PHP - While loop not showing table records - php

In below code it shows table row value outside while loop but not shows inside while loop. While loop not working; Please let me know whats wrong in it?
<?php
$sql = "SELECT * FROM cl_banner ORDER BY id;";
$res = q($sql) or die(mysql_error());
if($res && mysql_num_rows($res)>0)
{
while($row = mysql_fetch_assoc($res));
{
echo $row["title"];
echo "hi";
} // End While
} // End If
?>

<?php
$sql = "SELECT * FROM cl_banner ORDER BY id";
^^^^
$res = mysql_query($sql) or die(mysql_error());
^^^^^^^^^^
if($res && mysql_num_rows($res)>0)
{
while($row = mysql_fetch_assoc($res));
{
echo $row["title"];
echo "hi";
} // End While
} // End If
?>

Try this:
<?php
$sql = "SELECT * FROM cl_banner ORDER BY id";
$res = mysql_query($sql) or die(mysql_error());
if($res && mysql_num_rows($res)>0)
{
while($row = mysql_fetch_assoc($res))
{
echo $row["title"];
echo "hi";
} // End While
} // End If
?>

Related

Get data from category with php mysql

I try to get data from category using mysql and php.
Sql Structure:
Category
-cat_id
-name
Date
-id
-url
-category
Php code:
<?php
$sql = "select * from category";
$result = mysql_query($sql);
if(mysql_num_rows($result) > 0)
{
while ($row = mysql_fetch_assoc($result))
{
echo '<option value="'.$row["cat_id"].'">'.$row["name"].'</option>';
}
}
$sql = "select * from date WHERE category='1'";
$result = mysql_query($sql);
if(mysql_num_rows($result) > 0)
{
while ($row = mysql_fetch_assoc($result))
{
echo '.$row["url"].';
}
}
?>
when i select the category the data is not listed.
Any idea?
Try this Code . Just removed single quotations
$sql = "select * from date WHERE category='1'";
$result = mysql_query($sql);
if(mysql_num_rows($result) > 0)
{
while ($row = mysql_fetch_assoc($result))
{
echo $row["url"];
}
}
May this will work:
PHP Code
<?php
$sql = "select * from category";
$result = mysql_query($sql);
if(mysql_num_rows($result) > 0)
{
while ($row = mysql_fetch_assoc($result))
{
echo '<select id="category" name="category">';
echo '<option value="'.$row["cat_id"].'">'.$row["name"].'</option>';
echo '</select>';
}
}
$sql = "select * from date WHERE category='1'";
$result = mysql_query($sql);
if(mysql_num_rows($result) > 0)
{
while ($row = mysql_fetch_assoc($result))
{
echo 'url is: '.$row["url"];
}
}
?>
Try this
$sql = "select * from date WHERE category='1'";
$result = mysql_query($sql);
if(mysql_num_rows($result) > 0)
{
while ($row = mysql_fetch_assoc($result))
{
echo $row["url"];
}
}
if you want result with in single quotes
$sql = "select * from date WHERE category='1'";
$result = mysql_query($sql);
if(mysql_num_rows($result) > 0)
{
while ($row = mysql_fetch_assoc($result))
{
echo "'".$row["url"]."'";
}
}
Please avoid mysql_* because the mysql_* functions have been removed in PHP7. Use MySQLi instead.
PHP + Mysql :
<?php
$sql = "select * from category";
$result = mysql_query($sql);
echo "<select name='category'>";
if(mysql_num_rows($result) > 0)
{
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="'.$row["cat_id"].'">'.$row["name"].'</option>';
}
}
echo "</select>";
if(!empty($_POST['category'])) {
$category_id = $_POST['category'];
$sql = "select * from date WHERE category = '".$category_id."'";
$result = mysql_query($sql);
if(mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_assoc($result))
{
echo '.$row["url"].';
}
}
}
?>
PHP + Mysqli
<?php
$servername = "localhost";
$username = "username";
$password = "";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "select * from category";
$result = $conn->query($sql);
echo "<select name='category'>";
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo '<option value="'.$row["cat_id"].'">'.$row["name"].'</option>';
}
}
echo "</select>";
if(!empty($_POST['category'])) {
$category_id = $_POST['category'];
$sql = "select * from date WHERE category = '".$category_id."'";
$result = $conn->query($sql);
if($result->num_rows > 0) {
while ($row = $result->fetch_assoc())
{
echo '.$row["url"].';
}
}
}
$conn->close();
?>
Nothing changed.
I'm using this code:
<?php
$sql = "select * from category";
$result = mysql_query($sql);
echo "<select name='category'>";
if(mysql_num_rows($result) > 0)
{
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="'.$row["cat_id"].'">'.$row["name"].'</option>';
}
}
echo "</select>";
if(!empty($_POST['category'])) {
$sql = "select * from date WHERE category = '1'";
$result = mysql_query($sql);
if(mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_assoc($result))
{
echo $row["url"];
}
}
}
?>
If i delete if condition the data is listed but all the time.

