I have a php page that connects to a mysql database. I know that the connection to the database is good because I have a php code that displays info from the database onto the webpage. When I try to insert new data into the databse, the page refreshes and the data is not inserted. I have checked to insure that the insert into command has the correct values.
<?php
if (isset($_POST['User_Name']))
{
include "connect_to_mysql.php";
$name = mysql_real_escape_string($_POST["Name"]);
$sql = mysql_query("SELECT TestID FROM test WHERE Name='$name' LIMIT 1")or die (mysql_error());
$productMatch = mysql_num_rows($sql);
if ($productMatch > 0)
{
echo 'Sorry you tried to place a duplicate "User Account" into the system, click here';
exit();
}
else
{
$sql = mysql_query("INSERT INTO test (TestID,Name)
VALUES('', '$name')") or die (mysql_error());
$uid = mysql_insert_id();
header("location: index.php");
exit();
}
}
?>
<?php
include "connect_to_mysql.php";
$User_list = "";
$sql = mysql_query("SELECT * FROM test");
$UserCount = mysql_num_rows($sql);
if ($UserCount > 0)
{
while($row = mysql_fetch_array($sql))
{
$id = $row["TestID"];
$name = $row["Name"];
$User_list .= "Users ID: $id - <strong>$name</strong> <br />";
}
}
else
{
$User_list = "You have no users listed in the database.";
}
?>
<!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>Untitled Document</title>
</head>
<body>
<div align="center" id="mainWrapper">
<div id="pageContent"><br />
<div align="right" style="margin-right:32px;">+ Add New User</div>
<div align="left" style="margin-left:24px;">
<h2>User list</h2>
<?php echo $User_list; ?>
</div>
<hr />
<a name="UserForm" id="UserForm"></a>
<h3>
↓ Add New User Form ↓
</h3>
<form action="index.php" enctype="multipart/form-data" name="myForm" id="myform" method="post">
<table width="90%" border="0" cellspacing="0" cellpadding="6">
<tr>
<td width="20%" align="right">Name</td>
<td width="80%"><label>
<input name="name" type="text" id="name" size="50" />
</label></td>
</tr>
<tr>
<td> </td>
<td><label>
<input type="submit" name="button" id="button" value="Add This Name Now" />
</label></td>
</tr>
</table>
</form>
<br />
<br />
</div>
</div>
</body>
</html>
I see two problems right away (there may be more). First, PHP array keys are case sensitive. You are accessing $_POST['Name'] but your form input is name. Second, you are testing for $_POST['User_Name'] which doesn't appear to exist anywhere:
// Look for name in the $_POST
if (isset($_POST['name']))
{
include "connect_to_mysql.php";
// name is case-sensitive
$name = mysql_real_escape_string($_POST["name"]);
Later, if your table has an AUTO_INCREMENT id on TestID, you should either omit it or insert NULL in the insert statment:
// Don't include TestID if it is AUTO_INCREMENT. That will happen automatically
$sql = mysql_query("INSERT INTO test (Name)
VALUES('$name')") or die (mysql_error());
I think it will work if you change
if (isset($_POST['User_Name']))
to
if (isset($_POST['Name']))
You check the existence of something which doesn't exists in your form.
Addition:
If TestID is autoincrement, change the below
"INSERT INTO test (TestID,Name) VALUES('', '$name')"
to
"INSERT INTO test (Name) VALUES('$name')"
If you do not get any errors that means you have an error in your MySQL syntax two ways to test it would be to copy the syntax into PHPMyAdmin or whatever your native MySQL command line is and see if you get an output error. Or another thing you can do is to modify all your mysql_query(); functions by adding mysql_query()or die(mysql_error());
Related
Hey and thanks in advance, I'm trying to get data from a form and enter it into a database. I'm following a guide on how to do this, but the guide doesnt account for anything not working........
I've only just started out with PHP so my skills are quite limited, however I do understand all code I have written, it just doesn't work! The php script will not move past the if statement below and I can't figure out why:
// Check for existing user with the new id
$sql = "SELECT COUNT(*) FROM sessions.users WHERE userid = '$_POST[newid]'";
$result = mysqli_query($cxn, $sql);
if (!$result) {
error('A database error occurred in processing your '.
'submission.\nIf this error persists, please '.
