|
PHP - Mysql Php If Else Not Working
I know this is probably something very dumb but I am having trouble with my if else. I have a sql query that looks for a record and if nothing is found it echos a warning.
Code: [Select] <?php $sql7="SELECT * FROM `$tbl_name2` WHERE Mac = '$Mac' AND Type = '$Type'"; $result7=mysql_query($sql7); if($result7) { while($row=mysql_fetch_array($result7)) echo "The phone you are using with mac $row[0] $row[1] is a Cisco IP Phone! <br>" ; } else { echo "<img src=\"stop.png\"/> The phone you are trying to use with mac $row[0] $row[1] is not a correct Mac\Type. <br> Please confirm the mac and type again."; } ?> I believe my syntax is correct as my if echo works fine. However, when no records are found I do not see my last else echo. Any input? Similar TutorialsWhen I echo the POST, it echoes the correct value. The MySQL portion seems to just ignore it all together. I've tried changing the dropdown option to just a text field, same thing occurred. I have text fields right above this particular one that update just fine with the SAME exactly scripting. if POST, update query, done. Works. This one for some reason will not. MySQL portion: if ($_POST['bUpdate']){ mysql_query("UPDATE `Patients` SET `b` = '$_POST[bUpdate]' WHERE `id` = '".$_GET['id']."'"); } echo $_POST['bUpdate']; Form Portion: Code: [Select] <tr onmouseover="color(this, '#baecff');" onmouseout="uncolor(this);"> <td width="310" colspan="2" align="center"><span class="fontoptions">Postcard Status </span><br /> <? if ($data['b'] == 1){ echo '<select name="bUpdate"><option value="1" selected>Yes</option><option value="0">No</option></select>'; } else { echo '<select name="bUpdate"><option value="1">Yes</option><option value="0" selected>No</option></select>'; } ?> </td> </tr> hello, here is my query $this->sql_query = "SELECT * FROM ".$this->classifieds_table." WHERE category ".$this->in_statement." AND live=1 "; the above works....but when I add {AND location = "".$_COOKIE["State"]."\"} - it doesnt work no matter what location equals it wont read query. as soon as i remove the extra AND it works. help please I am a noob at php, looking to become great. My php loop only seems to draw 1 row from the database when listing it out. Here is the loop <?php require('header.php'); mysql_select_db('center'); $sql = "select username, password from users"; $result = mysql_query($sql); $row = mysql_fetch_array($result); $user = $row['username']; $pass = $row['password']; echo "<table border='1'>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>$user</td>"; echo "<td>$pass</td>"; echo "</tr>"; } echo "</table>"; ?> Hiya peeps! I have this code; $q3 = "SELECT `data` FROM `firm_text_strings` WHERE `id` = '100' AND `languages_id` = 'EN'"; $q4 = "SELECT `data_2` FROM `firm_text_strings` WHERE `id` = '100' AND `languages_id` = 'EN'"; $r3 = mysql_query($q3); $r4 = mysql_query($q4); if($r3 && $r4) { while($w3 = mysql_fetch_assoc($r3) && $w4 = mysql_fetch_assoc($r4)) { $resultStr = $w3['data'] . '<br />' . $w4['data_2']; } } else { $resultStr = 'error'; } everything is working up to the while function, it just wont do the loop for some reason. Anyone have any ideas. Many thanks, James. Hi there, I have this query, which outputs the records properly in PhpMyAdmin. But when using this query inside my PHP code, the result is incomplete. Instead of retrieving the expected 15 results, I only get the first one twice. Also, I don't get any errors. Code: (php) [Select] <?php $get_files = "SELECT `id` FROM `files` WHERE `category` = 