Finding library details based on user search input - using LIMIT and OFFSET - php

Basically I have 4 fields in a form. I want to the user search for books in a library by either title, or by author or both. I also want the user to set the length of the list of items and from the starting point, these are not restrictions though the user does not have to specify.
Here is the code:
require_once __DIR__.'/config.php';
session_start();
$dbh = new PDO('mysql:host=' . DB_HOST . ';dbname=' . DB_USERNAME, DB_USERNAME, DB_PASSWORD);
$title = $_GET["title"];
$authors = $_GET["authors"];
$st = $_GET["start"];
$ln = $_GET["length"];
$stmt = $dbh->prepare("SELECT title, authors, description, price FROM books WHERE title = :title LIMIT :length OFFSET :start");
$stmt->execute(array(':title' => $title,':start' => $st,':length' => $ln));
while ($row = $stmt->fetch(PDO::FETCH_ASSOC))
{
$title = $row['title'];
$authors = $row['authors'];
$description = $row['description'];
$price = $row['price'];
}
echo "<table>";
echo "<tr>";
echo "<td>Title</td>";
echo "<td>$title</td>";
echo "</tr>";
echo "<tr>";
echo "<td>Authors</td>";
echo "<td>$authors</td>";
echo "</tr>";
echo "<tr>";
echo "<td>Description</td>";
echo "<td>$description</td>";
echo "</tr>";
echo "<tr>";
echo "<td>Price</td>";
echo "<td>$price</td>";
echo "</tr>";
echo "</table>";
So far it literally just returns me what I have typed in the input - so nothing much at all! Does anyone know how I can do this?

Adapt your code this way:
echo "<table>";
while ($row = $stmt->fetch(PDO::FETCH_ASSOC))
{
$title = $row['title'];
$authors = $row['authors'];
$description = $row['description'];
$price = $row['price'];
echo "<tr>";
echo "<td>Title</td>";
echo "<td>$title</td>";
echo "</tr>";
echo "<tr>";
echo "<td>Authors</td>";
echo "<td>$authors</td>";
echo "</tr>";
echo "<tr>";
echo "<td>Description</td>";
echo "<td>$description</td>";
echo "</tr>";
echo "<tr>";
echo "<td>Price</td>";
echo "<td>$price</td>";
echo "</tr>";
}
echo "</table>";
to print all the rows returned by the query. Your code was printing just the last row.

Related

deleting a row with a particular id in while loop

<?php
$sql = "SELECT * from movieinfo";
$sql_data = mysqli_query($db,$sql);
echo "<table>";
echo "<tr>";
echo "<td><b>Movie Name</b></td>";
echo"<td><center><b>Delete</b></center></td>";
echo"</tr>";
while($row = mysqli_fetch_array($sql_data,MYSQLI_ASSOC)) {
echo "<tr>";
echo "<td style='color:white'>";
echo $row['title'];
echo "<td><center><a href='adminpage.php?mid={$row['movieid']}'><button
class='contact100-form-btn' name='deletem'>Delete Movie</button></a>
</center></td>";
$_SESSION['delete'] = $row['movieid'];
echo "<td>";
echo"</tr>";
}
echo "</table>";
?>
if (isset($_POST['deletem'])) {
$delete=$_SESSION['delete'];
$query6 = "DELETE FROM movieinfo WHERE movieid=$delete";
mysqli_query($db, $query6);
}
i need to delete the particular row with the id selected so i am storing in session the id and deleting it but when i press on the deleting button it deletes the last row only its because session is storing all the id's because of loop how should i fix it?
<?php
$sql = "SELECT * from movieinfo";
$sql_data = mysqli_query($db,$sql);
echo "<table>";
echo "<tr>";
echo "<td><b>Movie Name</b></td>";
echo"<td><center><b>Delete</b></center></td>";
echo"</tr>";
while($row = mysqli_fetch_array($sql_data,MYSQLI_ASSOC)) {
echo "<tr>";
echo "<td style='color:white'>";
echo $row['title'];
**echo "</td>";
$mid = isset($_GET['mid']) ? $_GET['mid'] : '';
$del = "DELETE FROM movieinfo WHERE movieid=$mid";
mysqli_query($db, $del);
{
echo "<td><center><a href='adminpage.php?mid={$row['movieid']}'><button
class='contact100-form-btn' name='deletem'>Delete Movie</button></a>
</center></td>";
}
echo"</tr>";
}
echo "</table>";
?>