'contact you#example.com. This is from error 1');
Entire code:
signup.php
<?php //signup.php
include 'common.php';
include 'db.php';
if(!isset($_POST['submitok'])):
// display user signup form
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>New User Registration</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
</head>
<body>
<h3>New User Registration Form</h3>
<p><font color="orangered" size="+1"><tt><b>*</b></tt></font> indicates a required field</p>
<form method="post" action="<?=$_SERVER['PHP_SELF']?>">
<table border="0" cellpadding="0" cellspacing="5">
<tr>
<td align="right">
<p>User ID</p>
</td>
<td>
<input name="newid" type="text" maxlength="100" size="25" />
<font color="orangered" size="+1"><tt><b>*</b></tt></font>
</td>
</tr>
<tr>
<td align="right">
<p>Full Name</p>
</td>
<td>
<input name="newname" type="text" maxlength="100" size="25" />
<font color="orangered" size="+1"><tt><b>*</b></tt></font>
</td>
</tr>
<tr>
<td align="right">
<p>E-Mail Address</p>
</td>
<td>
<input name="newemail" type="text" maxlength="100" size="25" />
<font color="orangered" size="+1"><tt><b>*</b></tt></font>
</td>
</tr>
<tr valign="top">
<td align="right">
<p>Other Notes</p>
</td>
<td>
<textarea wrap="soft" name="newnotes" rows="5" cols="30"></textarea>
</td>
</tr>
<tr>
<td align="right" colspan="2">
<hr noshade="noshade" />
<input type="reset" value="Reset Form" />
<input type="submit" name="submitok" value=" OK " />
</td>
</tr>
</table>
</form>
</body>
</html>
<?php
else:
//process sign up submission
dbConnect('sessions');
if ($_POST['newid']=='' or $_POST['newname']==''
or $_POST['newemail']=='') {
error('One or more required fields were left blank.\\n'.
'Please fill them in and try again.');
}
// Check for existing user with the new id
$sql = "SELECT COUNT(*) FROM sessions.users WHERE userid = '$_POST[newid]'";
$result = mysqli_query($cxn, $sql);
if (!$result) {
error('A database error occurred in processing your '.
'submission.\nIf this error persists, please '.
'contact you#example.com. This is from error 1');
}
if (#mysqli_result($result,0,0)>0) {
error('A user already exists with your chosen userid.\n'.
'Please try another.');
}
$newpass = substr(md5(time()),0,6);
$sql = "INSERT INTO sessions.users SET
userid = '$_POST[newid]',
password = PASSWORD('$newpass'),
fullname = '$_POST[newname]',
email = '$_POST[newemail]',
notes = '$_POST[newnotes]'";
if (!mysqli_query($sql))
error('A database error occurred in processing your '.
'submission.\nIf this error persists, please '.
'contact you#example.com. This is from error 2');
// Email the new password to the person.
$message = "G'Day!
Your personal account for the Project Web Site
has been created! To log in, proceed to the
following address:
http://www.example.com/
Your personal login ID and password are as
follows:
userid: $_POST[newid]
password: $newpass
You aren't stuck with this password! Your can
change it at any time after you have logged in.
If you have any problems, feel free to contact me at
<you#example.com>.
-Your Name
Your Site Webmaster
";
mail($_POST['newemail'],"Your Password for the Project Website",
$message, "From:Your Name <you#example.com>");
?>
<!DOCTYPE html PUBLIC "-//W3C/DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title> Registration Complete </title>
<meta http-equiv="Content-Type"
content="text/html; charset=iso-8859-1" />
</head>
<body>
<p><strong>User registration successful!</strong></p>
<p>Your userid and password have been emailed to
<strong><?=$_POST['newemail']?></strong>, the email address
you just provided in your registration form. To log in,
click here to return to the login
page, and enter your new personal userid and password.</p>
</body>
</html>
<?php
endif;
?>
db.php
<?php // db.php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
function dbConnect($db='') {
global $dbhost, $dbuser, $dbpass;
$cxn = mysqli_connect($dbhost,$dbuser,$dbpass);
mysqli_select_db($cxn,$db)
or die ("Couldn’t select database.");
return $cxn;
}
?>
common.php
<?php // common.php
function error($msg) {
?>
<html>
<head>
<script language="JavaScript">
<!--
alert("<?=$msg?>");
history.back();
//-->
</script>
</head>
<body>
</body>
</html>
<?php
exit;
}
?>
Any help as to why this if statement ALWAYS produces the error would be grand - thank you.
