At the model of MVC pattern of codeigniter, I make some codes that is same at below.
function getBoardNum($option){
foreach($option as $key=>$content){
$this->db->where($key,$content);
//echo "$key $content";
}
if(isset($this->db->get('parsing_tb')->row()->num)){
$num = $this->db->get('parsing_tb')->row()->num;
return $num;
} else {
return false;
}
}
The problem is the isset at the if condition seems to not work.
To find out what is wrong, I used var_dump and such like confirming the $option data. But everything is okay...
Please let me know what is wrong! Thank you for reading.
Idk why isset() is not working, but you can try this instead :
//...
$query = $this->db->get('parsing_tb');
if ($query->num_rows() > 0)
{
$num = $this->db->get('parsing_tb')->row()->num;
return $num;
}
else
return false;
You should use isset to check if some variable is initialized - that means you check if it has any value.
even this case will yield TRUE:
$var = "";
if( isset( $var )){
echo TRUE;
}
In your code it looks like you are first trying to initialize it and assign some value to it and you use isset. You should use a different method for whatever you are trying to accomplish.
Related
Im writing a page in HTML/PHP that connects to a Marina Database(boats,owners etc...) that takes a boat name chosen from a drop down list and then displays all the service that boat has had done on it.
here is my relevant code...
if(isset($_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
I have created a function that checks if the image is empty or image variable has no value or the image is not found then it returns default image, but on some products it gives results but not on all of them..
function image_check($image)
{
$no_image = "noimagefound.jpg";
if(isset($image) || !empty($image) || $image != " ")
{
if(file_exists('uploads/store/products/'.$image))
{
return 'uploads/store/products/'.$image;
}
else
{
return 'uploads/web_service/'.$no_image;
}
}
else
{
return 'uploads/web_service/'.$no_image;
}
}
Can anyone make it work properly? What am I missing?
function image_check($image)
{
$no_image = "noimagefound.jpg";
if( !empty($image) && file_exists('uploads/store/products/'.$image) )
{
return 'uploads/store/products/'.$image;
}
return 'uploads/web_service/'.$no_image;
}
As they pointed out in the comments, your condition failed because you were checking if it was empty, not if it wasn't empty. isset() and !empty() are redundant in this case.
You also don't need all of those else checks. Be careful complicating your code more than you need to. You only need one check, if that fails, then return your $no_image.
I have some code written in php to validate a postcode field in my form. The code is meant to check that the field is not empty (mandatory) and matches one of 5 usable postcodes, if not it displays an alert. The problem i am having is that when i leave the field empty and hit the submit button the proper alert is show but if i enter a wrong value and hit submit the form just loads to a blank screen, can anyone spot a mistake in my code? :
<?php
if (isset($_POST['submit'])) {
$post = $_POST["post"];
$words = array('2747','2750','2753','2760','2777');
$msgp = "";
if (!empty($post)) {
foreach ($words as $item)
{
if (strpos($post, $item) !== false)
return true;
}
$msgp = '<span class="error"><b>Please enter correct postcode</b></span>';
return false;
} else if(empty($post)) {
$msgp = '<span class="error"><b>Please enter postcode</b></span>';
}
}
?>
<form name="eoiform" method="POST" action="<?php echo $_SERVER["PHP_SELF"];?>" id="eoi">
<b>Post Code</b>
<br>
<input type="text" id="post" name="post" /><?php echo $msgp; ?>
</form>
return? Return where?
When you return in your main code, it's (nearly) the same as die()'ing.
So when you return, the remaining PHP won't be executed anymore.
I'd consider to set some variable like $success = true/false; instead of returning.
You return false after $msgp = '<span class="error"><b>Please enter correct postcode</b></span>'; therefor you do not continue to the form below... remove the returns from your code to be able to handle and show an error.
You are using return. Are you in a function() {} ? If so, all your variables are in function scope. You could do global $msgp; to make the variable accessible outside of the function.
If not... then you shouldn't use return.
Change php code to
<?php
if (isset($_POST['post'])) {
$post = intval($_POST["post"],0);
$words = array(2747,2750,2756,2760,2777);
$msgp = '<span class="error"><b>'.in_array($post,$words) ? "Please enter postcode" : "Please enter correct postcode" .'</b></span>';
}
There are a lot of things that can be simplified. For example codes as int array and using in_array() function. Return statement should be used to return something from method/function using it in global scope then execution of the current script file is ended. There is no need for it here. If there are more $words values you should consider using simple REGEX /[0-9]{4}/ and preg_match()
You want to display $msgp right? Use echo $msgp;. But you can return to nowhere in your code. (put it in a function).
Try this
<?php
$post = $_POST["post"];
$words = array('2747','2750','2753','2760','2777');
$msgp = "<span class='error'><b>Please enter correct postcode</b></span>";
function check($post){
global $words;
$return=true;
if (!empty($post)) {
foreach ($words as $item)
{
if (strpos($post, $item) !== false)
//$return=true;
}
} else if(empty($post)) {
$return=false;
}
return $result;
}
$return=check($post);
if($return === true){
// echo you are right!
}
else {
echo $msgp;
}
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
}
can php return a boolean like this:
return $aantal == 0;
like in java you can
public boolean test(int i)
{
return i==0;
}
or do you Have to use a if contruction?
because if i do this.
$foutLoos = checkFoutloos($aantal);
function checkFoutloos($aantal)
{
return $aantal == 0;
}
echo "foutLoos = $foutLoos";
it echo's
foutLoos =
so not true or false
thanks
matthy
It returns a boolean, but the boolean is not converted to a string when you output it. Try this instead:
$foutLoos = checkFoutloos($aantal);
function checkFoutloos($aantal)
{
return $aantal == 0;
}
echo "foutLoos = " . ( $foutLoos ? "true" : "false" );
Try it out!
function is_zero($n) {
return $n == 0;
}
echo gettype(is_zero(0));
The output:
boolean
yes ideed i found out that when you echo a false you get nothing and true echo's 1 thats why i was confused ;)
Yes. you can even through in the ternary operator.
function foo($bar == 0) {
return ($bar) ? true : false;
}
Yes, you can return a boolean test in the return function. I like to put mine in parenthesis so I know what is being evaluated.
function Foo($Bar= 0) {
return ($Bar == 0);
}
$Return = Foo(2);
$Type = var_export($Return, true);
echo "Return Type: ".$Type; // Return Type: boolean(true)
In fact, almost anything can be evaluated on the return line. Don't go crazy though, as it may make refactoring more difficult (if you want to allow plugins to manipulate the return, for instance).
You could just do it like this i think, with type casting:
return (bool) $aantal;