'W';"; include_once('connection.inc.php'); $con = mysql_connect(MYSQL_SERVER, MYSQL_USERNAME, MYSQL_PASSWORD) or die ("Connection failed: " . mysql_error()); $dbname = "myDB"; $select_db = mysql_select_db($dbname, $con) or die ("Can't open database: " . mysql_error()); $result_files = mysql_query($get_files) or die ("Query failed: " . mysql_error()); $files = mysql_fetch_array($result_files); $W_files = "'".implode("', '", $files)."'"; // echo $W_files; outputs the first number twice, instead of the expected 15 different ones $get_checklist = "SELECT * FROM `checklist` WHERE `id` IN ($W_files) ORDER BY `id` ASC;"; $result_checklist = mysql_query($get_checklist) or die ("Query failed: " . mysql_error()); // .... code for displaying ?> hi i am having a problem with a php script. i have 3 files index.php, home.html.php, form.html.php. home.html.php is working fine, if has a form that is submitting fine but when it submits i get a problem. maybe someone can help me out please i would be grateful as i am new. the is nothing wrong with anything else mysql database is running fine and there is no problem with connection. index.php (you can ignore what is commented out) <?php include $_SERVER['DOCUMENT_ROOT'] . '/includes/db.inc.php'; //$_GET['id'] is categoryid if (isset($_GET['id'])) { include $_SERVER['DOCUMENT_ROOT'] . '/includes/db.inc.php'; $id = mysqli_real_escape_string($link, $_GET['id']); $sql = "SELECT businessid FROM businesscategory WHERE categoryid LIKE '$id'"; $result = mysqli_query($link, $sql); if ($result !='') { $businessid = mysqli_fetch_array($result); $sql = "SELECT content FROM business WHERE id LIKE '$businessid'"; $result = mysqli_query($link, $sql); } if ($result !='') { $content = mysqli_fetch_array($result); include 'form.html.php'; exit(); } } /* // The basic SELECT statement $select = 'SELECT content'; $from = ' FROM business'; $where = ' WHERE TRUE'; $id = mysqli_real_escape_string($link, $_GET['id']); if ($category != '') // An owner is selected { $where .= " AND categoryid='$categoryid"; } $categoryid = mysqli_real_escape_string($link, $_GET['category']); if ($categoryid != '') // A category is selected { $from .= ' INNER JOIN businesscategory ON id = business'; $where .= " AND categoryid='$categoryid'"; } $text = mysqli_real_escape_string($link, $_GET['text']); if ($text != '') // Some search text was specified { $where .= " AND content LIKE '%$text%'"; } $result = mysqli_query($link, $select . $from . $where); if (!$result) { $error = 'Error fetching businesses.'; include 'error.html.php'; exit(); } while ($row = mysqli_fetch_array($result)) { $businesses[] = array('id' => $row['id'], 'text' => $row['content']); } include 'form.html.php'; exit(); } */ $result = mysqli_query($link, 'SELECT id, name FROM category ORDER BY name'); if (!$result) { $error = 'Error fetching categories: ' . mysqli_error($link); include 'error.html.php'; exit(); } while ($row = mysqli_fetch_array($result)) { $categories[] = array('id' => $row['id'], 'text' => $row['name']); } include 'home.html.php'; ?> form.html.php <?php include_once $_SERVER['DOCUMENT_ROOT'] . '/includes/helpers.inc.php'; ?> <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"> <html> <head> <meta http-equiv="content-type" content="text/html; charset=windows-1250"> <meta name="generator" content="PSPad editor, www.pspad.com"> <title>Business</title> </head> <body> it works <?php echo $content; ?> </body> </html> Hello,
I am trying to import a CSV file into an existing mysql table but it doesn't seem to work well.
