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PHP - Posted Value Not Showing In Ie6
Hi Friends,
I have form in that i am sending some values using POST method. But i cant able to get values using $_POST['var'] or $_REQEUST['var'] Data is not coming in IE6, But it is coming in firefox and other browsers even in IE8. What may be problem? any settings i have to make for IE6. ? Any suggestions Similar TutorialsI need to echo back the users drop down selection. How can I do this without storing the value? Code: [Select] $isset = $_POST['industry']; foreach ($industries as $industry) { if (isset($isset) && $industry['industry_id'] == $isset) { echo "<option value=\"{$industry['industry_id'] }\" selected=\"selected\">${industry['industry']}</option>\n"; } else { echo "<option value=\"{$industry['industry_id'] }\">${industry['industry']}</option>\n"; } } When I submit my form the variables are sent to the url in my address bar and not to the php post page (resersend.php). Here is the pertinent html page code (index.html)... Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <script type="text/javascript"> function showMonth(str) { if (str=="") { document.getElementById("txtHint").innerHTML=""; return; } if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("txtHint").innerHTML=xmlhttp.responseText; } } xmlhttp.open("GET","reserv.php?q="+str,true); xmlhttp.send(); } </script> </head> <body> <form> <p><font face="arial" size="2" color="#336600">Reservation month:</font> <select name="months" onchange="showMonth(this.value)"> <option value="" selected="selected">Choose One</option> <option value="May 2011">May 2011</option> <option value="June 2011">June 2011</option> <option value="July 2011">July 2011</option> <option value="August 2011">August 2011</option> <option value="September 2011">September 2011</option> <option value="October 2011">October 2011</option> <option value="November 2011">November 2011</option> <option value="December 2011">December 2011</option> <option value="January 2012">January 2012</option> <option value="February 2012">February 2012</option> <option value="March 2012">March 2012</option> <option value="April 2012">April 2012</option> </select> </form></p> </body> </html> And here is the php code from the page (reserv.php) that works with the above html page... Code: [Select] <?php $q=$_GET["q"]; $con = mysql_connect('My_host', 'my_user', 'my_password'); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("my_db", $con); $sql="SELECT * FROM daterange WHERE DEND > DATE(NOW()) AND STATUS='A' AND MONTH = '".$q."' ORDER BY RID, DATE, SITE"; $result = mysql_query($sql); // Determine the number of reservation dates $number = mysql_numrows($result); // Create drop-down menu of reservation dates print "<font size=\"3\" face=\"Arial\"><b>Select Your Reservation:</b><br> <form action=\"resersend.php\" method=\"post\"> <select name=\"RID\"> <option value=\"\">Choose One</option>"; for ($i=0; $i<$number; $i++) { $RID = mysql_result($result,$i,"RID"); $DATE = mysql_result($result,$i,"DATE"); $SITE = mysql_result($result,$i, "SITE"); $PRICE = mysql_result($result,$i, "PRICE"); print "<option value=\"$RID\">$DATE, $SITE, $PRICE</option>"; } print "</select><p align=left><label><font size=\"3\" face=\"Arial\">First Name: <input type=\"text\" name=\"FNAME\" size=\"50\" maxlength=\"50\" tabindex=\"1\"<br>"; print "<p align=left><label>Last Name: <input type=\"text\" name=\"LNAME\" size=\"50\" maxlength=\"50\" tabindex=\"2\"<br>"; print "<p align=left><label>Address Line 1: <input type=\"text\" name=\"ADDR1\" size=\"50\" maxlength=\"50\" tabindex=\"3\"<br>"; print "<p align=left><label>Address Line 2: <input type=\"text\" name=\"ADDR2\" size=\"50\" maxlength=\"50\" tabindex=\"4\"<br>"; print "<p align=left><label>City: <input type=\"text\" name=\"CITY\" size=\"50\" maxlength=\"50\" tabindex=\"5\"<br>"; print "<p align=left><label>State (abbrev.): <input type=\"text\" name=\"STATE\" size=\"2\" maxlength=\"2\" tabindex=\"6\"<br>"; print "<p align=left><label>Zip Code: <input type=\"text\" name=\"ZIP\" size=\"5\" maxlength=\"5\" tabindex=\"7\"<br>"; print "<p align=left><label>Contact Phone Number: (<input type=\"text\" name=\"PHONE1\" size=\"3\" maxlength=\"3\" tabindex=\"8\""; print "<label>)<input type=\"text\" name=\"PHONE2\" size=\"3\" maxlength=\"3\" tabindex=\"9\""; print "<label>-<input type=\"text\" name=\"PHONE3\" size=\"4\" maxlength=\"4\" tabindex=\"10\"<br>"; print "<p align=left><label>Email: <input type=\"text\" name=\"EMAIL\" size=\"50\" maxlength=\"50\" tabindex=\"11\"<br>"; print "<p align=left><input type=\"submit\" value=\"Book Now!\" name=\"submit\">"; print "<input type=\"reset\" value=\"reset\" name=\"reset\"></form>"; // Close the database connection mysql_close($con); ?> I am sure this is easy, but any help would be greatly appreciated. Thanks. I have a status system, built with ajax, and I need to get the unique id that the database assigns to the post and return it in the results that are appended to the page via javascript, however I'm not sure how the bessst way would be to do this? I have been doing this: ///variables that are sent to the database $session = $_COOKIE['id']; $uid = $_POST['id']; $post = $_POST['post']; $date = mktime(); //way I have got the unique id of this post $query = mysql_fetch_assoc(mysql_query("SELECT * FROM posts WHERE from_id='$session' AND to_id='$uid' AND post='$post' AND date='$date'")); $id = $query['id']; Although that does work, It feels like it might not always be accurate. How can I echo the $id? Every time I post approved it refreshes my page and I can see the echoed out ID. I need to know what the ID value is. Code: [Select] if (isset($_POST['approve'])) { if(is_array($_POST['approve'])) { $keys = array_keys($_POST['approve']); $id = $keys[0]; echo $id; die(); //$sql = "UPDATE `blog_comments` SET `approved` = 1 WHERE `post_id` = '$id'"; //header("Location: " . $_SERVER['PHP_SELF'] . "?" . $_SERVER['QUERY_STRING'] ); } } I need some help displaying the posted values onto a second page. I thought I was doing it right, but nothing is showing up on my second page. I have my form set to PHP_SELF to validate before sending the user to the second page, and it's working just fine. If no errors are present, the user is redirected to the second page, but it's supposed to be displaying their posted data, but it's not. What am I doing wrong? Code: [Select] <?php /*$name = ""; $address = ""; $city = ""; $state = ""; $zip = ""; $phone = ""; $position = ""; $display_errors = "";*/ if (@$_POST['submitted']){ $name = (@$_POST['name']); $address = (@$_POST['address']); $city = (@$_POST['city']); $state = (@$_POST['state']); $zip = (@$_POST['zip']); $phone = (@$_POST['phone']); $position = (@$_POST['position']); if (get_magic_quotes_gpc()){ $name = stripslashes($name); $address = stripslashes($address); $city = stripslashes($city); $state = stripslashes($state); $zip = stripslashes($zip); $phone = stripslashes($phone); $position = stripslashes($position); } $error_msg=array(); if ($name==""){ $error_msg[]="Please enter your name"; } if ($address==""){ $error_msg[]="Please enter your address"; } if ($city==""){ $error_msg[]="Please enter your city"; } if ($state==""){ $error_msg[]="Please enter your state"; } if ($zip==""){ $error_msg[]="Please enter your zip code"; } if ($phone==""){ $error_msg[]="Please enter your phone number"; } if ($position==""){ $error_msg[]="Please enter your desired position"; } if ($error_msg){ $display_errors = "<ul style=color:red>\n"; foreach ($error_msg as $err){ $display_errors .= "<li>".$err."</li>\n"; } $display_errors .= "</ul>\n"; } if (!