I'm trying to retrieve all data with the LIKE query from the users input and match it to the database, it works but only returns one record but I have many records in the table.
It returns the closest record it can find,
so say for example I have 2 records who's ItemDesc field contains the characters 'The', when I search for 'The' in my input box and click submit it returns the closest (earliest created) record when it is supposed to return both.
<?php
$username = "a3355896_guy";
$password = "++++++";
$hostname = "mysql5.000webhost.com";
$dbh = mysql_connect($hostname, $username, $password) or die("Unable to connect to MySQL");
mysql_select_db("a3355896_book") or die("Unable to connect to database");
$ItemDesc = $_POST['ItemDesc'];
$query = "select * from StockItems where ItemDesc LIKE '%$ItemDesc%'";
$result=mysql_query($query);
$num=mysql_num_rows($result);
mysql_close();
?>
Sorry was supposed to included the retrieval:
<?php
if ($num>0)
{
echo "<center><table border=1><tr><th>Item Code</th><th>Item Desc</th>";
echo "<th>Item Stock Qty</th>";
echo "<th>Item Unit Price</th><th>Item Category</th></tr>";
$ItemCode = mysql_result($result,$i,"ItemCode");
$ItemDesc = mysql_result($result,$i,"ItemDesc");
$ItemStockQty = mysql_result($result,$i,"ItemStockQty");
$ItemUnitPrice = mysql_result($result,$i,"ItemUnitPrice");
$ItemCategory = mysql_result($result,$i,"ItemCategory");
echo "<tr><td>$ItemCode</td><td>$ItemDesc</td><td align=right>";
echo "$ItemStockQty</td>";
echo "<td align=right>$ItemUnitPrice</td>";
echo "<td>$ItemCategory</td></tr>";
echo "</table></center>";
}
else
{
echo "<form name='DeleteStock2'>";
echo "<p> Sorry, $ItemDesc does not exist!<p>";
echo "<input type='button' value='Leave' onclick='history.go(-1)'>";
}
?>
You aren't actually accessing your data here- you need to iterate over the result set.
$setLength = mysql_num_rows($result);
for($i = 0; $i < $setLength; $i++){
//Here, mysql_fetch_assoc automatically grabs the next result row on each iteration
$row = mysql_fetch_assoc($result);
//do stuff with "row"
}
Unless you ARE doing that and you just chose to not include it in your snippit. Let us know :)
--Edit--
First off, I apologize- out of old habit I suggested that you use mysql_fetch_assoc instead of the mysqli set of functions.
Try using the fetch_assoc or fetch_array functions, it could solve your issue. I've never used the method you used, I think it has been deprecated for a while.
Check it out here:
http://php.net/manual/en/mysqli-result.fetch-assoc.php
Related
First I'm hitting on a wall here and I really could use your help. I coded the database so I have it all up and working plus all the data inside. I worked the HTML and the CSS media print query and I have it how I want it to look exactly. All I have to do now is:
for every row of the mysql select table I have to fill every specific input form
of the html page I made and print it
Can someone give me a hint of how I can do that?
Assuming you want to connect to your database and simply fetch the id you can do the following.
Ensure you change my_host, my_user, my-password, my_databse,my_tablewith your configuration settings. Then if you want to fetch anything else thanid` just change it to the column name you are looking for.
Be aware we are using PHP here.
// Open Connection
$con = #mysqli_connect('my_host', 'my_user', 'my-password', 'my_databse');
if (!$con) {
echo "Error: " . mysqli_connect_error();
exit();
}
// Some Query
$sql = 'SELECT * FROM my_table';
$query = mysqli_query($con, $sql);
while ($row = mysqli_fetch_array($query))
{
echo $row['id'];
}
// Close connection
mysqli_close ($con);
Check this link to get a in-depth explanation.
You can do this with two pages. One page gives you the overview and the other page gives you a print preview of your invoice.