showing results obtained after $_POST in a new page

I have been trying to work this out for a while.
I have created a form with a dropdown box that gets results from a database. from this i then $_POST that from to another page. From that second page i wish to get the ID number and then get the records and display them on screen.
I will then put them in a table to organise the results better.
can anyone help me in achieving this.
Here is the code for the form (which works and sends the $PlantID)
$sql = "SELECT DISTINCT * FROM PLANTS";
$result = mysqli_query($mysqli,$sql)or die(mysqli_error());
//********************* Botannical name drop down box
echo "<form name='selection' id='selection' action='profile.php' method='post'>";
echo "<select name='flower'>";
while($row = mysqli_fetch_array($result)) {
$plantid = $row['FlowerID'];
$plantname = $row['Botannical_Name'];
$plantcommon = $row['Common_Name'];
/* $plantheight = $row['Height'];
$plantav = $row['AV'];
$plantcolours = $row['Colours'];
$plantflowering = $row['Flower_Time'];
$plantspecial = $row['Special_Conditions'];
$plantfrost = $row['Frost_Hardy'];
$plantaspect = $row['Aspect'];
$plantspeed = $row['Growth_Speed'];*/
echo "<option value=".$plantid.">".$plantname." -> AKA -> ".$plantcommon."</option>";
}
echo "</select>";
echo "<br />";
//********************* End of form
echo "<input type='submit' name='submit' value='Submit'/>";
echo "</form>";
I have created this page to get the ID and display that ID on screen. AS you can tell i have probably doubled up on ways to try work this out.
$sql = "SELECT * FROM PLANTS";
$result = mysqli_query($mysqli,$sql)or die(mysqli_error());
if(isset($_POST['submit'])){
$selected_val = $_POST['Botannical_Name']; // Storing Selected Value In Variable
echo "You have selected :" .$selected_val; // Displaying Selected Value
}
echo "<br />";
echo "well:".$_POST["Botannical_Name"]."<br/>";
echo "now:".$plantquery."<br />";
echo $_POST;
echo "<table>";
foreach ($_POST as $key => $value) {
echo "<tr>";
echo "<td>";
echo $key;
echo "</td>";
echo "<td>";
echo $value;
echo "</td>";
echo "</tr>";
}
echo "</table>";
Any help would be greatly appreciated.
you should use following to get the selected value,
$selected_val = $_POST['flower'];
if(isset($_POST['submit'])){
$selected_val = $_POST['flower']; // Storing Selected Value In Variable
echo "You have selected :" .$selected_val; // Displaying Selected Value
$sql = "SELECT * FROM PLANTS WHERE FlowerID='.$selected_val.'";
$result = mysqli_query($mysqli,$sql)or die(mysqli_error());
while ($row=mysqli_fetch_assoc($result))
{
echo $row['Botannical_Name'];
}
}
echo "<br />";
print_r($_POST);
if(!empty(_POST)) {
echo "<table>";
foreach ($_POST as $key => $value) {
echo "<tr>";
echo "<td>";
echo $key;
echo "</td>";
echo "<td>";
echo $value;
echo "</td>";
echo "</tr>";
}
echo "</table>";
}

How can I arrange this code so I can get different entry from database for each table row

