I need help.
As I do?
String same other string with different number.
Example
if ($current_server == "Lobby-01") {
echo "visiting in Lobby"
} elif ($current_server == "Lobby-02"){
echo "visiting in Lobby"
} and more..
I thought about trying
if ($current_server == "Lobby-/[0-99]+/"){
echo "visiting in Lobby"
} else {
//is false, then it will not show the message
}
is correct?
How do I do it?
Can you help me?
Use preg_match:
if (preg_match("/\bLobby-[0-9]{2}\b/i", $current_server)) {
echo "visiting in Lobby";
}
else {
// no message
}
Demo
Related
The problem is with this line:
if $var LIKE '1800%';
and I'm not sure how to fix it. Thanks.
<?php
//check to see if account number is like 1800*
if (isset($_POST['acct_number'])) {
$var = $_POST['acct_number'];
if $var LIKE '1800%'; {
//stop the code
exit;
} else {
echo 'normal account number';
}
}
?>
You need PHP not MySQL. For 1800% just check that it is found at position 0:
if(strpos($var, '1800') === 0) {
//stop the code
exit;
} else {
echo 'normal account number';
}
If it can occur anywhere like %1800% then:
if(strpos($var, '1800') !== false) {
//stop the code
exit;
} else {
echo 'normal account number';
}
Use substr function to get first 4 characters and compare it with 1800.
if(substr($var, 0, 4) == '1800')
{
// your code goes here.
}
``
Another way could be to use strpos()
if (strpos($var, '1800') === 0) {
// var starts with '1800'
}
I would use a regular expression for this preg_match('/^1800.+/', $search, $matches);
Following is my code here actually o/p should be hi..but it is giving no
<?php
$arr=array('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday');
$c='xyz,ccc';
if(in_array(isset($c) && $c,$arr))
{
echo 'hi';
}
else
{
echo 'no';
}
?>
output:hi
actual result should be 'no'.
Side note, this is bad code:
in_array(isset($weekendArr) && $weekendArr,$arr)
do it like
isset($weekendArr) && in_array($weekendArr,$arr)
and in_array is not strict so this
in_array(true,array('w','s'))
will be allways TRUE
do it with:
in_array(true,array('w','s'),true)
and you see.
And you can't check an array with an array the $needle be an STRING here.
The only solution is to do splitt your STRING into two values and then check two times for TRUE
$c='Sunday,Monday';
foreach(explode(',',$c) as $check){
if(in_array($c,$arr,true))
{
echo $check.' is in array';
}
else
{
echo $check.' is NOT in array';
}
}
Hope that helps a little.
<?php
$listDays=array('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday');
$day='Sunday'; //You cant test both days ! Just one value at a time
if(true === in_array($day, $listDays))
{
echo 'hi';
}
else
{
echo 'no';
}
?>
Or option two if you want to test different days
<?php
$listDays=array('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday');
$dayToTest='Sunday, Monday'; //Here we have multiple days
$tabTest = preg_split(',', $day); //split into an array
//Then test for each string in tabTest
foreach($tabTest as $string)
{
if(true === in_array($string, $listDays))
{
echo $string.' is OK';
}
else
{
echo 'no';
}
}
?>
Change your code to:
<?php
$arr=array('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday');
$c='Sunday,Monday';
if(in_array(isset($c) && $c,$arr))
{
echo 'hi';
}
else
{
echo 'no';
}
my problem is, i have a form which i fill blabla and after i submit i need to check if the var '$number' contains only 9 numbers. which means that if it contains at least 1 letter or has less or more than 9 length it should return false, else it should return true;
this is what i got so far:
if (!is_numeric ($number) {
//do
} else {
}
1st problem: This code should take care of the only numbers part but it doesnt, it always returns false.
2nd: do you guys know of any way to take care of the 9 digits only verification?
thanks and sorry for my bad english, not my native language :P
Your number may contain unwanted whitespaces which cause the is_numeric() test not to work properly
So do the following: $number = trim($number); to remove them.
Then indeed this snippet is good to check if your variable is a number:
if (!is_numeric ($number)) {
//do
} else {
}
And for the number digits do a if statement to see if your number is between 100000000 and 999999999
So the full code will be:
$number = trim($number);
if (!is_numeric ($number)) {
//do
} else {
if ($number >= 100000000 && $number <= 999999999) {
// Everything is ok
} else {
}
}
Didn't understood your complete question coz of you native language :p, but i think you want this:
if (is_numeric($number) {
if(strlen($number) == 9){
return true;
} else {
return false;
}
} else {
echo 'Not a number';
}
Check if it contains digits and check whether its exactly contains 9.
$number = '123456789';
if(!preg_match('/^\d{9}$/', $number)) {
echo 'not ok';
} else {
echo 'ok';
}
I have this code:
<?php $url = JURI::getInstance()->toString();
if ($url == "http://example.com/news/latest/"){
echo "This is latest page";
} else {
echo "This is not latest page";
}
?>
What I'm trying to do is instead of 'http://example.com/news/latest/', how can I select the pages/items under /latest/. If it makes any more sense, here's a syntax:
if ($url == "http://example.com/news/latest/" + ANYTHING UNDER THIS)
I cannot use not equal to ($url !=) since it will include other parent pages not equal to /latest/. I just want what's under it. If anyone understands it, I need help on how to put it into code.
Update:
What I'm trying to do is if the page is example.com/news/latest, it will echo "Latest". And if for example, I am in example.com/news/latest/subpage1/subpage2, it will echo "You are in a page that is under Latest." Anything beyond "Latest" will echo that.
$str = 'example.com/news/latest/dfg';
preg_match('/example.com\/news\/([^\/]+)\/?(.*)/', $str, $page);
if(isset($page[2]) && $page[2])
echo 'You are under: ' , $page[1];
elseif(isset($page[1]))
echo 'At: ' , $page[1];
else
echo 'Error';
Edit: after clarification switched to regular expression.
Use a regular expression:
$matches = array();
if((preg_match('#http://example\.com/news/latest/(.*)#', $url, $matches)) === 1) {
if(strlen($matches[0]) > 0) {
echo "You're at page: $matches[0]";
} else {
echo "You're at the root";
}
} else {
// Error, incorrect URL (should not happen)
}
EDIT: Fixed, untested so you may have to tweak it a little
Now I want to check if this text box contains one word or two, for example
if ($_POST['mytext'] == two words){
echo "That is Perfect";
} else{
echo "We don't accept this";
}
and I tried
if ($_POST['mytext'] > 1){
echo "That is Perfect";
} else{
echo "We don't accept this";
}
and it didn't work
that what I mean so how to make it?
Hope to find a way to do that.
Thanks
If you define two words as "some characters followed by one space followed by some characters" then you can do something like:
$mytext = $_POST["mytext"];
$parts = explode(" ", $mytext);
if (count($parts) !== 2) {
throw new Exception("too many or too little!");
}
if (strlen($parts[0]) === 0 || strlen($parts[1]) === 0) {
throw new Exception("not enough characters!");
}
Keep in mind that this allows a string like "# !"
Use str_word_count():
if (str_word_count($_POST['mytext']) > 1){
echo "That is Perfect";
} else{
echo "We don't accept this";
}
you could use the
`substr_count('some text', ' ');
it will return the number of space,.
try this
$text= preg_split(" ",$_POST['mytext']);
if (count($text) > 1){
echo "That is Perfect";
} else{
echo "We don't accept this";
}