tables a location, firstname, lastname, username, extension, mac, type What I am trying to do is export the data in these 2 tables to CSV (which I have working now) but I need to have multiple CSVs for each. For example, for my exiting table, I need to have one CSV export all the fields in the table and I also need another CSV to export just the location and username field. I have a simple html form with a submit button which is currently working now: <form action="exportexiting.php" method="POST" style="width: 456px; height: 157px"> <div> <fieldset> <legend style="width: 102px; height: 25px"><strong>Entering CSV:</strong></legend> <input name="Export" type="submit" id="Export" value="Export"> </fieldset><br/> </div> </form> This links to my exportexiting.php file: <?php $host = 'localhost'; $user = 'root'; $pass = 'xxxx'; $db = 'PhoneInventory'; $table = 'exiting'; // Connect to the database $link = mysql_connect($host, $user, $pass); mysql_select_db($db); require 'exportcsv.inc.php'; exportMysqlToCsv($table); ?> Then to exportcsv.inc.php: <?php function exportMysqlToCsv($table,$filename = 'export.csv') { $csv_terminated = "\n"; $csv_separator = ","; $csv_enclosed = ''; $csv_escaped = "\\"; $sql_query = "select * from $table"; // Gets the data from the database $result = mysql_query($sql_query); $fields_cnt = mysql_num_fields($result); $schema_insert = ''; for ($i = 0; $i < $fields_cnt; $i++) { $l = $csv_enclosed . str_replace($csv_enclosed, $csv_escaped . $csv_enclosed, stripslashes(mysql_field_name($result, $i))) . $csv_enclosed; $schema_insert .= $l; $schema_insert .= $csv_separator; } // end for $out = trim(substr($schema_insert, 0, -1)); $out .= $csv_terminated; // Format the data while ($row = mysql_fetch_array($result)) { $schema_insert = ''; for ($j = 0; $j < $fields_cnt; $j++) { if ($row[$j] == '0' || $row[$j] != '') { if ($csv_enclosed == '') { $schema_insert .= $row[$j]; } else { $schema_insert .= $csv_enclosed . str_replace($csv_enclosed, $csv_escaped . $csv_enclosed, $row[$j]) . $csv_enclosed; } } else { $schema_insert .= ''; } if ($j < $fields_cnt - 1) { $schema_insert .= $csv_separator; } } // end for $out .= $schema_insert; $out .= $csv_terminated; } // end while header("Cache-Control: must-revalidate, post-check=0, pre-check=0"); header("Content-Length: " . strlen($out)); // Output to browser with appropriate mime type, you choose header("Content-type: text/x-csv"); //header("Content-type: text/csv"); //header("Content-type: application/csv"); header("Content-Disposition: attachment; filename=$filename"); echo $out; exit; } ?> Again, this is working perfectly for only exporting all the data from the exiting table into one CSV file. My question is, how can I make my one submit button export the 2 CSV files that I need? Thanks for the help... Am new here - looks like a great foru! I would sincerely appreciate any help anyone can give me. I have been trying to solve my problem for hours and I am not having any luck, so I thought I would post and see if anyone can help. I am very stuck and am not making much progress on this project, and I am certain the answer is very simple. I am constructing a form to collect data for a specialized purpose. The form and program actually work for its intended function, but I am trying to enhance the user experience by preventing customers from having to reenter all of their data should there be a problem with any of the data submitted. I have been able to do that with the contact form portion, but what I am having trouble with is the portion which has as many as 400 possible entries. So, in a nutshell, if the customers contact data is incomplete or in error, the form will ask them to return to the page and correct things. The previous data entered has been saved in the session and the input value will equal the previous entry. i.e. <tr> <td align="right" class="infoBox"><?php echo ENTRY_EMAIL_ADDRESS; ?></td> <td align=left><?php echo "<input type=text name='cemail' value=\"$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 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. 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 { } 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, I've currently started to modify a chat script of mine to output a moderation panel but the moderation page seems empty(blank) every time I load it. What im trying to do is to take the ID part in my URL via the $_GET and look it up in my database table in the column named id, then select that specific row to be able to retrieve the StringyChat_ip and place it into another table to ban the IP and the second thing im trying to do is to be able to delete the specific row from my table.
My Http link look something like
http://imagecrab.fre.../ban.php?id=159
and my ban.php page where I want to lookup the 159 part and do the banning etc looks like
<?
include("admin_code_header.php");
if ($_POST["DeletePost"]) {
$id = $_POST['id'];
$query = "DELETE FROM ".$dbTable." WHERE id='".$id."'";
mysql_query($query);
echo "ID removed from system: ".$id;
}
if ($_POST["BanIP"]) {
$IP_To_Add = $_POST["ip"];
$sql = "INSERT INTO ".$IPBanTable." (ip) VALUES (\"$IP_To_Add\")";
$result = mysql_query($sql);
}
$result = mysql_query("SELECT * FROM ".$dbTable." WHERE id='".$id."'",$db);
while ($myrow = mysql_fetch_array($result)) {
$msg = $myrow["StringyChat_message"];
$idm = $myrow["id"];
?>
<html>
<form name="form<? echo $myrow["id"];?>" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input name="DeletePost" type="submit" id="DeletePost" value="Delete">
<input name="BanIP" type="submit" id="BanIP" value="Ban <? echo $myrow["StringyChat_ip"];?>">
</form>
</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? |
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