Here is how my mysql is looking like:
1 a_id int(11) AUTO_INCREMENT 2 a_lobecoid int(7) 3 a_code varchar(30) utf8_general_ci 4 a_omschint varchar(60) utf8_general_ci 5 a_beveiligingniv int(5) 6 a_type varchar(5) utf8_general_ci 7 a_assortiment int(5) 8 a_discipline varchar(30) utf8_general_ci 9 a_brutoprijs varchar(50) utf8_general_ci 10 a_status varchar(5) utf8_general_ci 11 a_levcode varchar(10) utf8_general_ci 12 a_omschr_nl varchar(60) utf8_general_ci 13 a_omschr_fr varchar(60) utf8_general_cithese are some lines from my CSV file 16158|-H|Factory installed heater|10|S|400|CCTV|45.0|E| 1829|Factory installed heater|Factory installed heater 16159|-IR|Factory installed IR LED ring|10|S|400|CCTV|50.0|E| 1829|Factory installed IR LED ring|Factory installed IR LED ring 9001|00-SBN2|Smoke box niet geaspireerd,230VAC|10|S|267|BRAND|1587.03|D| 642|Smoke box niet geaspireerd,230VAC|Smoke box pas aspiré,230VCA 9003|00-TP1|Telescopische verlengstok voor Smoke box,2,4-4,6m|10|S|267|BRAND|644.09000000000003|D| 642|Telescopische verlengstok voor Smoke box,2,4-4,6m|Rallonge téléscopique pour Smoke box,2,4-4,6m 9004|00-TP2|Telescopische verlengstok voor Smoke box,2,4-9,2m|10|S|267|BRAND|944.64999999999998|D| 642|Telescopische verlengstok voor Smoke box,2,4-9,2m|Rallonge téléscopique pour Smoke box,2,4-9,2m 12161|001-0081|Thermistor probe,rood, high temp. 0-+150°C|10|S|52|INBRAAK|136.91|D| 1731|Thermistor probe,rood, high temp. 0-+150°C|Sonde température,rouge,high temp. 0-+150°CHere you have the code for the import: $upload_article_query = "LOAD DATA INFILE 'ARTIKELS.CSV' INTO TABLE artikelen FIELDS TERMINATED BY '|' LINES TERMINATED BY '\\r\\n' (a_lobecoid, a_code, a_omschint, a_beveiligingniv, a_type, a_assortiment, a_discipline, a_brutoprijs, a_status, a_levcode, a_omschr_nl, a_omschr_fr)"; $upload_article_stmt = $dbh->prepare($upload_article_query); $upload_article_stmt->execute();If i use the code then in the MYSQL table the first line is filled in but where the second line has to start it just writes it into the last column and doesn't start a new line. Also if i edit that first line it shows a lot of "?" (questionmarks) into a window symbol. Anyone has an idea what i am doing wrong? In attachment some printscreens of my table after the insert. Apologies for the Dutch language. Thanks in advance Attached Files pic1.jpg 6.13KB 0 downloads pic2.jpg 37.9KB 0 downloads How do I get LIKE to check multiple values? It works if you type in a single word, but not multiple, which I want it to in order to get an accurate search result. Here is what I am trying to do: $target = "siemens 6s" ; $multis = explode(" ", $target) ; $query = "SELECT parts_table.*, manufacturers.* FROM parts_table, manufacturers WHERE parts_table.ManufacturerID = manufacturers.ManufacturerID AND parts_table.PartNumber LIKE" ; foreach($multis as $current) { $queryext .= " '%$current%' OR" ; } $queryext = substr($queryext, 0, -3) ; // Remove the last 4 charecters i.e. " AND". $query .= $queryext ; echo $query .= " OR manufacturers.ManufacturerID = parts_table.ManufacturerID AND manufacturers.ManufacturerName LIKE" . $queryext ; include("dbconnect.php") ; $result = mysql_query($query) ; while($row = mysql_fetch_assoc($result)){ echo $row['ManufacturerName'] ; echo $row['PartNumber'] ; } I'm experimenting out premade codes and trying to make them some how work the way I want to. This is "my" image uploader. For the uploader is suppose to find the image AND the code stored in mysql, but it doesn't seem to find it. any problem? <?php $link = mysql_connect("localhost", "mcd", "whatanicepassword"); mysql_select_db("mcd", $link); if (!isset($_POST["code"])) { die ("Error: Not all fields complete"); } $limit_size=5120; $target = "skin/"; $target = $target . basename( $_FILES['uploaded']['name']); $ok=1; $filelol = $_FILES['uploaded']['name']; $file_size=$_FILES['uploaded']['size']; $filecheck = "skin/".