You're not assigning your db connection within a usable scope. You are returning it from your dbConnect() function, you're just not doing anything with the returned value.
dbConnect('sessions'); should be $cxn = dbConnect('sessions');
You would be alerted to the cause of the problem if you used mysqli_error() to let MySQL tell you what it didn't like.
Finally, you should be using bound parameters in your query instead of injecting user-provided data directly. Search something like "mysqli bound parameters" to learn how to do this. What you have now is open to attack.
For your INSERT statement, use bound parameters:
$sql = "INSERT INTO sessions.users (userid, password, fullname, email, notes) VALUES (?, ?, ?, ?, ?)";
$stmt = mysqli_prepare($sql);
mysqli_stmt_bind_param($stmt, 'issss', $newID, PASSWORD($newpass), $fullName, $email, $notes);
if(!mysqli_stmt_execute($stmt))
{
// use this for debugging, but do NOT leave it in production code
die(mysqli_error($cxn));
}
mysqli_stmt_close($stmt);
Above is untested so it's possible you'll have to make small adjustments. It should give you a pretty good idea of what to do at least.
Also, make sure that PASSWORD() is really what you want to be using. I have a feeling it's not, but I don't want to assume.
I am trying to input and display the data on same page through mysqli but it shows an error "Couldn't fetch mysqli on line" and not unable to display the records.
<?php
$db=mysqli_connect("localhost","root","","abc") or die("Not connected".mysqli_error());
$database=mysqli_select_db($db,'abc') or die("Database not found".mysqli_error());
if(isset($_POST['submit'])){
$roll=$_POST['roll'];
$name=$_POST['name'];
$ins=mysqli_query($db,"insert into abc1 (roll,name)values('$roll','$name')");
}
mysqli_close($db);
?>
<!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>Untitled Document</title>
</head>
<body>
<form method="post">
Roll No <input type="text" name="roll" />
Name <input type="text" name="name" />
<input type="submit" name="submit" value="Submit" />
</form>
<table><tr>
<td>Roll No.</td>
<td>Name</td>
</tr>
<?php $query=mysqli_query($db,"select * from abc1");
$result=mysqli_query($db,$query);
$id=0;
while($row=mysqli_fetch_array($result,MYSQLI_ASSOC))
{
?>
<tr><td><?php echo $row['roll']; ?></td>
<td><?php echo $row['name']; ?></td>
<?php } ?>
</tr></table>
</body>
</html>
Any help would be appreciated.
Thanks in Advance
You are doing mysqli_query twice.
Try this.
$query = "select * from abc1";
$result = mysqli_query($db, $query);
You are calling mysqli_query function twice:
$query = mysqli_query($db,"select * from abc1");
$result = mysqli_query($db,$query);
Error: Couldn't fetch mysqli on line
Means, you are passing a a resource id in mysqli_fetch_array function instead of Query Statement. so you just need to use this as:
$query = "select * from abc1";
$result = mysqli_query($db,$query);
Side Note:
There is no need to connect db again:
$database=mysqli_select_db($db,'abc')
mysqli_connect() already connected your db.
Going to try to keep it short. I have a while loop in grid.php file to fill up a table as such...
<?php while($product = $products->fetch_assoc()) { ?>
<tr>
<td><?php echo $product['cd_id']?></td>
<td><?php echo $product['cd_title']?></td>
<td><?php echo $product['cd_musician_fname']?></td>
<td><?php echo $product['cd_musician_lname']?></td>
<td><?php echo $product['cd_price']?></td>
<td>Edit</td>
<td>Delete</td>
</tr>
<?php } ?>
If I click the first anchor tag takes me to a edit.php file and here is the head code for that file.