The overview:
// DB select stuff here
while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
echo "<tr>\n";
echo " <td>".htmlspecialchars($row['what'])."</td>\n";
echo " <td>".htmlspecialchars($row['ever'])."</td>\n";
echo " <td>Detail</td>\n";
echo "</tr>\n";
}
The detail page:
$sql = 'SELECT your, columns FROM tab WHERE id = ?';
$stmt = $db->prepare($sql);
$stmt->execute(array($_GET['id']));
$row = $stmt->fetch(PDO::FETCH_ASSOC);
if (!$row) {
echo "There is no data for the given Id\n";
return;
}
echo "What: ".htmlspecialchars($row['what'])."<br />\n";
echo "Ever: ".htmlspecialchars($row['ever'])."<br />\n";
I had a very random problem in an IIS class, it has stumped my tutor so here I am and I'll try my best to explain it well!
I'm running Xampp with Apache and MySQL, I run the query I want and get the expected output to a table, but I have a problem with the outputting of pictures. I have the right file type, extention and path selected, because I can get pictures to show up, but as long as at least one picture in the query result has the extension removed, and this picture will not load.
Database
Website
If I have each query result with the correct name and extension, which is the same as when it shows up, none of them show up at all!
PHP:
<?php
// set server access variables
include 'db2.inc';
// open connection
$connection = mysql_connect($hostname, $username, $password) or die ("Unable to connect!");
// select database
mysql_select_db($databaseName) or die ("Unable to select database!");
// create query
//$query = "SELECT * FROM products";
$query = "SELECT * FROM products WHERE CategoryName = 'Surfboards'";
//Check initial letter
//if (!$initialLetter=="")
//{
// $query = $query." Where country like '$initialLetter%' ";
//}
// execute query
$result = mysql_query($query) or die ("Error in query: $query. ".mysql_error());
// see if any rows were returned
if (mysql_num_rows($result) > 0) {
// yes
// print them one after another
echo "<table cellpadding=20 border=1>";
while($row = mysql_fetch_assoc($result)) {
echo "<tr>";
echo "<td>".$row['ProductID']."</td>";
echo "<td>".$row['Name']."</td>";
echo "<td>".$row['Description']."</td>";
echo "<td>".$row['Brand']."</td>";
echo "<td>".$row['Model']."</td>";
echo "<td>".$row['BoardLength']."</td>";
echo "<td>".$row['BoardType']."</td>";
echo "<td>".$row['Colour']."</td>";
echo "<td>"."<img src=images/".$row['Image']."> </td>";
echo "<td>".$row['UnitPrice']."</td>";
echo "<td>".$row['CategoryName']."</td>";
echo "</tr>";
}
echo "</table>";
}
else {
// no
// print status message
echo "No rows found!";
}
// free result set memory
mysql_free_result($result);
// close connection
mysql_close($connection);
?>
Any help or direction would be greatly appreciated!
Thanks
Will
On this line:
echo "<td>"."<img src=images/".$row['Image']."> </td>";
Your image src is not enclosed in quotes. If this is a direct copy/paste of your code then that is definitely a problem, but may not be the only one.
As other people have said, look at pdo, or mysqli. The mysql_ functions you are using are deprecated for multiple reasons.
In this application I have a drop down so you can select what table you want to view. Using Ajax I push the variable $tbl as an integer and the code below prints out the table that corresponds with that integer. Something is not working though and I need an extra pair of eyes to help debug.
if($tbl == 9){$table = "person";}
$conn = new mysqli(DBHOST, DBUSER, DBPASS, DBNAME);
$query = mysqli_query($conn, "SELECT * FROM $table ") or die(mysqli_error($conn));
if($query){
echo '<table border="1"><thead><tr>';
$colnames = array();
while ($finfo = mysqli_fetch_field($query)) {
$name=$finfo->name;
array_push($colnames, $name);
echo "<th>".$name."</th>";
}
mysqli_free_result($query);
echo '</tr></thead><tbody>';
$count = 0;
echo $colnames[$count]; // this test prints the correct column name but it ends up being printed OUTSIDE the table for some reason
while ($row = mysqli_fetch_array($query, MYSQLI_BOTH)){
echo '<td>'.$row[$colnames[$count]].'</td>';
$count++;
}
echo "</tbody></table>";
}
The last while loop is suppose to echo out each field value but it isn't executing <tbody> is left blank.
You called mysqli_free_result($query) too early. So by the time you called mysqli_fetch_array, the sql result was already gone. Move mysqli_free_result after the last loop.