$filelol; //This is our size condition if ($file_size >= $limit_size) { echo "Your file is too large.<br>"; $ok=0; } if (file_exists($filecheck)) { } else {echo $filelol." does not exist in the database. Please upload at the main site.<br>"; $ok=0;} // username and password sent from Form $uploaded=mysql_real_escape_string($_POST['uploaded']); $code=mysql_real_escape_string($_POST['code']); $checkquery = mysql_query("SELECT * FROM user_list WHERE uploaded = '$uploaded' AND code = '$code'") or die("Error : " . mysql_error()); $num_rows=mysql_num_rows($checkquery); if ($num_rows < 1 ) { echo 'File not found in MySQL<br><br><br>'; $ok=0; } else {echo 'YES!';} //This is our limit file type condition if (($_FILES["uploaded"]["type"] != "image/png")) { echo "You may only upload PNG files.<br>"; $ok=0; } if ($ok==0) { Echo "Sorry your file was not uploaded<br>"; } //If everything is ok we try to upload it else { if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target)) { echo "The file ".$filelol." has been uploaded"; } else { echo "Sorry, there was a problem uploading your file."; } } ?> Hello all, I am using the Lynda.com PHP course to further learn php. On one of the examples, they are using a SQL Query that works for them but is erroring out for me. It is the query below for the $page_set variable, particularly the use of the WHERE subject_id = {$subject["id"]}". Code: [Select] <?php require_once ('includes/conf.php'); ?> <?php include ('includes/header.php'); ?> <?php require_once ('includes/functions.php'); ?> <table id="stucture"> <tr> <td id="navigation"> <ul class="subjects"> <?php $subject_set = mysql_query("SELECT * FROM subjects", $connection); if (!$subject_set) { die ("Database query failed: " . mysql_error()); } while ($subject = mysql_fetch_array($subject_set)) { echo "<li>{$subject["menu_name"]}</li>"; } $page_set = mysql_query("SELECT * FROM pages WHERE subject_id = {$subject["id"]}"); // This appears to be the problem area, unless it is fooling me. It works with this exact statement in their code. if (!$page_set) { die ("Database query failed: " . mysql_error()); } echo "<ul class='pages'>"; while ($page = mysql_fetch_array($page_set)) { echo "<li>{$page['menu_name']}</li>"; } echo "</ul>"; ?> </ul> </td> <td id="page"><h2>Content Area</h2> </td> </tr> </table> <?php require ('includes/footer.php'); ?> The error listed is this: Database query failed: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 Thanks, you all are the best! BtD I have a mysql insert statement generated with php that is not populating the table. I've echoed the statement and if I copy and paste into phpmyadmin it works fine. The result of the mysql_query function is true. I've emptied the table so there are no primary key conflicts. I've put the statement in a try catch and it does not display a exception. What else can I try? Here's the statement INSERT INTO `wp_term_relationships` (object_id, term_taxonomy_id, term_order) VALUES (1597,83,0) Works absolute fine if I copy and paste into phpmyadmin. Does not populated the table if run through mysql_query Hi guys,
Does anyone know why my code isn't working?:
$servername="127.0.0.1";
$username="root";
$password="";
$dbname="testdb";
$connection = new mysqli($servername,$username,$password);
$result = "SELECT 'Country' from 'w'";
echo '<select>';
foreach($result as $res) {
echo '<option value="'.$res['Country'].'">' . $res['Country'] . '</option>';
}
echo '</select>';
}
There's not a credential issue, the table is called 'w' and the collumn I'm trying to get the values from is 'Country'
Any help is appreciated.