$error_msg){ //echo $name . "<br>"; //echo $address . "<br>"; //echo $city . "<br>"; //echo $state . "<br>"; header('Location: process_U2IP.php'); exit(); } } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Forms1.php</title> </head> <body> <div id="content"> <form action="<?php $_SERVER['PHP_SELF']; ?>" method="post"> <h3>Please fill out the form below and upload your resume to apply for a position within our company:</h3> <label for="name">Name*</label><br /> <input name="name" type="text" size="20" id="name" value="<?php echo $name; ?>" /><br /> <label for="address">Address*</label><br /> <input name="address" type="text" size="20" id="address" value="<?php echo $address; ?>" /><br /> <label for="city">City*</label><br /> <input name="city" type="text" size="20" id="city" value="<?php echo $city; ?>" /><br /> <label for="state">State*</label><br /> <input name="state" type="text" size="20" id="state" value="<?php echo $state; ?>" /><br /> <label for="zip">Zip*</label><br /> <input name="zip" type="text" size="20" id="zip" value="<?php echo $zip; ?>" /><br /> <label for="phone">Phone*</label><br /> <input name="phone" type="text" size="20" id="phone" value="<?php echo $phone; ?>" /><br /> <label for="position">Position*</label><br /> <input name="position" type="text" size="20" id="position" value="<?php echo $position; ?>" /><br /> <p><i>Your information will not be sold or shared with others.</i></p> <p style="color: red;">* denotes required field</p> <input type="hidden" name="submitted" value="1" /> <input type="submit" value="Submit" /> <input type="reset" name="reset" value="Reset" /> </form> <?php echo $display_errors; ?> </div> </body> </html> Code: [Select] <?php $name = (@$_POST['name']); $address = (@$_POST['address']); $city = (@$_POST['city']); $state = (@$_POST['state']); $zip = (@$_POST['zip']); $phone = (@$_POST['phone']); $position = (@$_POST['position']); $display_block = " <table> <tr> <td><b>Name:</b></td> <td>$name</td> </tr> <tr> <td><b>Address:</b></td> <td>$address</td> </tr> <tr> <td><b>City:</b></td> <td>$city</td> </tr> <tr> <td><b>State:</b></td> </tr> </table>"; echo $display_block; ?> It's just displaying the table but with no values or blank values for the posted variables. Can anyone help? am coming across the problem. When I use copied text in my input form it gives some characters this � character.
I am not sure how to post text so this doesn't happen.
Example:
This line copied:
The 976cc, 101-hp Rotax® 1000R V-Twin engine is known for its reliable power.
Produces:
The 976cc, 101-hp Rotax� 1000R V-Twin engine is Can-Am's most powerful side-by-side engine.
My code:
if (get_magic_quotes_gpc()) {
$description_clean_magic = stripslashes($description);
} else {
$description_clean_magic = $description;
}
Hi everyone,
I've been running a self-written php shopping cart for the last 8 years, with lots of success. Recently however, the credit card processor (Linkpoint) I use upgraded their gateway, and now I have a problem.
Basically, the last page of my checkout does a _POST to a specific url for their gateway. I send through a bunch of <input> variables, some of them hidden, but all of them check out okay. I have combed over their specs, made sure I'm not sending any bad characters, made sure I've got the right format for the fields they want, etc...
Right now the transaction is successful, the customer's card is charged, but Linkpoint's system is returning an error message: We are unable to process your request due to invalid data.
Long story short, their support staff has asked me to "send them the xml string" via email. I have no idea how to do this (even after spending a few hours online trying to figure it out).
Is there a simple way to generate the xml string that's being passed during the _POST? Some kind of php code I can incorporate into my final checkout page, run a test transaction, and maybe write the string out somewhere so I can get it to them?