<?php include '_includes/db.php';
$cd_id = trim($_GET['id']);
$message = '';
include '_includes/connection.php';
if($db->connect_error){
$message = $db->connect_error;
}else{
$sql = "SELECT * FROM CD WHERE cd_id = $cd_id";
$result = $db->query($sql);
$row = $result->fetch_assoc();
if($db->error){
$message = $db->error;
}
}
?>
Now I will show the html of edit.php
<!-- Product Musician last name-->
<fieldset class="form-group">
<label for="cd_musician_lname">Musician's lirst name</label>
<input type="text" class="form-control" id="cd_musician_lname" name="cd_musician_lname" value="<?php echo $row['cd_musician_lname'];?>">
</fieldset>
<!-- End of Musician last name-->
<!-- Product price-->
<fieldset class="form-group">
<label for="cd_price">Product price</label>
<input type="text" class="form-control" id="cd_price" name="cd_price" value="<?php echo $row['cd_price'];?>">
</fieldset>
<!-- End of Product price-->
<!-- Form submit button-->
Update Record
<a class="btn btn-primary" href="index.php" role="button">Go Back Home</a>
I have the edit.php page working just fine but if I make changes in the fields and click the submit anchor tag I get all the fields of the row empty but the PK. Here is the code for the final edit_confirm.php file
<?php
include '_includes/db.php';
$cd_id = trim($_GET['id']);
$cd_title = $_POST['cd_title'];
$cd_musician_fname = $_POST['cd_musician_fname'];
$cd_musician_lname = $_POST['cd_musician_lname'];
$cd_price = $_POST['cd_price'];
$message = '';
include '_includes/connection.php';
if($db->connect_error){
die("Connection failed: ".$db->connect_error);
} else {
$sql = "UPDATE CD SET cd_title='".$cd_title."', cd_musician_fname='".
$cd_musician_fname."', cd_musician_lname='".
$cd_musician_lname."', cd_price='".$cd_price."' WHERE cd_id = $cd_id ";
$db->query($sql);
var_dump($sql);
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<?php include '_includes/main-head.php';?>
</head>
<body>
<?php include '_includes/main-navbar.php';?>
<div class="container">
<hr>
<?php
if($db->query($sql) === TRUE){ ?>
<h1>Record updated successfully.</h1>
<?php echo $cd_title; ?>
<?php echo $record->affected_rows ?>
<p> record was updated in the database.</p></br>
<?php } else { ?>
<p>Error updating the record: </p> <?php $db->error; ?>
<?php }; ?>
<hr>
<a class="btn btn-primary" href="index.php" role="button">Go Back Home</a>
</div>
<?php include '_includes/main-script.php';?>
</body>
</html>
If you notice in the edit_confirm.php I did a var_dump to see what are the values in the variables and it shows empty.
I need help with this.
Thank you in advance.
Man the better way to do this is make it simple to test if the record is updating or not
formsample.php
<!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>Untitled Document</title>
</head>
<body>
<?php
include("connection.php");
$id = $_GET['id'];
$query= "select * from clients where id = '$id'";
$sql = mysqli_query($connect, $query);
$result = mysqli_fetch_assoc($sql);
?>
<form action="process.php" method="post">
<input type="text" name="name" id="name" value="<?php echo $result['name'] ?>" />
<input type="text" name="email" id="email" value="<?php echo $result['email'] ?>" />
<input type="hidden" name="id" id="id" value="<?php echo $id?>" />
<input type="submit" />
</form>
</body>
</html>
process.php
<?php
include("connection.php");
$id = $_POST['id'];
$name = $_POST['name'];
$email= $_POST['email'];
$query = "UPDATE clients set name= '$name', email ='$email' where id = '$id'";
$sql = mysqli_query($connect, $query);
?>
Update Record
This is not the proper way to submit a form - it won't work at all.
You need to have a form opening and closing tag, the target address is in the action attribute of the form element, and the method is on there too and should be post for this form (method="POST"). In your code you have a link where you should have a submit input so it won't submit the data, it will just redirect you to that URL. You should have something like this:
<input type="submit" value="Update Record" />
http://www.w3schools.com/html/html_forms.asp
I am working on PHP CRUD operations and I have created a basic edit form in PHP. I have not used any field validations and all I want is simply editing information.
I am following this tutorial
Once a user is clicked on Edit link he is directed to the following form on which the user is supposed to edit his data.
