I want to have if the variable = null then he makes it to 1.
if the variable exist do nothing and dont make it again to 1.
i got this code:
if (isset($_POST["register"])) {
if(!$l_NextPage){
$l_NextPage = 1;
echo "helaas" . "</br>";
}
if($l_NextPage == 1){
echo "hoi";
$l_NextPage = 2;
}else if($l_NextPage == 2){
echo "doei";
}
}
only the code dont work i tried empty, isset, $var == FALSE but everytime he makes $l_NextPage to 1. is there any solution i tried this too with session but even it don't work!
What am I doing wrong?
what happen when you refresh page, it assign $l_NextPage = 1 every time, thats why all the time hoi printed
you can use sessions for preserving value of variable after page refresh
try this code
// write this line of code at top of php block
session_start();
if (isset($_POST["register"]))
{
if (!isset($_SESSION["l_NextPage"]))
{
$_SESSION["l_NextPage"] = 1;
echo "helaas" . "</br>";
}
if($_SESSION["l_NextPage"] == 1)
{
echo "hoi";
$_SESSION["l_NextPage"] = 2;
}
else if($_SESSION["l_NextPage"] == 2)
{
echo "doei";
//unset( $_SESSION['l_NextPage'] ); unset varibale
}
}
after reaching at prefixed condition you can unset varible using
unset( $_SESSION['l_NextPage'] );
i have not tested code but this should work
you should try:
if(!isset($l_NextPage)) {
$l_NextPage = 1;
echo "helaas" . "</br>";
}
elseif($l_NextPage == 1) {
(...)
try like this,
if(!empty(trim($l_NextPage)))
Related
Im writing a page in HTML/PHP that connects to a Marina Database(boats,owners etc...) that takes a boat name chosen from a drop down list and then displays all the service that boat has had done on it.
here is my relevant code...
if(isset($_POST['form1'])){//if there was input data submitted
$form1 = $_POST['form1'];
$sql1 = 'select Status from ServiceRequest,MarinaSlip where MarinaSlip.SlipID = ServiceRequest.SlipID and BoatName = "'.$form1.'"';
$form1 = null;
$result1 = $conn->query($sql1);
$test = 0;
while ($row = mysqli_fetch_array($result1, MYSQLI_ASSOC)) {
$values1[] = array(
'Status' => $row['Status']
);
$test = 1;
}
echo '<p>Service Done:</p><ol>';
if($test = 1){
foreach($values1 as $v1){
echo '<li>'.$v1['Status'].'</li>';
}
echo '</ol>';
}else{
echo 'No service Done';
}
the issue im having is that some of the descriptions of sevice are simply Open which i do not want displayed as service done, or there is no service completed at all, which throws undefined variable: values1
how would I stop my script from adding Open to the values1 array and display a message that no work has been completed if values1 is empty?
Try this
$arr = array();
if (empty($arr))
{
echo'empty array';
}
We often use empty($array_name) to check whether it is empty or not
<?php
if(!empty($array_name))
{
//not empty
}
else
{
//empty
}
there is also another way we can double sure about is using count() function
if(count($array_name) > 0)
{
//not empty
}
else
{
//empty
}
?>
To make sure an array is empty you can use count() and empty() both. but count() is slightly slower than empty().count() returns the number of element present in an array.
$arr=array();
if(count($arr)==0){
//your code here
}
try this
if(isset($array_name) && !empty($array_name))
{
//not empty
}
You can try this-
if (empty($somelist)) {
// list is empty.
}
I often use empty（$arr) to do it.
Try this instead:
if (!$values1) {
echo "No work has been completed";
} else {
//Do staffs here
}
I think what you need is to check if $values1 exists so try using isset() to do that and there is no need to use the $test var:
if(isset($values1))
foreach($values1 as $v1){
echo '<li>'.$v1['Status'].'</li>';
}
Or try to define $values1 before the while:
$values1 = array();
then check if it's not empty:
if($values1 != '')
foreach($values1 as $v1){
echo '<li>'.$v1['Status'].'</li>';
}
All you have to do is get the boolean value of
empty($array). It will return false if the array is empty.
You could use empty($varName) for multiple uses.
For more reference : http://php.net/manual/en/function.empty.php
First thing first, i'm new to php.
I'm trying to check if two variables are both empty, and if so merge them togheter to display only one result, but if one of them is not empty, than i have to display is value.
I currently have:
$customerNotesWC = "";
$DYSPrintableOrderNotes = "test note";
Here's the code i tried so far:
function displayCustomerOrderNotes($customerNotesWC,$DYSPrintableOrderNotes){
if ($customerNotesWC == "") {
$customerNotesWC = "(none)";
}
if ($DYSPrintableOrderNotes ==""){
$DYSPrintableOrderNotes = "(none)";
}
if ($DYSPrintableOrderNotes == "(none)" && $customerNotesWC == "(none)"){
$NotesToDisplay = "(none)";
}
else {
$NotesToDisplay = $customerNotesWC . "<br/>" . $DYSPrintableOrderNotes;
}
}
Unfortunately this doesn't work, as the results is the following:
(none)
Pre-sale order note
I know that there must be a better way to achieve this, and any suggestions will be really appreciated.