Any help you guys could give me would be greatly appreciated. Thanks in advance! Danny Hi, I only want to display some included PHP code only if a value is displayed from form input. The value is a zip code which is used to display weather information, but the app produces an error when you first enter the page, because their is no value for the zip variable. I know I need an if/else statement, but not sure what to put in it. The form code is this <h3>Enter your zip code to receive weather info.</h3> <form method="post" action="<?php echo $PHP_SELF;?>"> Enter Zip Code:<input type="text" size="12" maxlength="12" name="zip"><br /> <input id="submit" type="submit" name="Submit" value="Enter Zip"/> </form> <?php // this is how we bring in the weather file include 'weather_report.php' ?> the zip variable is set-up in the weather_report.php file as follows $zip = $_POST["zip"]; but like I said until that value is set, an error is thrown by the weather app code. How can I prioritize the data posted. Right now this if statements are being posted at different times. If I hit submit and all the if statements post to mysql it is saying that they all receive the same time stamp, but the data is never posted in the same order. Can I order the posting process in php? Code: [Select] if ((isset($_POST['companywebsiteas'])) && ($_POST['companywebsiteas'] != $r['companywebsite'])) { $details = $_POST['companywebsiteas']; update_company_actions($info['companyid'], 18, $details); } if ((isset($_POST['companyindustryas'])) && ($_POST['companyindustryas'] != $info['industry'])) { $details = $_POST['companyindustryas']; update_company_actions($info['companyid'], 17, $details); } if ((isset($_POST['companystageas'])) && ($_POST['companystageas'] != $info['stage'])) { $details = $_POST['companystageas']; update_company_actions($info['companyid'], 10, $details); } if ((!isset($_POST['companystateas'])) && (($_POST['companycountryas'] != $info['country']) || ($_POST['companycityas'] != $info['city']))) { $details = $_POST['companycityas'].', '.$_POST['companycountryas']; update_company_actions($info['companyid'], 11, $details); } else if ((isset($_POST['companystateas'], $_POST['companycityas'])) && (($_POST['companystateas'] != $info['state']) || ($_POST['companycityas'] != $info['city']))) { $details = $_POST['companycityas'].', '.$_POST['companystateas']; update_company_actions($info['companyid'], 11, $details); }
<?php
include("add_p2.php");
//basic pdo connection (with added option for error handling)
if (isset($_POST['submit'])) {
try {
$stmt = $dbh->prepare("INSERT INTO passwords (description, username, password ) VALUES (:description, :username, :AES_ENCRYPT)");
$stmt->bindParam(':description', $description);
$stmt->bindParam(':username', $username);
$stmt->bindParam(':password', $password);
try {
$statement->execute(array(
"description" => "description",
"username" => "username",
"password" => "password"
));
} catch(PDOException $e) {
echo "Exception caught: $e";
}
//insert one row
$name = 'one';
$value = 1;
$stmt->execute();
header( 'Location: localhost/password_added.php' ); exit();
?>
<?php
$dsn = "mysql:host=127.0.0.1;dbname=pword_db;charset=utf8";
$opt = array(
PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC
);
$pdo = new PDO($dsn,'root','', $opt);
?>