Cheers,
Broken
hi - i'm trying to update mysql databse but not working... any ideas??? Code: [Select] [php] <?php ob_start(); require_once('../xconfig.php'); if(isset ($_GET['quote'])&& $_GET['quote']!=""){ $quote_id=$_GET['quote']; mysql_select_db($database, $makeconnection); $sql_find_quote = "SELECT * FROM tbl_quote WHERE quote_id = $quote_id"; $find_quote = mysql_query($sql_find_quote, $makeconnection) or die(mysql_error()); $row = mysql_fetch_assoc($find_quote); $totalRows = mysql_num_rows($find_quote); $quote_author = $_POST['quote_author']; $quote_desc = $_POST['quote_desc']; if (isset($_POST['submitted'])&&($_POST['submitted'] == "yes")) { $register_query = "SELECT quote_desc FROM tbl_quote WHERE quote_desc='$quote_desc'"; mysql_select_db($database, $makeconnection); $register_check=mysql_query($register_query, $makeconnection); $register_found_quote = mysql_num_rows($register_check); if($register_found_quote>1){ header ("Location: quote_modify.php?error=quoteexists"e=$quote_id"); }else{ $sql_modify = "UPDATE tbl_quote SET quote_desc = '$quote_desc' quote_author = '$quote_author' WHERE quote_id = '$quote_id'"; } mysql_select_db($database, $makeconnection); $Result1 = mysql_query($sql_modify, $makeconnection) or die(mysql_error()); header ("Location: quote.php"); } } ob_flush(); ?> [/php] here's my HTML form Code: [Select] <form action="" method="post" enctype="multipart/form-data" name="form1" id="form1"> <p> <input name="quote_author" type="text" class="active-field" id="quote_author" value="<? echo $row['quote_author'];?>"/> <br /> <input name="quote_desc" type="text" class="active-field" id="quote_desc" value="<? echo $row['quote_desc'];?>" /> <br /> <input name="submitted" type="hidden" id="submitted" value="yes" /> <input name="submit" type="submit" id="submit" class="submit-button" value="modiy"/> <br /> </p> </form> I'm a beginner here and i am learning the basic in converting from MySQL to MySQLi. I am currently working on this registration page which I would want to convert to MySQLi. Please advise me how to modify this script, I would prefer the procedural style.
The MySQLi coding is not working because it would notg insert into the database like the MySQL coding would, would appreciate if your can help me.
MYSQL
<?php
error_reporting(1);
$submit = $_POST['submit'];
//form data
$name = mysql_real_escape_string($_POST['name']);
$name2 = mysql_real_escape_string($_POST['name2']);
$email = mysql_real_escape_string($_POST['email']);
$password = mysql_real_escape_string($_POST['password']);
$password2 = mysql_real_escape_string($_POST['password2']);
$email2 = mysql_real_escape_string($_POST['email2']);
$address = mysql_real_escape_string($_POST['address']);
$address2 = mysql_real_escape_string($_POST['address2']);
$address3 = mysql_real_escape_string($_POST['address3']);
$address4 = mysql_real_escape_string($_POST['address4']);
$error = array();
if ($submit) {
//open database
$connect = mysql_connect("localhost", "root", "Passw0rd") or die("Connection Error");
//select database
mysql_select_db("logindb") or die("Selection Error");
//namecheck
$namecheck = mysql_query("SELECT * FROM users WHERE email='{$email}'");
$count = mysql_num_rows($namecheck);
if($count==0) {
}
else
{
if($count==1) {
$error[] = "<p><b>User ID taken. Try another?</b></p>";
}
}
//check for existance
if($name&&$name2&&$email&&$password&&$password2&&$email2&&$address&&$address2&&$address3&&$address4) {
if(strlen($password)<8) {
$error[] = "<p><b>Password must be least 8 characters</b></p>";
}
if(!preg_match("#[A-Z]+#",$password)) {
$error[] = "<p><b>Password must have at least 1 upper case characters</b></p>";
}
if(!preg_match("#[0-9]+#",$password)) {
$error[] = "<p><b>Password must have at least 1 number</b></p>";
}
if(!preg_match("#[\W]+#",$password)) {