I want every different record of database to be display on each table rows, but I'm unable to retrieve different record for each row and column. Please give me suggestion where I can paste than while block so it will give different result for each row and column.
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$rec_limit = 10;
$scriptname=$_SERVER['PHP_SELF'];
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db('online_shopping'); // include your code to connect to DB.
$tbl_name="mobile_db"; //your table name
$start = 0;
$limit = 5;
$sql = "SELECT id,company,model,price,availability,image FROM $tbl_name LIMIT $start, $limit";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result))
{
$company=$row['company'];
$model=$row['model'];
$available=$row['availability'];
$price=$row['price'];
}
echo "<table border='2'>";
$j = 0;
for($j = 0; $j<5; $j++)
{
echo "<tr>";
for($i = 0; $i<3; $i++)
{
echo "<td>";
echo "<table border='2'>";
echo "<tr>";
echo "<td><img src='abc.jpg' height='250' width='250'/></td>";
echo "</tr>";
echo "<tr>";
echo "<td>";
echo "<table border='2'>";
echo "<tr>";
echo "<td><b>Brand : </b></td>";
echo '<td>'.$company.'</td>';
echo "</tr>";
echo "<tr>";
echo "<td><b>Model : </b></td>";
echo '<td>'.$model.'</td>';
echo "</tr>";
echo "<tr>";
echo "<td><b>Availability : </b></td>";
echo '<td>'.$available.'</td>';
echo "</tr>";
echo "<tr>";
echo "<td><b>Price : </b></td>";
echo '<td>'.$price.'</td>';
echo "</tr>";
echo "</table>";
echo "</td>";
echo "</tr>";
echo "</table>";
echo "</td>";
}
echo "</tr>";
}
echo "</table>";
?>
You need to move the table rows inside the fetch loop or store the row in an array. I have simplified your tables to make the example clearer:
$result = mysql_query($sql);
if (!$result) {
/* Error */
}
echo '<table>';
while ($row = mysql_fetch_array($result)) {
echo '<tr><td><img src="', htmlspecialchars ($row['image']), '">';
echo '<tr><td><table>';
echo ' <tr><th>Brand<td>', htmlspecialchars ($row['company']);
echo ' <tr><th>Model<td>', htmlspecialchars ($row['model']);
echo ' <tr><th>Availability<td>', htmlspecialchars ($row['availability']);
echo ' <tr><th>Price<td>', htmlspecialchars ($row['price']);
echo ' </table>';
}
echo "</table>\n";
Some notes about the code:
Test the return value of mysql_query(). The query might fail.
Escape your output using htmlspecialchars().
You should use <th> elements for your headings and style those, instead of using inline <b> elements.
I added output of $row['image'] which might not do what you want.
And do not use the deprecated mysql extension. Use PDO or mysqli instead.
In your while loop you always rewrite the same variables, after loop you have only last record saved.
In the loop, you have to save records into array.
In your code you have nested tables, but in the first one, there is only one row and one table cell which contains another table. I use just nested table.
<?php
...
$result = mysql_query($sql);
$data = array();
while($row = mysql_fetch_array($result)) {
$data[] = $row;
}
if (count($data) > 0) {
echo '<table>';
foreach ($data as $row) {
echo '<tr>';
echo '<td>Brand: ' . $row['company'];
echo '<td>Model: ' . $row['model'];
echo '<td>Availability: ' . $row['availability'];
echo '<td>Price: ' . $row['price'];
}
echo '</table>';
} else {
echo 'no records';
}
?>
Not sure I fully understand what you are trying to accomplish but try this.
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$rec_limit = 10;
$scriptname=$_SERVER['PHP_SELF'];
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db('online_shopping'); // include your code to connect to DB.
$tbl_name="mobile_db"; //your table name
$start = 0;
$limit = 5;
$sql = "SELECT id,company,model,price,availability,image FROM $tbl_name LIMIT $start, $limit";
$result = mysql_query($sql);
echo "<table border='2'>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>".$row['company']."</td>";
echo "<td>".$row['model']."</td>";
echo "<td>".$row['availability']."</td>";
echo "<td>".$row['price']."</td>";
echo "</tr>";
}
echo "</table>";
?>