<form action="add_p.php" method="post"> <span class="cursive">Description: </span><br> <textarea name="description" rows="20" cols="50"></textarea> <br> <span class="cursive">Username: </span><br> <input type="text" name="username" /> <br> <span class="cursive">Password: </span><br> <input type="text" name="password" /> <br> <input type="submit"/> </form>Hi, above is the 3 components that should add form data into MySQL database using PDO. all it's doing is entering empty values, I'm really not sure why this is? please could someone help me with pointing me in the right direction, offer hint's making this work Thanks Attached Files PHPmyadmin.JPG 64.83KB 0 downloads To anyone, Apache 2.2.21/PHP5.3.5/MySQL 5.2.36. The following code returns the below errors. I am trying to get the POSTed values into MySQL. There are no other errors. It's like the code stopped at the ?> closing statement. Why would the code not return the 'echo' statements? Any ideas????? Much appreciated....Buchberg Notice: Use of undefined constant id - assumed 'id' in C:\website\do_addrecord_auth_users.php on line 2 Notice: Undefined index: id in C:\website\do_addrecord_auth_users.php on line 2 <? if ((!$_POST[id]) || (!$_POST[f_name]) || (!$_POST[l_name]) || (!$_POST[username]) || (!$_POST[password]) || (!$_POST[firm]) || (!$_POST[email])) { ("Location: /show_addrecord_auth_users.html"); exit; } $db_name="booster"; $table_name="auth_users"; $connection=mysql_connect("localhost", "username", "password") or die (mysql_error()); $db=mysql_select_db($db_name, $connection) or die (mysql_error()); $sql="INSERT INTO $table_name (id, f_name, l_name, username, password, firm, email) VALUES ('$_POST[id]', '$_POST[f_name]', '$_POST[l_name]', '$_POST[username]', '$_POST[password]', '$_POST[firm]', '$_POST[email]')"; $result=mysql_query($sql, $connection) or die (mysql_error()); ?> <HTML> <HEAD> <TITLE>Add a record to Auth Users</TITLE> </HEAD> <BODY> <H1>Adding a record to <? echo "$table_name"; ?></H1> <TABLE CELLSPACING=3 CELLPADDING=3> <TR> <TD VALIGN=TOP> <P>id<BR> <? echo "$_POST[id]"; ?></P> </TD> <TD VALIGN=TOP> <P>First Name<BR> <? echo "$_POST[f_name]"; ?></P> </TD> <TD VALIGN=TOP> <P>Last Name<BR> <? echo "$_POST[l_name]"; ?></P> </TD> <TD VALIGN=TOP> <P>Username<BR> <? echo "$_POST[username]"; ?></P> </TD> <TD VALIGN=TOP> <P>Password<BR> <? echo "$_POST[password]"; ?></P> </TD> <TD VALIGN=TOP> <P>Firm<BR> <? echo "$_POST[firm]"; ?></P> </TD> <TD VALIGN=TOP> <P>email<BR> <? echo "$_POST[email]"; ?></P> </TD> <P><a href="show_addrecord_auth_users.html">Add Another</a></P> </TD> </TR> </TABLE> </BODY> </HTML> Here there,
This is a post for all of you ajax/php pimps who don't mind helping a beginner. I have been working on a little ajax script to filter a table called "recipes2" in a database called "recipes" using a dropdown menu. Once I get this script to work, I can run queries without having to refresh the page.
The problem, something is not right, the php script does not seem to be getting the posted data. even though the ajax script returns success when i check for it. What am I doing wrong?
Also please note that, if possible, I'd like all the code to stay on one page.
This is the SQL script if you want to create the database and table:
[attachment=3349:recipesDB.txt]
Here is the code, the file should be named 'filterRecipes.php'.