I'm not sure whether it's a typo or that the following is where the problem lies, but you're missing opening and closing <tr> tags around the column names after <tbody> and before </tbody>. This may help explain why stuff is being printed apparently "outside" the table.
I'm having a problem with some specific PHP code that I've been working on for a few days. It's meant to be a reporting code where I can input a day and a month and it will list the total sales of that particular day.
But, I can't seem to make the last statement whereby, if there are no values(there are no data) in the query, it will display 'No Sales on this particular day'. Here's the code I've been working on. But the last echo statement is not executing. Any ideas?
<?php
session_start();
if ((isset($_SESSION["admin"])) ){
$day=#$_POST['day'];
$month=#$_POST['month'];
echo "<center><h2>Sales report on " .$day. "." .$month. ".2013</h2></center>";
echo "<center><table style='border:2px solid black;' align=center width=600>";
echo "<tr><th colspan=12><center><h2>Sales Report</h2><hr size='2' color='black' /></center></th></tr>";
echo " <th width=400> Amount Collected</th>";
?>
<br>
<?php
$x = 1; //counter
//open a connection to a MySQL server using function mysql_connect
//returns a MySQL link identifier on success, or FALSE on failure.
$conn= mysql_connect("localhost","root","");
if (!$conn)
die ("Connection error: ".mysql_error());
else {
//select a MySQL database
//returns TRUE on success or FALSE on failurue.
$db=mysql_select_db("cqfos");
if(!$db)
die ("DB not found: ".mysql_error());
else {
//put query in a variable $query
$query= "select ROUND(sum(orderdetails.tprice),2)
from orders JOIN orderdetails ON orders.orderID = orderdetails.orderID WHERE DAY(orders.date) = '$day' AND MONTH(orders.date) = '$month'";
$result=mysql_query($query);
if(!$result)
die ("Invalid query: ".mysql_error());
//if record exists
else {
//fetch a result row as both associative array and numeric array
if(mysql_num_rows($result)== 1){
while ($row=mysql_fetch_array($result,MYSQL_BOTH)){
echo "<tr>";
echo "<td align='center'>RM ".$row[0]."</td></tr>";
$x++; //increase the counter
}
}
else {
echo "<tr><th colspan=12>No sales made.</td></tr>";}
}
}
}
echo"</table></center>";
?>
Several problems here
your HTML table syntax is incorrect, and your using an old sql library - and it dose not look like your SQL syntax is right... try this code (not tested as I don't have your data)
<?php
session_start();
if ((isset($_SESSION["admin"])) ){
echo '<div style="margin:auto; textalign:center;">';
echo "<h2>Sales report on " .$_POST['day']. "." .$_POST['month']. ".2013</h2>";
echo "<h2>Sales Report</h2>"
echo "<table style='border:2px solid black;' align=center width=600>";
echo "<tr><th width=400> Amount Collected</th></tr>";
?>
<br>
<?php
$conn = new mysqli("localhost","root","","cqfos");///use mysqli, not mysql : mysql is depricated
if ($conn->mysqli)
exit ("Connection error: ".$conn->errno ." : " $conn->error);
else {
//put query in a variable $query
$eDay = $conn->mysql_real_escape_string($_POST['day']);//escape these to protect the database
$eMonth = $conn->mysql_real_escape_string($_POST['month']);;//escape these to protect the database
//your column name is probably not a rounded value, replaced it with * (return all columns)
$query= "select * from orders JOIN orderdetails ON orders.orderID = orderdetails.orderID WHERE DAY(orders.date) = '"
.$eDay."' AND MONTH(orders.date) = '".$eMonth."'";
$result=$con->query($query);
if($conn->errno)
exit ("Invalid query: ".$conn->errno ." : " $conn->error);
//if record exists
else {
$numericArray = $result->fetch_array(MYSQLI_NUM); //fetch a result row as numeric array
$associativeArray = $result->fetch_array(MYSQLI_ASSOC); //fetch as an associtive array this is not used, just an example
$bothArray = $result->fetch_array(MYSQL_BOTH); //both associtive and numeric this is not used, just an example
}
if(!empty($numericArray))
{
foreach ($numericArray as $value) {
echo "<tr><td>RM ".$value[0]."</td><tr>";//is there more then 1 col? if not you should consider an html list
}
} else {
echo "<tr><td>No sales made</td><tr>";
}
echo"</table></center>";
}
?>
Your SQL (likely) returns more than only one row, so change the line I mentioned before to this:
if(mysql_num_rows($result)>0){
Just letting you know your code is vulnerable to SQLi because you have not sanitized $day and $month. Also please consider using PDO.