$error[] = "<p><b>Password must have at least 1 symbol</b></p>";
}
//encrypt password
$password = sha1($password);
$password2 = sha1($password2);
if($_POST['password'] != $_POST['password2']) {
$error[] = "<p><b>Password does not match</b></p>";
}
//rescue email match check
if($_POST['email2'] == $_POST['email']) {
$error[] = "<p><b>Rescue Email must not be the same as User ID</b></p>";
}
//generate random code
$random = rand(11111111,99999999);
//check for error messages
if(isset($error)&&!empty($error)) {
implode($error);
}
else
{
//Registering to database
$queryreg = mysql_query("INSERT INTO users VALUES ('','$name','$name2','$email','$password','$password2','$email2','$address','$address2','$address3','$address4','$random','0')");
$lastid = mysql_insert_id();
echo "<meta http-equiv='refresh' content='0; url=Activate.php?id=$lastid&code=$random'>";
die ();
}
}
}
?>
MYSQLi (NOT WORKING AFTER CONVERTING)
<?php
error_reporting(1);
$submit = $_POST['submit'];
//form data
$name = mysqli_real_escape_string($connect, $_POST['name']);
$name2 = mysqli_real_escape_string($connect, $_POST['name2']);
$email = mysqli_real_escape_string($connect, $_POST['email']);
$password = mysqli_real_escape_string($connect, $_POST['password']);
$password2 = mysqli_real_escape_string($connect, $_POST['password2']);
$email2 = mysqli_real_escape_string($connect, $_POST['email2']);
$address = mysqli_real_escape_string($connect, $_POST['address']);
$address2 = mysqli_real_escape_string($connect, $_POST['address2']);
$address3 = mysqli_real_escape_string($connect, $_POST['address3']);
$address4 = mysqli_real_escape_string($connect, $_POST['address4']);
$error = array();
if ($submit) {
//open database
$connect = mysqli_connect("localhost", "root", "Passw0rd", "logindb") or die("Connection Error");
//namecheck
$namecheck = mysqli_query($connect, "SELECT * FROM users WHERE email='{$email}'");
$count = mysqli_num_rows($namecheck);
if($count==0) {
}
else
{
if($count==1) {
$error[] = "<p><b>User ID taken. Try another?</b></p>";
}
}
//check for existance
if($name&&$name2&&$email&&$password&&$password2&&$email2&&$address&&$address2&&$address3&&$address4) {
if(strlen($password)<8) {
$error[] = "<p><b>Password must be least 8 characters</b></p>";
}
if(!preg_match("#[A-Z]+#",$password)) {
$error[] = "<p><b>Password must have at least 1 upper case characters</b></p>";
}
if(!preg_match("#[0-9]+#",$password)) {
$error[] = "<p><b>Password must have at least 1 number</b></p>";
}
if(!preg_match("#[\W]+#",$password)) {
$error[] = "<p><b>Password must have at least 1 symbol</b></p>";
}
//encrypt password
$password = sha1($password);
$password2 = sha1($password2);
if($_POST['password'] != $_POST['password2']) {
$error[] = "<p><b>Password does not match</b></p>";
}
//rescue email match check
if($_POST['email2'] == $_POST['email']) {
$error[] = "<p><b>Rescue Email must not be the same as User ID</b></p>";
}
//generate random code
$random = rand(11111111,99999999);
//check for error messages
if(isset($error)&&!empty($error)) {
implode($error);
}
else
{
//Registering to database
$queryreg = mysqli_query($connect, "INSERT INTO users VALUES ('','$name','$name2','$email','$password','$password2','$email2','$address','$address2','$address3','$address4','$random','0')");
$lastid = mysqli_insert_id();
echo "<meta http-equiv='refresh' content='0; url=Activate.php?id=$lastid&code=$random'>";
die ();
}
}
}
?>
Hello. i am writing an API for lua... that sends POST requests to a php script. this php script is malfunctioning. below is the code and error. The code below worked perfectly on my old hosting service. I've had to move to a new host and I get this error when I try to run the code. QuoteWarning: Cannot modify header information - headers already sent by (output started at ../user/dbconfig.php:11) This is the code:-
<?php
// connect to DATABASE!