[attachment=3350:filterRecipes.txt]
Hi, I have a feeling that this is going to be an easy miss but I've been over is a good few times and cannot see what it could be. Essentially, when I post the form in the first set of code (I've got rid of a load of unecessary html) the SESSION variables are not loaded into the second piece of code. Any ideas? Code: [Select] <?php session_start(); require ("connect.php"); $timeout = 1800; $logout_redirect_url = "login.php"; if (isset($_SESSION['start_time'])) { $elapsed_time = time() - $_SESSION['start_time']; if ($elapsed_time >= $timeout) { session_destroy(); header("Location: $logout_redirect_url"); } } $_SESSION['start_time'] = time(); ?> <div id="leftlink"> <a href="logout.php">Logout</a> </div> <table align="center" width="600px" cellpadding="0" cellspacing="0"> <tr> <td> <h2 align="center"><a href="index.php"><img src="images/logo.jpg" border="0" /></a> <br /> Emergency</h2> <hr /> </td> </tr> </table> </head> <body> <form action="post_reset.php" method="post" name"passre"> <table width="500" border="0" align="center" cellpadding="2" cellspacing="1"> <tr> <td width="230"><div align="right">E-mail Address : </div></td> <td width="40"> </td> <td width="230"><div align="left"> <input name="email" readonly="readonly" value="<?php echo ($_SESSION['email']) ; ?>" /></div></td> </tr> <tr> <td colspan="3"><div align="center"><input type="submit" name="submit" value="Submit" /></div></td> </tr> </table> </form> </body> </html> Code: [Select] <?php session_start; echo ($_SESSION['start_time']); require ("connect.php"); if (isset($_POST['submit'])){ $email = $_POST['email']; $getde = "SELECT * FROM users WHERE email = '$email'"; $getder = mysql_query($getde, $conn); $getdere = mysql_fetch_array($getder) Thanks Gareth hi there, basically in my database i have DATETIME for when an article was posted, using this, I'd like to make it possible for viewers of the article to see when it was posted in a "Posted Today", "Posted Yesterday" or Posted 23 hours ago" format. If it was posted longer ago than yesterday, then I would like it to just be the normal date. I know how to work out if it was posted today, just through: $date = date('j F', strtotime($rows['date'])); $today=date('j F'); but not sure how to do the hours. thanks! I have a basicc form that takes users info and when they submit it it show on a new web page in the proper order. I need to know how do I skip a blank field. This form takes a users First, Middle, and Last name but many times they leave the middle name blank so I would want to skip this field from showing up when the form is posted. Here is what I have for the basic php script ..Would like to know how to skip a field if it is left blank on the form. echo $_POST["First"], print "."; ?><?php echo $_POST["Middle"], print "."; ?><?php echo $_POST["Last"], print "."; ?><?php echo $_POST["Persona1"], print ".";<br /> this code is to create a new user based on posted variables. it works fine for creating the necessary tables and granting privileges. but then when i log out as root and try to log back in as the user it's failing. why is it failing? Thanks, G. Code: [Select] <?php extract( $_POST ); include( '../php/userobject.php' ); @ $db = mysqli_connect('localhost', 'root', 'xxxxxx', 'mealchamp'); if ( mysqli_connect_errno() ) { echo 'Error: could not connect to database as root.<br />'; exit; } else echo 'Connected to database as root.<br />'; $query = "SELECT username FROM userlist WHERE username = $uname"; $result = mysqli_query($db, $query); if ($result) echo 'That user name is already taken.<br/>Please hit back on your browser, and try a different user name.<br/>'; else { echo 'That username is available.<br/>'; echo "$fname<br/>"; echo "$lname<br/>"; echo "$add1<br/>"; echo "$add2<br/>"; echo "$city<br/>"; echo "$prov<br/>"; echo "$ctry<br/>"; echo "$post<br/>"; echo "$email<br/>"; echo "$uname<br/>"; echo "$pass1<br/>"; echo "$sex<br/>"; echo "$birth<br/>"; echo "$height<br/>"; echo "$weight<br/>"; $query = "INSERT INTO userlist (firstname, lastname, address1, address2, city, province, country, postal, email, username, password, sex, birth, height, weight) VALUES ('$fname', '$lname', '$add1', '$add2', '$city','$prov', '$ctry', '$post','$email','$uname', '$pass1', '$sex', '$birth', '$height', '$weight')"; $result = mysqli_query($db, $query); if ($result) echo "Userlist was updated successfully<br />"; else echo "Could not update userlist"; $query = "CREATE TABLE ".$uname."