Here is the code
<?php
include_once './functions.php';
include_once './database.php';
function renderForm($firstName,$lastName,$age){
?>
<!DOCTYPE html>
<html>
<head>
<title>Edit</title>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width">
</head>
<body>
<form action="edit.php" method="post">
First Name<input type="text" name="firstname" value="<?php $firstName ;?>"><br/>
Last Name<input type="text" name="lastname" value="<?php $lastName ;?>"><br/>
Age<input type="text" name="age" value="<?php $age ;?>"><br/>
<input type="submit" name="submit" value="Edit">
Cancel
</form>
<?php
}
?>
<?php
if (isset($_POST['submit'])) {
$firstName = cleanData($_POST['firstname']);
$lastName = cleanData($_POST['lastname']);
$age = (int) $_POST['age'];
$id = $_GET['id'];
$query = "UPDATE basic ";
$query.="SET first_name='$firstName',last_name='$lastName',age=$age ";
$query.="WHERE id=$id";
confirmQuery($query);
closeDatabase();
}else{
$id=cleanData($_GET['id']);
$query="SELECT * FROM basic WHERE id= {$id} ";
$result=confirmQuery($query);
$rows= mysqli_fetch_assoc($result);
$firstName=$rows['first_name'];
$lastName=$rows['last_name'];
$age=$rows['age'];
renderForm($firstName, $lastName, $age);
}
?>
</body>
</html>
//Additional information
//functions included in other files
function cleanData($input){
global $connection;
return mysqli_real_escape_string($connection,$input);
}
function confirmQuery($query){
global $connection;
$result=mysqli_query($connection, $query);
if(!$result){
return "Query failed : ".mysqli_error($connection);
}
else{
return $result;
}
}
function closeDatabase(){
global $connection;
mysqli_close($connection);
}
//I have not included the file which I am using to
//connect to the DB. I am sure there is no error with that file since it works
//properly with other php files
The problem that I have with my edit form is it does not show previously entered data and just shows only a blank form (similar to create form). (It does not happen when I run the demo in the above mentioned tutorial)
Netbenas IDE says variables which are inside HTML input tags seems to be unused in its scope. I have googled this question and found that warning can be simply ignored.
But Where have I gone wrong?
I am grateful to anyone who can kindly go through my code and show me the error.
Thank You :)
I have change your PHP code to below code use in your edit.php.if u get any issue put comment.
<?php
include_once './functions.php';
include_once './database.php';
if (isset($_POST['submit'])) {
$firstName = cleanData($_POST['firstname']);
$lastName = cleanData($_POST['lastname']);
$age = (int) $_POST['age'];
$id = $_GET['id'];
$query = "UPDATE basic ";
$query.="SET first_name='$firstName',last_name='$lastName',age=$age ";
$query.="WHERE id=$id";
$r=mysql_query($query);
if($r)
{
echo "Record updated";
}
}
$id=$_GET['id'];
$query="SELECT * FROM basic WHERE id='$id' ";
$result=confirmQuery($query);
$rows= mysqli_fetch_assoc($result);
$firstName=$rows['first_name'];
$lastName=$rows['last_name'];
$age=$rows['age'];
?>
<!DOCTYPE html>
<html>
<head>
<title>Edit</title>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width">
</head>
<body>
<form action="edit.php" method="post">
First Name<input type="text" name="firstname" value="<?php echo $firstName ;?>"><br/>
Last Name<input type="text" name="lastname" value="<?php echo $lastName ;?>"><br/>
Age<input type="text" name="age" value="<?php echo $age ;?>"><br/>
<input type="submit" name="submit" value="Edit">
Cancel
</form>
</body>
</html>
I'm trying to create a set of webpages that work together to allow users to view, delete, and edit rows of a MS Access database using PHP.
Membership.php shows a list of the names of members in the Access database. Their names are also hyperlinks that, when clicked, take users to another page EditRecord.php where all of information on the member whose name was clicked on Membership.php is displayed in text boxes with the option to completely delete the record, or just update certain fields.
Membership.php and EditRecord.php are displayed below. The error code is for line 91 of my source for EditRecord.php, but I cut some things out of this post for privacy. Instead, the line has been marked like so:
//--------This is the error line----------
code
[Membership.php]
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<link rel="stylesheet" href="Accounts.css">
<style type="text/javascript" src="Validate.js"></style>
<style type="text/javascript" scr="Redirect.js"></style>
<style type="text/javascript" src="Utilities.js"></style>
<title>Member Information Input</title>
</head>
<body>
<div id="content">
<?php
//Establish data connection using external file
require("connection.php");
//Issue SQL SELECT Statement
$sql = "SELECT * FROM Membership";
//Stores any results that match the search term.