include_once('../user/dbconfig.php');
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);
$options = [
\PDO::ATTR_ERRMODE => \PDO::ERRMODE_EXCEPTION,
\PDO::ATTR_DEFAULT_FETCH_MODE => \PDO::FETCH_ASSOC,
\PDO::ATTR_EMULATE_PREPARES => false,
];
$dsn = "mysql:host=$host;dbname=$dbname;charset=$charset;port=$port";
try {
// create a PDO connection with the configuration data
$conn = new PDO($dsn, $user, $dbpassword, $options);
} catch (PDOException $e) {
// report error message
echo $e->getMessage();
}
$ans = $_POST['answer'];
//Setup the filename that our CSV will have when it is downloaded.
if($ans == 'yes'){
$fileName = 'answers-right.csv';
} elseif ($ans == 'no') {
$fileName = 'answers-wrong.csv';
}
// Create our SQL query.
$sql = 'SELECT ID,email,firstname,lastname,message FROM checks WHERE answer = ?';
$stmt = $conn->prepare($sql);
$stmt->execute([$ans]);
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC); // NB: now using fetchAll because there's going to be more than one record
//Get the column names.
$columnNames = array();
if(!empty($rows)){
//We only need to loop through the first row of our result
//in order to collate the column names.
$firstRow = $rows[0];
foreach($firstRow as $colName => $val){
$columnNames[] = $colName;
}
}
//If the file exists and is writeable then delete it
if(is_writable($fileName)){
//Delete the file
$deleted = unlink($fileName);
}
//Set the Content-Type and Content-Disposition headers to force the download.
header('Content-Type: application/excel');
header('Content-Type: application/excel Content-Disposition: attachment; filename="' . $fileName . '"');
//Open up a file pointer
$fp = fopen('php://output', 'w');
//Start off by writing the column names to the file.
fputcsv($fp, $columnNames);
//Then, loop through the rows and write them to the CSV file.
foreach ($rows as $row) {
fputcsv($fp, $row);
}
//Close the file pointer.
fclose($fp);
?>
Would appreciate some help please? view_leader.php
Hi guys need some help.I created a simple Update Customer Details page Where you Enter the Customer ID ,The Customer Details Get Displayed inside a Form and you make changes within the Form.But the update query I wrote is not working as it should be.It Executes the Query ,when I hardcode the Customer ID inside the Update query but fails when I need to Enter the Customer ID directly from the form using post.In This Code I tried to Update only the Customer's Firstname.Thanks Here Is The Code Code: [Select] <?php $db=mysql_connect("localhost","root","") or die('Unable to Connect To Database.Please check the Database Parameters'); mysql_select_db('ecommerce') or die(mysql_error()); if(isset($_POST['enterid'])) { $_POST['inputid']; } $query="SELECT * FROM customer WHERE Cid='$_POST[inputid]'"; $result=mysql_query($query) or die(mysql_error()); $row=mysql_fetch_array($result); $new_cid=$row['Cid']; $new_firstname=$row['Cfname']; $new_lastname=$row['Clname']; $new_email=$row['Email_id']; $new_address=$row['Address']; $new_pincode=$row['Pincode']; $new_payment=$row['Mode_of_payment']; $new_city=$row['City']; $new_state=$row['State']; $new_phone=$row['Phone']; $html1=<<<HTML1 <form action="updatecustomer.php" method="post"> <p class="inputidentifier">Please Enter The Customer's ID</p><input type="text" style="width:375px;height:40px;font-size:30px;" name="inputid" size="20"><br><br> <input type="submit" name="enterid" style="height:40px"> HTML1; print($new_cid); if(isset($_POST['update'])) { $query1="UPDATE customer SET Cfname='$_POST[firstname]' WHERE Cid='$_POST[inputid]'"; mysql_query($query1) or die(mysql_error()); if(mysql_affected_rows()==1) print("Query sucessful"); else print("Something went wrong"); } ?> <html> <head> <style type="text/css"> .inputtext {width:300px; height:40px;font-size:30px;} .inputidentifier{font-size:25px;font-family:"Arial"} .h1type{font-family:"Arial"} </style> <title>Test</title> </head> <body> <?php print($html1); ?