_persfood (persfoodid INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY, food VARCHAR(30) NOT NULL)"; $result = mysqli_query($db, $query); if ($result) echo "Personal food table was created successfully<br />"; else echo "Could not create personal food table<br />"; $query = "GRANT select ON mealchamp.userlist TO ".$uname." IDENTIFIED BY '$pass1'"; $result = mysqli_query($db, $query); if ($result) echo "User priveleges were granted successfully<br />"; else echo "Could not grant user priveleges<br />"; $query = "GRANT select, insert, update, delete, index ON mealchamp.".$uname."_persfood TO ".$uname." IDENTIFIED BY '$pass1'"; $result = mysqli_query($db, $query); if ($result) echo "User priveleges were granted successfully<br />"; else echo "Could not grant user priveleges<br />"; mysqli_close($db); @ $db = mysqli_connect('localhost', $uname, $pass1, 'mealchamp'); if ( mysqli_connect_errno() ) { echo 'Error: could not connect to database as user.<br />'; exit; } else { echo 'Connected to database as user.<br />'; ?> This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=320461.0 My issue in my php page where i need to show all the news that have the status only active. in my mysql table i have 3 fiels news,posted date and expiry date. on page load i check the status of the news and displays only the active news.For that i am doing the following code,which gives me error. that is on page load, first i select all the news ,update the status inactive for the news which has crossed expiry time i check the status of the news and displays only the active news.For that i am doing the following code,which gives me error. $query1="select time(date_posted),time(expiry_time) from muh_title where status='Active'"; $res=mysql_query($query1); $row=mysql_fetch_row($res); $expiry_time=$row[0];$dateposted=$row[1]; $query11="select addtime(time(date_posted),time(expiry_time)) from muh_title"; $res11=mysql_query($query11); $row11=mysql_fetch_row($res11); $newtime=$row11[0]; $query2="SELECT TIMEDIFF(time(date_posted) ,time(NOW())) FROM muh_title"; $res2=mysql_query($query2); $row2=mysql_fetch_row($res2); $diff=$row2[0]; //while($newtime){ $query5="select *from muh_main_title where date_posted> date_sub(now(),interval (expiry_time) hour-minute )"; //$query3="update muh_title set status='Inactive' "; $res3=mysql_query($query5); } $q="select * from muh_title where status='Active' "; $r= mysql_query($q); $row= mysql_fetch_row($r); echo "<font size='6' color='white'><b><marquee direction='left' loop='-1' align='absmiddle' behaviour='scroll' scrolldelay='90'>$row[1]</marquee></b></font>"; ?> i am sure some where i have lost the flow of code. I am pretty new to PHP and am trying to create a simple (so I assumed) page to takes data from one html page(works fine) and updates a MYSQL Database. I am getting no error message, but the connect string down to the end of the body section is showing up as plain text in my browser window. I do not know how to correct this. I have tried using two different types of connect strings and have verified my names from the HTML page are the same as listed within the php page. Suggestions on what I need to look for to correct would be great. I have looked online, but so far all I am getting is how to connect, or how to create a comment, so I thought I would try here. Thank you for any assistance I may get!! - Amy - Code: [Select] <body><font color="006600"> <div style="background-color:#f9f9dd;"> <fieldset> <h1>Asset Entry Results</h1> <?php // create short variable names $tag=$_POST['tag']; $serial=$_POST['serial']; $category=$_POST['category']; $status=$_POST['status']; $branch=$_POST['branch']; $comments=$_POST['comments']; if (!$tag || !$serial || !$category || !$status || !$branch) { echo "You have not entered all the required details.<br />" ."Please go back and try again."; exit; } if (!get_magic_quotes_gpc()) { $tag = addslashes($tag); $serial = addslashes($serial); $category = addslashes($category); $status = addslashes($status); $branch = addslashes($branch); $comments = addslashes($comments); } //@ $db = new mysqli('localhost', 'id', 'pw', 'inventory'); $db = DBI->connect("dbi:mysql:inventory:localhost","id","pw") or die("couldnt connect to database"); $query = "insert into assets values ('".$serial."', '".$tag."', '".$branch."', '".$status."', '".$category."', '".$comments."')"; $result = $db->query($query); if ($result) { echo $db->affected_rows." asset inserted into Inventory."; } else { echo "An error has occurred. The item was not added."; } $db->close(); ?> </fieldset> </div> </body> |
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