Conditional Logic on MySQL Query with PHP

I am doinga MySQL query to retreive data using PHP. I need to have a logic that if the data set returned is empty it shows a warning message else it displays the results:
$searchQuery = mysql_escape_string($_POST['searchQuery']);
$sql="SELECT * FROM db.tblname WHERE column1 = '".$searchQuery."'";
$result = mysqli_query($conn,$sql);
while($row = mysqli_fetch_array($result)){
echo $row["column1"];
}
mysqli_close($conn);
You need num_rows on the object
$result = mysqli_query($conn,$sql);
if ($result->num_rows == 0) {
echo 'result empty';
} else {
while($row = mysqli_fetch_array($result)){
echo $row["column1"];
}
}
http://php.net/manual/en/mysqli-result.num-rows.php
Change mysql_escape_string to mysqli_escape_string here
$where = "";
if(isset($_POST['searchQuery']) && trim($_POST['searchQuery'])){
$searchQuery = mysqli_escape_string($conn,$_POST['searchQuery']);
$where =" WHERE column1 = '".$searchQuery."'";
}
$sql="SELECT * FROM db.tblname ".$where;
It's as simple as this:
if ($result) {
//Your code
} else {
echo "Error: ".mysqli_error($conn);
}
First check with mysqli_num_rows(). Here is how to
$searchQuery = mysql_escape_string($_POST['searchQuery']);
$sql="SELECT * FROM db.tblname WHERE column1 = '".$searchQuery."'";
$result = mysqli_query($conn,$sql);
if(mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_array($result)){
echo $row["column1"];
}
}
else {
echo "No records found";
}
mysqli_close($conn);
$searchQuery = mysql_escape_string($_POST['searchQuery']);
$sql="SELECT * FROM db.tblname WHERE column1 = '".$searchQuery."'";
$result = mysqli_query($conn,$sql);
if(mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_array($result)){
echo $row["column1"];
}
} else {
echo "No results found";
}
mysqli_close($conn);

How to access while array outside the loop in PHP

I want to access while array $row2 outside the loop so that I can compare in if else condition, because $row2 array contains many number. Is there any solution make $row array accessible outside the loop or any other method?
<?php
$sql = (" SELECT * FROM result ");
$query = mysqli_query($db, $sql);
while($row = mysqli_fetch_array($query)) {
$row2 $row['id'];
}
for($i=1;$i<=10;$i++) {
if ($i<= $row2) {
echo "<font color='red'><font size='50px'>".$i."</font></font>"."<br>";
continue;
} else {
echo $i.'<br>';
}
}
?>
If you want to loop over two dependent things you need to nest them or else $row2 will always contain the last value in the while loop:
<?php
$sql = (" SELECT * FROM result ");
$query = mysqli_query($db, $sql);
while ($row = mysqli_fetch_array($query)) {
$row2 = $row['id']; // I added an = sign here.
for($i=1;$i<=10;$i++) {
if ($i<= $row2) {
echo "<font color='red'><font size='50px'>".$i."</font></font>"."<br>";
continue;
} else {
echo $i.'<br>';
}
}
}
If this isn't what you want please clarify your question.
Pass your variable to a separate function that performs whatever task you want.
<?php
while($row = mysqli_fetch_array($query))
{
MyFunction($row2);
}
function MyFunction($passed_row2_value){
// perform whatever task you want here.
}
?>
This shows with some format the ids retrieved by the query and fills the blanks without format.
<?php
$sql = (" SELECT * FROM result ");
$query = mysqli_query($db, $sql);
$row2 = [];
while($row = mysqli_fetch_array($query)) {
$row2[$row['id']] = true;
}
for($i=1;$i<=10;$i++) {
if ($row2[$i]) {
echo "<font color='red'><font size='50px'>".$i."</font></font>"."<br>";
} else {
echo $i.'<br>';
}
}
?>