> <h1 align="center" class="h1type">Update Customer Details</h1> <form action="updatecustomer.php" method="POST"> <table align="center" cellspacing="10" cellpadding="10" border="0" width="60%"> <tr> <td align="right" class="inputidentifier">First Name</td> <td align="left"><input type="text" class="inputtext" name="firstname" placeholder="eg:Kevin" value="<?php print($new_firstname) ?>"></td> </tr> <tr> <td align="right" class="inputidentifier">Last Name</td> <td><input type="text" class="inputtext" name="lastname" placeholder="eg:Aloysius" value="<?php print($new_lastname) ?>"></td> </tr> <tr> <td align="right" class="inputidentifier">E-mail</td> <td align="left"><input type="email" style="width:500" class="inputtext" name="email" placeholder="yourname@email.com" value="<?php print($new_email) ?>"> </tr> <tr> <td align="right" class="inputidentifier">Phone Number</td> <td align="left"><input type="text" class="inputtext" name="phone" placeholder="How Do We Call You?" value="<?php print($new_phone) ?>"></td> </tr> <tr> <td align="right" class="inputidentifier">Address</td> <td><textarea style="width:500;height:150" wrap="virtual" class="inputtext" name="address" placeholder="Where is your Crib?"><?php print($new_address) ?></textarea></td> </tr> <tr> <td align="right" class="inputidentifier">State</td> <td align="left"><input type="text" style="width:500" class="inputtext" name="state" placeholder="State" value="<?php print($new_state) ?>"> </tr> <tr> <td align="right" class="inputidentifier">City</td> <td align="left"><input type="text" class="inputtext" name="city" placeholder="City" value="<?php print($new_city) ?>"> </tr> <tr> <td align="right" class="inputidentifier">Pin Code</td> <td><input type="text" class="inputtext" name="pincode" placeholder="Mulund 400080" maxlength="6" value="<?php print($new_pincode) ?>"></td> </tr> <tr> <td align="right" class="inputidentifier">How do Pay for your Bling?</td> <td align="left"> <input type="text" class="inputtext" name="payment" value="<?php print($new_payment)?>"> </tr> <tr> <td></td> <td><input type="submit" name="update" value="Update!" style="width:100px;height:60px;"></td> </tr> </table> </form> </body> </html> Hi everyone, I have been trying to get this code working but no matter what I try it will not work, I am trying to allow a user to update their personal details when they login to my website. I have created a form and submit button, the code is shwn below: <?php echo $_SESSION['myusername']; ?> <br><br> <form action="updated.php" method="post"> Firstname: <input type="text" name="firstname" /><br><br> Surname : <input type="text" name="surname" /><br><br> Date Birth: <input type="text" name="dob" /><br><br> Total Wins: <input type="text" name="wins" /> Total Loses: <input type="text" name="loses" /><br><br> Email Add: <input type="text" name="email" /><br><br> Country : <input type="text" name="born" /><br><br> Other Info: <input type="text" name="other" /><br><br> <input type="submit" name="Submit" value="Update" align="right"></td> </form> The updated.php file <?php mysql_connect ("localhost","root","") or die("Cannot connect to Database"); mysql_select_db ("test"); $sql=mysql_query("UPDATE memberdetails SET firstname='{$_POST['firstname']}', surname='{$_POST['surname']}', dob='{$_POST['dob']}', totalwins='{$_POST['wins']}', totalloses='{$_POST['loses']}', email='{$_POST['email']}', country='{$_POST['born']}', info='{$_POST['other']} WHERE username=$_SESSION['myusername']); if (!mysql_query($sql)) { die('Error: ' . mysql_error()); } echo "Details Updated"; ?> This is the error i recieve: Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in C:\wamp\www\website\updated.php on line 45 Line 45 is $sql=mysql_query line If anybody can help it will be very much appreciated! |
|||||||