Thanks a lot
In order to get your desired functionality of only printing "(none)" when both values are blank, the simplest thing to do to fix your code is remove the code that's setting either value to "(none)" and change your main if-statement to check for "":
function displayCustomerOrderNotes($customerNotesWC,$DYSPrintableOrderNotes){
if ($DYSPrintableOrderNotes == "" && $customerNotesWC == ""){
$NotesToDisplay = "(none)";
}
else {
$NotesToDisplay = $customerNotesWC . "<br/>" . $DYSPrintableOrderNotes;
}
//shouldn't there be an echo $NotesToDisplay; or return $NotesToDisplay; ? or something?
}
The variable in the session only changes if i update the file meaning:
<?php
if(isset($_GET['row']) && isset($_GET['quantity'])) {
if($_GET['quantity'] != 0) {
$_SESSION['basket'][$_GET['row']]['barcode']['quantity'] = $_GET['quantity'];
} else {
unset($_SESSION['basket'][$_GET['row']]);
}
}
?>
Changes the quantity first time and then in other tries doesn't.
If I change it to something like:
<?php
if(isset($_GET['row']) && isset($_GET['quantity'])) {
if($_GET['quantity'] != 0) {
$num = $_GET['quantity'];
$_SESSION['basket'][$_GET['row']]['barcode']['quantity'] = $num;
} else {
unset($_SESSION['basket'][$_GET['row']]);
}
}
?>
I save the file in works once and then it stops changing the quantity.
When echoing the variable in the code in shows that it's changed but in the rest of the site it doesn't. the part where it shows doesn't manipulate the variables only prints them.
It seems as if the code works only if the file's been saved and only for the first time.
Thanks in advance.
I am trying to check if the mysql_fetch_array() function returns an empty array or not. But my code doesn't seem to work. Here I want to ensure that if the array is empty I want to display under construction message.
Code :
$queryContents= queryMembers();
$exeQuery = mysql_query($queryContents);
while($fetchSet = mysql_fetch_array($exeQuery)) {
if(count($fetchSet) == 0) {
echo "This Page is Under Construction";
}else{
// something else to display the content
}
}
How do I check to acheive such feature ?
use mysql_num_rows to count number of rows. try this.
$exeQuery = mysql_query($queryContents);
if(mysql_num_rows($exeQuery)== 0){
echo "This Page is Under Construction";
}
else{
while($fetchSet = mysql_fetch_array($exeQuery)) {
// something else to display the content
}
}
You really should be using mysql_num_rows http://us2.php.net/manual/en/function.mysql-num-rows.php
However, on a side note, you should use php empty() instead. http://us2.php.net/empty
When you use mysql_fetch_array(), it returns the rows from the data
set one by one as you use the while loop.
If there will be no record, while loop wont execute. In this case, declare a boolean variable and make it true if it enters the while loop. Like:
$queryContents= queryMembers();
$exeQuery = mysql_query($queryContents);
$recordExists = 0;
while($fetchSet = mysql_fetch_array($exeQuery)) {
if($recordExists == 0 )
$recordExists = 1;
// something else to display the content
}
if($recordExists == 0 ){
echo "This Page is Under Construction";
}
Hope this works!
You can do it this way:
while($r[]=mysql_fetch_array($sql));
// now $r has all the results
if(empty($r)){
// do something
}
source: php doc
Your code inside the while loop never runs if there are no results. mysql_fetch_array returns null/false if there are no more results. What you need yo do is check with mysql_num_rows first, before the while.
$queryContents= queryMembers();
$exeQuery = mysql_query($queryContents);
if(mysql_num_rows ($exeQuery) == 0) {
echo "This Page is Under Construction";
}
while($fetchSet = mysql_fetch_array($exeQuery)) {
// something else to display the content
}
Try this
if(empty($fetchSet)
{
echo "This Page is Under Construction";
}
else
{
// something else to display the content
}
Good day guys,
I've made a sweet favorites function with php mysql and ajax, and its working great. Now I want to show 'favorite' when favorite = 0 and show 'unfavorite' when favorite = 1
if ($favorites == 0) {
$favorite = 'Favorite';
}
if ($favorites == 1) {
$unfavorite = 'unFavorite';
}
and echo it in the row as :
<div id="favorites">' .($favorite). ' ' .($unfavorite). '</div>
The problem is: when favorite = 0, both $favorite and $unfavorite are being shown. When favorite = 1 only $unfavorite is being shown correctly. Of course it should be $favorite OR $unfavorite. I assume the problem is clear and simple to you, please assist :)
Thanks in advance
It's easier to use just one variable:
$text = ''
if ($favorites == 0) {
$text = 'Favorite';
} else {
$text = 'unFavorite';
}
...
echo $text;
If you want to check $favorite, you are using the wrong variable in your control statement. Also, it is better coding practice to use elseif rather than if for that second if. One more thing: it's easier to manage one resulting variable.
$output = "";
if ($favorite == 0) {
$output = 'Favorite';
}
elseif ($favorite == 1) {
$output = 'unFavorite';
}
...
echo $output; // Or whatever you want to do with your output
Is $favorites an integer?
Anyway try using three equal signs (===) or else instead of the second if:
if ( $favorites === 0 )
{
// ...
}
else // or if ($favorites === 1)
{
// ...
}
You're making a toggle, so you only need one variable:
if(empty($favourites)){
$fav_toggle = 'Favorite';
} else {
$fav_toggle = 'unFavorite';
}
echo $fav_toggle;
Same code is working on me if I assigned $favorites = 0; or $favorites = 1;
You can also use if else
$favorites = 1;
if ($favorites == 0) {
$favorite = 'Favorite';
}
else if ($favorites == 1) {
$unfavorite = 'unFavorite';
}