If you haven't already - Try running the SQL statement into PHPMyAdmin and see where it outputs the error (if there is one), else it will output the data.*
*Manually inputting the day/month substituting for the variables.
I have the following code which gets a users data from a table based on their log in details
//==========================================
// CONNECT TO THE LOCAL DATABASE
//==========================================
$user_name = "xxxx";
$pass_word = "xxxxx";
$database = "xxxx";
$server = "xxxxxx";
$db_handle = mysql_connect($server, $user_name, $pass_word);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
$SQL = "SELECT * FROM students WHERE L1 = '$uname' AND L2 = '" .md5 ($_POST['password'])."'";
$result = mysql_query($SQL);
$num_rows = mysql_num_rows($result);
//====================================================
// CHECK TO SEE IF THE $result VARIABLE IS TRUE
//====================================================
if ($result) {
if ($num_rows > 0) {
$color="1";
$result = mysql_query("SELECT * FROM entry, students WHERE entry.studentName = students.studentName AND students.L1='$uname' ") or die(mysql_error());
echo "<p>Welcome "; echo $row[studentName];
echo "<p>You records as of ";
echo date('l jS \of F Y h:i:s A');
echo "<table border='1' cellpadding='2' cellspacing='0'>";
echo "<tr> <th>Date</th><th>Student Name</th> <th>Tutor name</th> <th>Procedure name</th> <th>Grade</th><th>Student Reflection</th><th>Tutor Comments</th><th>Professionalism</th> <th>Communication</th> <th>Alert</th> <th>Dispute</th> </tr>";
// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array( $result )) {
if($color==1){
echo "<tr bgcolor=#DDD ><td>".$row['date']."</td><td>".$row['studentName']."</td><td>".$row['tutorName']."</td><td>".$row['procedureName']."</td><td>".$row['grade']."</td><td>".$row['studentReflection']."</td><td>".$row['tutorComments']."</td><td>".$row['professionalism']."</td><td>".$row['communication']."</td><td>".$row['alert']."</td><td>".$row['dispute']."</td></tr>";
// Set $color==2, for switching to other color
$color="2";
}
// When $color not equal 1, use this table row color
else {
echo "<tr bgcolor='#CCC'><td>".$row['date']."</td><td>".$row['studentName']."</td><td>".$row['tutorName']."</td><td>".$row['procedureName']."</td><td>".$row['grade']."</td><td>".$row['studentReflection']."</td><td>".$row['tutorComments']."</td><td>".$row['professionalism']."</td><td>".$row['communication']."</td><td>".$row['alert']."</td><td>".$row['dispute']."</td></tr>";
// Set $color back to 1
$color="1";
}
}
echo '</table>';
}
What I want to do it is echo or print out the users name above the table but I am having difficulty doing so. The only want I can get it it to work is to add it to the while loop but then it prints it out as many times as the user has records.
Can you help?
You will need to fetch the first row in order to gain access to the user information. That means you will have to call mysql_fetch_array once before the loop.
That will get you in trouble, because you will also need to call it in the loop. You could hack around that by using all kinds of boolean flags or copies of the row, but the best way is to change the structure of your code a little.
Use a do-while loop, combined with an if statement. This allows you to fetch a single row first, and take special action if none is found. After that, you got a loop that does what it does now, only it checks if there is a next row after the iteration instead of before, otherwise the first row would be skipped in the table output.
if ($row = mysql_fetch_array( $result )) {
// Print user info
// Print table header
do {
// Print table row
} while ($row = mysql_fetch_array( $result ));
// Print table footer
}
else
{
// User not found. Print error or whatever.
}
You have mistake at your logic...
You are making table of studens and you want to echo Welcome to the student which is logged_in
But you use echo "<p>Welcome "; echo $row[studentName];
Where $row is not set... This name should come from a previous select ..