how to select two columns from two tables in mysql

I am trying to execute this query but i got error " Undefined index:
lname".I want to count row from one column(fname) from table a and
select column(lname) from other table b. so please help me.
$result = mysql_query("SELECT COUNT(fname),lname FROM a,b");
while ($row = mysql_fetch_array($result))
{
echo "<tr><td>";
echo $row['lname'];
echo "</td>";
echo "<td>";
echo $row['COUNT(fname)'];
echo "</td></tr>";
}
If you still get an error you can try to fetch both separately:
$result = mysql_query("SELECT COUNT(fname) FROM a");
while ($row = mysql_fetch_array($result))
{
echo "<tr><td>";
echo $row['COUNT(fname)'];
echo "</td></tr>";
}
$result1 = mysql_query("SELECT lname FROM b");
while ($row = mysql_fetch_array($result1))
{
echo "<tr><td>";
echo $row['lname'];
echo "</td></tr>";
}
You need to use an alias. Use this:
$result = mysql_query("SELECT COUNT(fname) AS countfname,lname FROM a,b");
while ($row = mysql_fetch_array($result))
{
echo "<tr><td>";
echo $row['lname'];
echo "</td>";
echo "<td>";
echo $row['countfname'];
echo "</td></tr>";
}
Try this code:
$result = mysql_query("SELECT COUNT(a.fname) as fname,b.lname as lname FROM a,b");
while ($row = mysql_fetch_array($result))
{
echo "<tr><td>";
echo $row['lname'];
echo "</td>";
echo "<td>";
echo $row['COUNT(fname)'];
echo "</td></tr>";
}

Php undefine variable issue

I've a html form which is insert data to mysql database and then get those data with following php code (From supplier_jv table)
<?php
include("include/address2.php");
include("include/menu.php");
$uname_ad = $_SESSION['uname_ad'];
$id = $_GET['id'];
$sql = mysql_query("SELECT * FROM supplier_jv WHERE jv_id = '$id'");
$num = mysql_num_rows($sql);
if($num == 0)
{
echo "<p><font color=red>Accounts is emtpy</font></p>";
}
else
{
$re_name = mysql_fetch_array($sql);
echo "<center><h2>";
echo "<strong>Accounts of </strong>";
echo $re_name['jv_name'];
echo "</h2></center>";
echo "<center>";
echo "<table>";
echo "<table border='0' cellpadding='5' cellspacing='5' width='1000'>";
echo "<tr/>";
echo "<td><strong>Date</strong></td>";
echo "<td><strong>Particular</strong></td>";
echo "<td><strong>Folio(C)</strong></td>";
echo "<td><strong>Folio(J)</strong></td>";
echo "<td><strong>Debit</strong></td>";
echo "<td><strong>Credit</strong></td>";
echo "<td><strong>Balance</strong></td>";
echo "</tr>";
while($re= mysql_fetch_array($sql))
{
$day = $re['day'];
$month $re['month'];
$year = $re['year'];
$parti = $re['particulars'];
$folio = $re['folio'];
$folio2 = $re['folio2'];
$debit = $re['debit'];
$credit = $re['credit'];
$balance = $re['balance'];
$b = $debit - $credit;
$total_debit = mysql_query("SELECT SUM(debit) FROM supplier_jv");
$re_t = mysql_fetch_array($total_debit);
$t_d = $re_t['SUM(debit)'];
$total_credit = mysql_query("SELECT SUM(credit) FROM supplier_jv");
$re_t2 = mysql_fetch_array($total_credit);
$t_c = $re_t2['SUM(credit)'];
$b = $t_d - $t_c;
echo "<tr>";
echo "<td>$day/$month/$year</td>";
echo "<td>$parti</td>";
echo "<td>$folio</td>";
echo "<td>$folio2</td>";
echo "<td>";
echo number_format($debit);
echo "</td>";
echo "<td>";
echo number_format($credit);
echo "</td>";
echo "<td></td>";
echo "</tr>";
}
echo "<tr bgcolor='#f3f3f3'>";
echo "<td></td>";
echo "<td></td>";
echo "<td></td>";
echo "<td></td>";
echo "<td></td>";
echo "<td><strong>Total Balance-</strong></td>;
echo "<td><strong>";
echo number_format($b);
echo "</strong></td>";
echo "</tr>";
echo "</table>";
echo "</center>";
}
?>
Well, After insert data then it's show:
Notice: Undefined variable: b in E:\xampp\htdocs\Accounts\admin\content
\supplier_account.php on line 108
But if i insert data in second time then it's OK!!.
Any idea or solution.
Thanks
shibbir.
You need to use isset to see if it is set, not null and also avoid the Notice: Undefined variable error:
if (isset($b))
{
// your code here
}
Where you have:
$b = $t_d - $t_c;
Make sure that there is some value coming up for $t_d and $t_c

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