$rs = odbc_exec($conn, $sql);
//Set counter for search results to zero
$results = 0;
//Iterates through search results and prints information on records that match
while($row = odbc_fetch_array($rs))
{
$results += 1;
echo '<p>' . $row['FirstName'] . " " . $row['LastName'] . "</p>";
}
?>
</div>
</body>
</html>
[EditRecord.php]
<?php
//Retrieve ID value - if the page is loading for the first time, use $_GET[]. If the
//delete or edit button has been clicked, use $_POST[]
if (isset($_GET['ID'])) {
$userID = $_GET['ID'];
}
else {
$userID=$_POST['ID'];
}
//Establish data connection
require("connection.php");
//If the Delete Button is clicked
if (isset($_POST['DelBtn'])) {
//Issue SQL Statement to Delete Selected Record
$sqlDelete = "DELETE FROM Membership WHERE ID = $userID";
//Execute the SQL Delete Query
$rsDelete = odbc_exec($conn,$sqlDelete);
if(odbc_num_rows($rsDelete) == 1) {
echo "Record successfully deleted!";
}
}
//If the Edit Button is clicked
else if (isset($_POST['EditBtn'])) {
//Collect form field values in scalar variables
$FirstName = $_POST['FirstName'];
$LastName = $_POST['LastName'];
$Address = $_POST['Address'];
$City = $_POST['City'];
$State = $_POST['State'];
$Email = $_POST['Email'];
$Gender = $_POST['Gender'];
$Comments = $_POST['Comments'];
//Issue SQL Statement to Update Selected Record
$sqlUpdate = "UPDATE Membership SET FirstName = '$FirstName', LastName = '$LastName', Address = '$Address', City = '$City', State = '$State'" .
"Email='$Email', Gender = '$Gender', Comments = '$Comments' WHERE ID = $userID";
//Execute the SQL UPDATE Query
$rsEdit = odbc_exec($conn,$sqlUpdate);
if(odbc_num_rows($rsEdit) == 1) {
echo "Record successfully updated!";
}
}
//Issue SQL SELECT Statement to Select Record to Edit or Delete
$sql = "SELECT * FROM Membership WHERE ID = $userID";
//Execute the SQL Query
$rs = odbc_exec($conn, $sql);
odbc_close($conn);
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<link rel="stylesheet" href="Accounts.css">
<style type="text/javascript" src="Validate.js"></style>
<style type="text/javascript" src="Utilities.js"></style>
<title>Member Information Input</title>
</head>
<body>
<div id="content">
<form method="post" action="EditMember.php" name="EditForm">
<?php
// Loop through and display the recordset returned by SELECT statement. Display the record values in HTML Text Boxes
**//--------This is the error line----------
while ($row = odbc_fetch_array($rs)) {
?>**
First Name: <input type="text" name="FirstName" value="<?php echo $row['FirstName']?>"><br>
Last Name: <input type="text" name="LastName" value="<?php echo $row['LastName']?>"><br>
Address: <input type="text" name="Address" value="<?php echo $row['Address']?>"><br>
City: <input type="text" name="Telephone" value="<?php echo $row['City']?>"><br>
State: <input type="text" name="Telephone" value="<?php echo $row['State']?>"><br>
Email: <input type="text" name="Email" value="<?php echo $row['Email']?>"><br>
Gender: <input type="text" name="Telephone" value="<?php echo $row['Gender']?>"><br>
Comments: <input type="text" name="Comments" value="<?php echo $row['Comments']?>"><br><br>
<input type="hidden" name="ID" value="<?php echo $row['ID']?>" >
<?php
}
?>
<input type="submit" name="EditBtn" value="Edit Record"> <input type="submit" name="DelBtn" value="Delete Record">
</form>
</div>
<div id="footer">
<?php require("Footer.php"); ?>
</div>
</body>
</html>
I also find this strange, because there are five records in my database, not four. Is that because it starts counting at zero?
Any insight or advice would be greatly appreciated.
Your problem is that you are calling odbc_close() and closing the connection before your loop calls odbc_fetch_array(). You need to leave the connection open until after you've fetched all of the rows.
Also, the "4" in the error message does not refer to a number of rows or anything like that; it's just the numeric representation of result identifier for the resource created by the odbc_exec() call.