SQL query not showing more than 1 row

This is the entire code the error must be in query 3 or 4. As you can see query 3 just gets info to build query 4 which should return the results.
I've got query 1 & 2 working ok which have the correct data showing.
<?php
session_start();
//printf('<pre>%s</pre>', print_r($_SESSION, true));
require('includes/config.inc.php');
require('includes/session.php');
//WORKING
DB_Connect();
$query = "SELECT * FROM food_delivery_orders_items WHERE food_delivery_orders_items.type = 'product' AND food_delivery_orders_items.order_id=".$_SESSION['pdf_quote']['id']."";
$result = mysql_query( $query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$hash = $row['hash'];
$foreignid = $row['foreign_id'];
$qty = $row['cnt'];
}
$query2 = "SELECT * FROM food_delivery_products WHERE food_delivery_products.id = ".$foreignid."";
$result2 = mysql_query( $query2) or die(mysql_error());
while($row = mysql_fetch_array($result2))
{
$name = $row['name'];
$description = $row['description'];
}
//---------------------------------------------------------------------------------------------------------------
$query3 = "SELECT * FROM food_delivery_orders_items WHERE food_delivery_orders_items.type = 'extra' AND food_delivery_orders_items.order_id=".$_SESSION['pdf_quote']['id']."";
$result3 = mysql_query( $query3)or die(mysql_error()) ;
while($row = mysql_fetch_array($result3))
{
$foreignidextra = $row['foreign_id'];
$qtyextra = $row['cnt'];
}
$query4 = "SELECT * FROM food_delivery_extras WHERE food_delivery_extras.id = ".$foreignidextra."";
$result4 = mysql_query( $query4) or die(mysql_error());
while($row = mysql_fetch_array($result4))
{
$nameextra = $row['name'];
}
echo $nameextra;
echo $qtyextra;
DB_Disconnect();
//$products = implode(", ",$_SESSION['pdf_quote']['product_arr']);;
//print $products
?>
print $row; is only valid inside the loop if you want to print all of them. Otherwise it will print only the last $row. Make it
while($row = mysql_fetch_array($result3))
{
$foreignidextra = $row['foreign_id'];
$qtyextra = $row['cnt'];
print $row;
}
its because you are printing the $row after the while loop gets executed
So it only prints the final result set
Use something like this instead:
$query3 = "SELECT * FROM food_delivery_orders_items WHERE food_delivery_orders_items.type = 'extra' AND food_delivery_orders_items.order_id=".$_SESSION['pdf_quote']['id']."";
$result3 = mysql_query( $query3) ;
while($row = mysql_fetch_array($result3))
{
$foreignidextra = $row['foreign_id'];
$qtyextra = $row['cnt'];
print_r($row);
}
you can also use the var_dump($row); instead of print_r($row);
this will output all the values stored in the $row array each time the loop iterates

Why does this query show only one result?

The query I have below will only show me one result even if there are multiple matching entries (completely or partially matching). How do I fix it so it will return all matching entries:
//$allowed is a variable from database.
$sql = "SELECT `users`.`full_name`, `taglines`.`name`, `users`.`user_id` FROM
`users` LEFT JOIN `taglines` ON `users`.`user_id` = `taglines`.`person_id`
WHERE ( `users`.`user_settings` = '$allowed' ) and ( `users`.`full_name`
LIKE '%$q%' ) LIMIT $startrow, 15";
$result = mysql_query($sql);
$query = mysql_query($sql) or die ("Error: ".mysql_error());
$num_rows1 = mysql_num_rows($result);
if ($result == "")
{
echo "";
}
echo "";
$rows = mysql_num_rows($result);
if($rows == 0)
{
}
elseif($rows > 0)
{
while($row = mysql_fetch_array($query))
{
$person = htmlspecialchars($row['full_name']);
}
}
}
print $person;
Because your overwriting $person on each iteration.
Hold it in a $person[] array if your expecting more then one. Then loop through it with a foreach loop when you intend to output.
Not related but your also querying twice, you only need 1 $result = mysql_query($sql);
Update (Simple Outputting Example):
<?php
$person=array();
while($row = mysql_fetch_array($query)){
$person[] = array('full_name'=>$row['full_name'],
'email'=>$row['email'],
'somthing_else1'=>$row['some_other_column']);
}
//Then when you want to output:
foreach($person as $value){
echo '<p>Name:'.htmlentities($value['full_name']).'</p>';
echo '<p>Eamil:'.htmlentities($value['email']).'</p>';
echo '<p>FooBar:'.htmlentities($value['somthing_else1']).'</p>';
}
?>
Or an alternative way to is to build your output within the loop using concatenation.
<?php
$person='';
while($row = mysql_fetch_array($query)){
$person .= '<p>Name:'.$row['full_name'].'</p>';
$person .= '<p>Email:'.$row['email'].'</p>';
}
echo $person;
?>
Or just echo it.
<?php
while($row = mysql_fetch_array($query)){
echo '<p>Name:'.$row['full_name'].'</p>';
echo '<p>Email:'.$row['email'].'</p>';
}
?>

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